3.348 \(\int \frac{(e+f x)^3 \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=1021 \[ \text{result too large to display} \]

[Out]

(2*(e + f*x)^3*ArcTan[E^(c + d*x)])/(b*d) - (2*a^2*(e + f*x)^3*ArcTan[E^(c + d*x)])/(b*(a^2 + b^2)*d) - (a*(e
+ f*x)^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/((a^2 + b^2)*d) - (a*(e + f*x)^3*Log[1 + (b*E^(c + d*
x))/(a + Sqrt[a^2 + b^2])])/((a^2 + b^2)*d) + (a*(e + f*x)^3*Log[1 + E^(2*(c + d*x))])/((a^2 + b^2)*d) - ((3*I
)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])/(b*d^2) + ((3*I)*a^2*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])
/(b*(a^2 + b^2)*d^2) + ((3*I)*f*(e + f*x)^2*PolyLog[2, I*E^(c + d*x)])/(b*d^2) - ((3*I)*a^2*f*(e + f*x)^2*Poly
Log[2, I*E^(c + d*x)])/(b*(a^2 + b^2)*d^2) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b
^2]))])/((a^2 + b^2)*d^2) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b
^2)*d^2) + (3*a*f*(e + f*x)^2*PolyLog[2, -E^(2*(c + d*x))])/(2*(a^2 + b^2)*d^2) + ((6*I)*f^2*(e + f*x)*PolyLog
[3, (-I)*E^(c + d*x)])/(b*d^3) - ((6*I)*a^2*f^2*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)])/(b*(a^2 + b^2)*d^3) -
((6*I)*f^2*(e + f*x)*PolyLog[3, I*E^(c + d*x)])/(b*d^3) + ((6*I)*a^2*f^2*(e + f*x)*PolyLog[3, I*E^(c + d*x)])/
(b*(a^2 + b^2)*d^3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^
3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^3) - (3*a*f^2*(e
+ f*x)*PolyLog[3, -E^(2*(c + d*x))])/(2*(a^2 + b^2)*d^3) - ((6*I)*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(b*d^4) +
((6*I)*a^2*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(b*(a^2 + b^2)*d^4) + ((6*I)*f^3*PolyLog[4, I*E^(c + d*x)])/(b*d^
4) - ((6*I)*a^2*f^3*PolyLog[4, I*E^(c + d*x)])/(b*(a^2 + b^2)*d^4) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a
- Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^4) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2
+ b^2)*d^4) + (3*a*f^3*PolyLog[4, -E^(2*(c + d*x))])/(4*(a^2 + b^2)*d^4)

________________________________________________________________________________________

Rubi [A]  time = 1.32959, antiderivative size = 1021, normalized size of antiderivative = 1., number of steps used = 39, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {5567, 4180, 2531, 6609, 2282, 6589, 5573, 5561, 2190, 6742, 3718} \[ -\frac{6 i \text{PolyLog}\left (4,-i e^{c+d x}\right ) f^3}{b d^4}+\frac{6 i a^2 \text{PolyLog}\left (4,-i e^{c+d x}\right ) f^3}{b \left (a^2+b^2\right ) d^4}+\frac{6 i \text{PolyLog}\left (4,i e^{c+d x}\right ) f^3}{b d^4}-\frac{6 i a^2 \text{PolyLog}\left (4,i e^{c+d x}\right ) f^3}{b \left (a^2+b^2\right ) d^4}-\frac{6 a \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) f^3}{\left (a^2+b^2\right ) d^4}-\frac{6 a \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) f^3}{\left (a^2+b^2\right ) d^4}+\frac{3 a \text{PolyLog}\left (4,-e^{2 (c+d x)}\right ) f^3}{4 \left (a^2+b^2\right ) d^4}+\frac{6 i (e+f x) \text{PolyLog}\left (3,-i e^{c+d x}\right ) f^2}{b d^3}-\frac{6 i a^2 (e+f x) \text{PolyLog}\left (3,-i e^{c+d x}\right ) f^2}{b \left (a^2+b^2\right ) d^3}-\frac{6 i (e+f x) \text{PolyLog}\left (3,i e^{c+d x}\right ) f^2}{b d^3}+\frac{6 i a^2 (e+f x) \text{PolyLog}\left (3,i e^{c+d x}\right ) f^2}{b \left (a^2+b^2\right ) d^3}+\frac{6 a (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) f^2}{\left (a^2+b^2\right ) d^3}+\frac{6 a (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) f^2}{\left (a^2+b^2\right ) d^3}-\frac{3 a (e+f x) \text{PolyLog}\left (3,-e^{2 (c+d x)}\right ) f^2}{2 \left (a^2+b^2\right ) d^3}-\frac{3 i (e+f x)^2 \text{PolyLog}\left (2,-i e^{c+d x}\right ) f}{b d^2}+\frac{3 i a^2 (e+f x)^2 \text{PolyLog}\left (2,-i e^{c+d x}\right ) f}{b \left (a^2+b^2\right ) d^2}+\frac{3 i (e+f x)^2 \text{PolyLog}\left (2,i e^{c+d x}\right ) f}{b d^2}-\frac{3 i a^2 (e+f x)^2 \text{PolyLog}\left (2,i e^{c+d x}\right ) f}{b \left (a^2+b^2\right ) d^2}-\frac{3 a (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) f}{\left (a^2+b^2\right ) d^2}-\frac{3 a (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) f}{\left (a^2+b^2\right ) d^2}+\frac{3 a (e+f x)^2 \text{PolyLog}\left (2,-e^{2 (c+d x)}\right ) f}{2 \left (a^2+b^2\right ) d^2}+\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (\frac{e^{c+d x} b}{a-\sqrt{a^2+b^2}}+1\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (\frac{e^{c+d x} b}{a+\sqrt{a^2+b^2}}+1\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

