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

Optimal. Leaf size=421 \[ \frac{i a^2 f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{i a^2 f \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}-\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}+\frac{a f \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}-\frac{i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2}+\frac{i f \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2}-\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}-\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}+\frac{a (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}-\frac{2 a^2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d \left (a^2+b^2\right )}+\frac{2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.596636, antiderivative size = 421, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5567, 4180, 2279, 2391, 5573, 5561, 2190, 6742, 3718} \[ \frac{i a^2 f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{i a^2 f \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}-\frac{a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}+\frac{a f \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}-\frac{i f \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2}+\frac{i f \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2}-\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}-\frac{a (e+f x) \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}+\frac{a (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}-\frac{2 a^2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d \left (a^2+b^2\right )}+\frac{2 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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

Mathematica [A]  time = 2.63113, size = 438, normalized size = 1.04 \[ \frac{-2 a f \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )-2 a f \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+a f \text{PolyLog}(2,-\sinh (2 (c+d x))-\cosh (2 (c+d x)))-2 i b f \text{PolyLog}(2,-i (\sinh (c+d x)+\cosh (c+d x)))+2 i b f \text{PolyLog}(2,i (\sinh (c+d x)+\cosh (c+d x)))-2 a c f \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )-2 a c f \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )-2 a d f x \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )-2 a d f x \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )-2 a d e \log (a+b \sinh (c+d x))+2 a c f \log (a+b \sinh (c+d x))+2 a c^2 f+2 a d e \log (\sinh (2 (c+d x))+\cosh (2 (c+d x))+1)-2 a c d e+2 a c d f x+2 a d f x \log (\sinh (2 (c+d x))+\cosh (2 (c+d x))+1)-2 a d^2 e x+4 b d e \tan ^{-1}(\sinh (c+d x)+\cosh (c+d x))+4 b d f x \tan ^{-1}(\sinh (c+d x)+\cosh (c+d x))}{2 d^2 \left (a^2+b^2\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*a*c*d*e + 2*a*c^2*f - 2*a*d^2*e*x + 2*a*c*d*f*x + 4*b*d*e*ArcTan[Cosh[c + d*x] + Sinh[c + d*x]] + 4*b*d*f*
x*ArcTan[Cosh[c + d*x] + Sinh[c + d*x]] - 2*a*c*f*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*a*d*f*x*L
og[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])] - 2*a*c*f*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] - 2*a*d
*f*x*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])] - 2*a*d*e*Log[a + b*Sinh[c + d*x]] + 2*a*c*f*Log[a + b*Sin
h[c + d*x]] + 2*a*d*e*Log[1 + Cosh[2*(c + d*x)] + Sinh[2*(c + d*x)]] + 2*a*d*f*x*Log[1 + Cosh[2*(c + d*x)] + S
inh[2*(c + d*x)]] - 2*a*f*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] - 2*a*f*PolyLog[2, -((b*E^(c + d*
x))/(a + Sqrt[a^2 + b^2]))] - (2*I)*b*f*PolyLog[2, (-I)*(Cosh[c + d*x] + Sinh[c + d*x])] + (2*I)*b*f*PolyLog[2
, I*(Cosh[c + d*x] + Sinh[c + d*x])] + a*f*PolyLog[2, -Cosh[2*(c + d*x)] - Sinh[2*(c + d*x)]])/(2*(a^2 + b^2)*
d^2)

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Maple [B]  time = 0.154, size = 1287, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d^2*f/(2*a^2+2*b^2)*ln(1-I*exp(d*x+c))*a*c-2/d*f/(2*a^2+2*b^2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2
+b^2)^(1/2)))*a*x-2/d^2*f/(2*a^2+2*b^2)*ln((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*a*c-2/d*f/(
2*a^2+2*b^2)*ln((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*a*x-2/d^2*f/(2*a^2+2*b^2)*ln((b*exp(d*x+
c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*a*c+2/d*f/(2*a^2+2*b^2)*ln(1+I*exp(d*x+c))*a*x-4/d^2*f*c/(2*a^2+2*b
^2)*b*arctan(exp(d*x+c))+2/d^2*f*c/(2*a^2+2*b^2)*a*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b)-2/d^2*f*c/(2*a^2+2*b^
2)*(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))-2/d^2*f*c/(2*a^2+2*b^2)*a*ln(1+exp(2*d*x+
2*c))-2/d*e/(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))*a^2+2*I/d^2*f/(2*a
^2+2*b^2)*dilog(1-I*exp(d*x+c))*b-2*I/d^2*f/(2*a^2+2*b^2)*dilog(1+I*exp(d*x+c))*b+2/d^2*f/(2*a^2+2*b^2)*ln(1+I
*exp(d*x+c))*a*c+2/d*f/(2*a^2+2*b^2)*ln(1-I*exp(d*x+c))*a*x+2/d^2*f*c/(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arctanh(1/
2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))*a^2+