(2*(e + f*x)^3*ArcTan[E^(c + d*x)])/(b*d) - (2*a^2*(e + f*x)^3*ArcTan[E^(c + d*x)])/(b*(a^2 + b^2)*d) - (a*(e
+ f*x)^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/((a^2 + b^2)*d) - (a*(e + f*x)^3*Log[1 + (b*E^(c + d*
x))/(a + Sqrt[a^2 + b^2])])/((a^2 + b^2)*d) + (a*(e + f*x)^3*Log[1 + E^(2*(c + d*x))])/((a^2 + b^2)*d) - ((3*I
)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])/(b*d^2) + ((3*I)*a^2*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])
/(b*(a^2 + b^2)*d^2) + ((3*I)*f*(e + f*x)^2*PolyLog[2, I*E^(c + d*x)])/(b*d^2) - ((3*I)*a^2*f*(e + f*x)^2*Poly
Log[2, I*E^(c + d*x)])/(b*(a^2 + b^2)*d^2) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b
^2]))])/((a^2 + b^2)*d^2) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b
^2)*d^2) + (3*a*f*(e + f*x)^2*PolyLog[2, -E^(2*(c + d*x))])/(2*(a^2 + b^2)*d^2) + ((6*I)*f^2*(e + f*x)*PolyLog
[3, (-I)*E^(c + d*x)])/(b*d^3) - ((6*I)*a^2*f^2*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)])/(b*(a^2 + b^2)*d^3) -
((6*I)*f^2*(e + f*x)*PolyLog[3, I*E^(c + d*x)])/(b*d^3) + ((6*I)*a^2*f^2*(e + f*x)*PolyLog[3, I*E^(c + d*x)])/
(b*(a^2 + b^2)*d^3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^
3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^3) - (3*a*f^2*(e
+ f*x)*PolyLog[3, -E^(2*(c + d*x))])/(2*(a^2 + b^2)*d^3) - ((6*I)*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(b*d^4) +
((6*I)*a^2*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(b*(a^2 + b^2)*d^4) + ((6*I)*f^3*PolyLog[4, I*E^(c + d*x)])/(b*d^
4) - ((6*I)*a^2*f^3*PolyLog[4, I*E^(c + d*x)])/(b*(a^2 + b^2)*d^4) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a
- Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^4) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2
+ b^2)*d^4) + (3*a*f^3*PolyLog[4, -E^(2*(c + d*x))])/(4*(a^2 + b^2)*d^4)

Rule 5567

Int[(((e_.) + (f_.)*(x_))^(m_.)*Tanh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sech[c + d*x]*Tanh[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sec
h[c + d*x]*Tanh[c + d*x]^(n - 1))/(a + b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0]
&& IGtQ[n, 0]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 5573

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[b^2/(a^2 + b^2), Int[((e + f*x)^m*Sech[c + d*x]^(n - 2))/(a + b*Sinh[c + d*x]), x], x] + Dist[1/(
a^2 + b^2), Int[(e + f*x)^m*Sech[c + d*x]^n*(a - b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && I
GtQ[m, 0] && NeQ[a^2 + b^2, 0] && IGtQ[n, 0]

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rubi steps

\begin{align*} \int \frac{(e+f x)^3 \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x)^3 \text{sech}(c+d x) \, dx}{b}-\frac{a \int \frac{(e+f x)^3 \text{sech}(c+d x)}{a+b \sinh (c+d x)} \, dx}{b}\\ &=\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{a \int (e+f x)^3 \text{sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{b \left (a^2+b^2\right )}-\frac{(a b) \int \frac{(e+f x)^3 \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}-\frac{(3 i f) \int (e+f x)^2 \log \left (1-i e^{c+d x}\right ) \, dx}{b d}+\frac{(3 i f) \int (e+f x)^2 \log \left (1+i e^{c+d x}\right ) \, dx}{b d}\\ &=\frac{a (e+f x)^4}{4 \left (a^2+b^2\right ) f}+\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{a \int \left (a (e+f x)^3 \text{sech}(c+d x)-b (e+f x)^3 \tanh (c+d x)\right ) \, dx}{b \left (a^2+b^2\right )}-\frac{(a b) \int \frac{e^{c+d x} (e+f x)^3}{a-\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}-\frac{(a b) \int \frac{e^{c+d x} (e+f x)^3}{a+\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}+\frac{\left (6 i f^2\right ) \int (e+f x) \text{Li}_2\left (-i e^{c+d x}\right ) \, dx}{b d^2}-\frac{\left (6 i f^2\right ) \int (e+f x) \text{Li}_2\left (i e^{c+d x}\right ) \, dx}{b d^2}\\ &=\frac{a (e+f x)^4}{4 \left (a^2+b^2\right ) f}+\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}+\frac{6 i f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{6 i f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{a \int (e+f x)^3 \tanh (c+d x) \, dx}{a^2+b^2}-\frac{a^2 \int (e+f x)^3 \text{sech}(c+d x) \, dx}{b \left (a^2+b^2\right )}+\frac{(3 a f) \int (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac{(3 a f) \int (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}-\frac{\left (6 i f^3\right ) \int \text{Li}_3\left (-i e^{c+d x}\right ) \, dx}{b d^3}+\frac{\left (6 i f^3\right ) \int \text{Li}_3\left (i e^{c+d x}\right ) \, dx}{b d^3}\\ &=\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{6 i f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{6 i f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{(2 a) \int \frac{e^{2 (c+d x)} (e+f x)^3}{1+e^{2 (c+d x)}} \, dx}{a^2+b^2}+\frac{\left (3 i a^2 f\right ) \int (e+f x)^2 \log \left (1-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}-\frac{\left (3 i a^2 f\right ) \int (e+f x)^2 \log \left (1+i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}+\frac{\left (6 a f^2\right ) \int (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^2}+\frac{\left (6 a f^2\right ) \int (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^2}-\frac{\left (6 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^4}+\frac{\left (6 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^4}\\ &=\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{6 i f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{6 i f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 i f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac{6 i f^3 \text{Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac{(3 a f) \int (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d}-\frac{\left (6 i a^2 f^2\right ) \int (e+f x) \text{Li}_2\left (-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^2}+\frac{\left (6 i a^2 f^2\right ) \int (e+f x) \text{Li}_2\left (i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^2}-\frac{\left (6 a f^3\right ) \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^3}-\frac{\left (6 a f^3\right ) \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^3}\\ &=\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 a f (e+f x)^2 \text{Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{6 i f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac{6 i f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 i f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac{6 i f^3 \text{Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac{\left (3 a f^2\right ) \int (e+f x) \text{Li}_2\left (-e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d^2}-\frac{\left (6 a f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{b x}{-a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}-\frac{\left (6 a f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{b x}{a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}+\frac{\left (6 i a^2 f^3\right ) \int \text{Li}_3\left (-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^3}-\frac{\left (6 i a^2 f^3\right ) \int \text{Li}_3\left (i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^3}\\ &=\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 a f (e+f x)^2 \text{Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{6 i f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac{6 i f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{3 a f^2 (e+f x) \text{Li}_3\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}-\frac{6 i f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac{6 i f^3 \text{Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}+\frac{\left (6 i a^2 f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}-\frac{\left (6 i a^2 f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}+\frac{\left (3 a f^3\right ) \int \text{Li}_3\left (-e^{2 (c+d x)}\right ) \, dx}{2 \left (a^2+b^2\right ) d^3}\\ &=\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 a f (e+f x)^2 \text{Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{6 i f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac{6 i f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{3 a f^2 (e+f x) \text{Li}_3\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}-\frac{6 i f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac{6 i a^2 f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}+\frac{6 i f^3 \text{Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac{6 i a^2 f^3 \text{Li}_4\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}+\frac{\left (3 a f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{4 \left (a^2+b^2\right ) d^4}\\ &=\frac{2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{3 i a^2 f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 a f (e+f x)^2 \text{Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac{6 i f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac{6 i f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{6 i a^2 f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{3 a f^2 (e+f x) \text{Li}_3\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}-\frac{6 i f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac{6 i a^2 f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}+\frac{6 i f^3 \text{Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac{6 i a^2 f^3 \text{Li}_4\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}+\frac{3 a f^3 \text{Li}_4\left (-e^{2 (c+d x)}\right )}{4 \left (a^2+b^2\right ) d^4}\\ \end{align*}

Mathematica [B]  time = 25.5273, size = 2333, normalized size = 2.29 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

(8*b*Sqrt[-(a^2 + b^2)^2]*d^3*e^3*ArcTan[E^(c + d*x)] - 8*a^2*Sqrt[a^2 + b^2]*d^3*e^3*ArcTan[(a + b*E^(c + d*x
))/Sqrt[-a^2 - b^2]] - 8*a^2*Sqrt[-a^2 - b^2]*d^3*e^3*ArcTanh[(a + b*E^(c + d*x))/Sqrt[a^2 + b^2]] + (12*I)*b*
Sqrt[-(a^2 + b^2)^2]*d^3*e^2*f*x*Log[1 - I*E^(c + d*x)] + (12*I)*b*Sqrt[-(a^2 + b^2)^2]*d^3*e*f^2*x^2*Log[1 -
I*E^(c + d*x)] + (4*I)*b*Sqrt[-(a^2 + b^2)^2]*d^3*f^3*x^3*Log[1 - I*E^(c + d*x)] - (12*I)*b*Sqrt[-(a^2 + b^2)^
2]*d^3*e^2*f*x*Log[1 + I*E^(c + d*x)] - (12*I)*b*Sqrt[-(a^2 + b^2)^2]*d^3*e*f^2*x^2*Log[1 + I*E^(c + d*x)] - (
4*I)*b*Sqrt[-(a^2 + b^2)^2]*d^3*f^3*x^3*Log[1 + I*E^(c + d*x)] + 4*a*Sqrt[-(a^2 + b^2)^2]*d^3*e^3*Log[1 + E^(2
*(c + d*x))] + 12*a*Sqrt[-(a^2 + b^2)^2]*d^3*e^2*f*x*Log[1 + E^(2*(c + d*x))] + 12*a*Sqrt[-(a^2 + b^2)^2]*d^3*
e*f^2*x^2*Log[1 + E^(2*(c + d*x))] + 4*a*Sqrt[-(a^2 + b^2)^2]*d^3*f^3*x^3*Log[1 + E^(2*(c + d*x))] - 4*a*Sqrt[
-(a^2 + b^2)^2]*d^3*e^3*Log[2*a*E^(c + d*x) + b*(-1 + E^(2*(c + d*x)))] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^3*e^2*f*
x*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^3*e*f^2*x^2*Log
[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])] - 4*a*Sqrt[-(a^2 + b^2)^2]*d^3*f^3*x^3*Log[1 + (b*
E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^3*e^2*f*x*Log[1 + (b*E^(2*c
+ d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^3*e*f^2*x^2*Log[1 + (b*E^(2*c + d*x
))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])] - 4*a*Sqrt[-(a^2 + b^2)^2]*d^3*f^3*x^3*Log[1 + (b*E^(2*c + d*x))/(a*E^
c + Sqrt[(a^2 + b^2)*E^(2*c)])] - (12*I)*b*Sqrt[-(a^2 + b^2)^2]*d^2*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)]
+ (12*I)*b*Sqrt[-(a^2 + b^2)^2]*d^2*f*(e + f*x)^2*PolyLog[2, I*E^(c + d*x)] + 6*a*Sqrt[-(a^2 + b^2)^2]*d^2*e^
2*f*PolyLog[2, -E^(2*(c + d*x))] + 12*a*Sqrt[-(a^2 + b^2)^2]*d^2*e*f^2*x*PolyLog[2, -E^(2*(c + d*x))] + 6*a*Sq
rt[-(a^2 + b^2)^2]*d^2*f^3*x^2*PolyLog[2, -E^(2*(c + d*x))] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^2*e^2*f*PolyLog[2, -
((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] - 24*a*Sqrt[-(a^2 + b^2)^2]*d^2*e*f^2*x*PolyLog[2, -(
(b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^2*f^3*x^2*PolyLog[2, -((
b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^2*e^2*f*PolyLog[2, -((b*E
^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] - 24*a*Sqrt[-(a^2 + b^2)^2]*d^2*e*f^2*x*PolyLog[2, -((b*E^
(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] - 12*a*Sqrt[-(a^2 + b^2)^2]*d^2*f^3*x^2*PolyLog[2, -((b*E^(
2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] + (24*I)*b*Sqrt[-(a^2 + b^2)^2]*d*e*f^2*PolyLog[3, (-I)*E^(c
+ d*x)] + (24*I)*b*Sqrt[-(a^2 + b^2)^2]*d*f^3*x*PolyLog[3, (-I)*E^(c + d*x)] - (24*I)*b*Sqrt[-(a^2 + b^2)^2]*
d*e*f^2*PolyLog[3, I*E^(c + d*x)] - (24*I)*b*Sqrt[-(a^2 + b^2)^2]*d*f^3*x*PolyLog[3, I*E^(c + d*x)] - 6*a*Sqrt
[-(a^2 + b^2)^2]*d*e*f^2*PolyLog[3, -E^(2*(c + d*x))] - 6*a*Sqrt[-(a^2 + b^2)^2]*d*f^3*x*PolyLog[3, -E^(2*(c +
d*x))] + 24*a*Sqrt[-(a^2 + b^2)^2]*d*e*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])
)] + 24*a*Sqrt[-(a^2 + b^2)^2]*d*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] +
24*a*Sqrt[-(a^2 + b^2)^2]*d*e*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] + 24*a*
Sqrt[-(a^2 + b^2)^2]*d*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] - (24*I)*b*S
qrt[-(a^2 + b^2)^2]*f^3*PolyLog[4, (-I)*E^(c + d*x)] + (24*I)*b*Sqrt[-(a^2 + b^2)^2]*f^3*PolyLog[4, I*E^(c + d
*x)] + 3*a*Sqrt[-(a^2 + b^2)^2]*f^3*PolyLog[4, -E^(2*(c + d*x))] - 24*a*Sqrt[-(a^2 + b^2)^2]*f^3*PolyLog[4, -(
(b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] - 24*a*Sqrt[-(a^2 + b^2)^2]*f^3*PolyLog[4, -((b*E^(2*c
+ d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/(4*Sqrt[-a^2 - b^2]*(a^2 + b^2)^(3/2)*d^4)

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Maple [F]  time = 0.353, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3}\tanh \left ( dx+c \right ) }{a+b\sinh \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

int((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -e^{3}{\left (\frac{2 \, b \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{a \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac{a \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d}\right )} + \int \frac{2 \, f^{3} x^{3}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} + \frac{6 \, e f^{2} x^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} + \frac{6 \, e^{2} f x{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-e^3*(2*b*arctan(e^(-d*x - c))/((a^2 + b^2)*d) + a*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^2 + b^2
)*d) - a*log(e^(-2*d*x - 2*c) + 1)/((a^2 + b^2)*d)) + integrate(2*f^3*x^3*(e^(d*x + c) - e^(-d*x - c))/((b*(e^
(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c))) + 6*e*f^2*x^2*(e^(d*x + c) - e^(-d*x - c))/((b*
(e^(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c))) + 6*e^2*f*x*(e^(d*x + c) - e^(-d*x - c))/((b
*(e^(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c))), x)

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Fricas [C]  time = 3.12044, size = 4177, normalized size = 4.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(6*a*f^3*polylog(4, (a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)
/b^2))/b) + 6*a*f^3*polylog(4, (a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((
a^2 + b^2)/b^2))/b) + 3*(a*d^2*f^3*x^2 + 2*a*d^2*e*f^2*x + a*d^2*e^2*f)*dilog((a*cosh(d*x + c) + a*sinh(d*x +
c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + 3*(a*d^2*f^3*x^2 + 2*a*d^2*e*f^2*
x + a*d^2*e^2*f)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^
2)/b^2) - b)/b + 1) - (3*a*d^2*f^3*x^2 + 3*I*b*d^2*f^3*x^2 + 6*a*d^2*e*f^2*x + 6*I*b*d^2*e*f^2*x + 3*a*d^2*e^2
*f + 3*I*b*d^2*e^2*f)*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) - (3*a*d^2*f^3*x^2 - 3*I*b*d^2*f^3*x^2 + 6*a*d^
2*e*f^2*x - 6*I*b*d^2*e*f^2*x + 3*a*d^2*e^2*f - 3*I*b*d^2*e^2*f)*dilog(-I*cosh(d*x + c) - I*sinh(d*x + c)) + (
a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqr
t((a^2 + b^2)/b^2) + 2*a) + (a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*log(2*b*cosh(d*x + c)
+ 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + (a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x
+ 3*a*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) +
b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) + (a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + 3*a*c*
d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d
*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) - (a*d^3*e^3 + I*b*d^3*e^3 - 3*a*c*d^2*e^2*f - 3*I*b*c*d^2*e^2*f + 3*a*
c^2*d*e*f^2 + 3*I*b*c^2*d*e*f^2 - a*c^3*f^3 - I*b*c^3*f^3)*log(cosh(d*x + c) + sinh(d*x + c) + I) - (a*d^3*e^3
- I*b*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*I*b*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - 3*I*b*c^2*d*e*f^2 - a*c^3*f^3 + I*b*c
^3*f^3)*log(cosh(d*x + c) + sinh(d*x + c) - I) - (a*d^3*f^3*x^3 - I*b*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 - 3*I*b*
d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x - 3*I*b*d^3*e^2*f*x + 3*a*c*d^2*e^2*f - 3*I*b*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 +
3*I*b*c^2*d*e*f^2 + a*c^3*f^3 - I*b*c^3*f^3)*log(I*cosh(d*x + c) + I*sinh(d*x + c) + 1) - (a*d^3*f^3*x^3 + I*b
*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*I*b*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + 3*I*b*d^3*e^2*f*x + 3*a*c*d^2*e^2*f
+ 3*I*b*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 - 3*I*b*c^2*d*e*f^2 + a*c^3*f^3 + I*b*c^3*f^3)*log(-I*cosh(d*x + c) - I
*sinh(d*x + c) + 1) - (6*a*f^3 + 6*I*b*f^3)*polylog(4, I*cosh(d*x + c) + I*sinh(d*x + c)) - (6*a*f^3 - 6*I*b*f
^3)*polylog(4, -I*cosh(d*x + c) - I*sinh(d*x + c)) - 6*(a*d*f^3*x + a*d*e*f^2)*polylog(3, (a*cosh(d*x + c) + a
*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2))/b) - 6*(a*d*f^3*x + a*d*e*f^2)*pol
ylog(3, (a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2))/b) + (
6*a*d*f^3*x + 6*I*b*d*f^3*x + 6*a*d*e*f^2 + 6*I*b*d*e*f^2)*polylog(3, I*cosh(d*x + c) + I*sinh(d*x + c)) + (6*
a*d*f^3*x - 6*I*b*d*f^3*x + 6*a*d*e*f^2 - 6*I*b*d*e*f^2)*polylog(3, -I*cosh(d*x + c) - I*sinh(d*x + c)))/((a^2
+ b^2)*d^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e + f x\right )^{3} \tanh{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral((e + f*x)**3*tanh(c + d*x)/(a + b*sinh(c + d*x)), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

Timed out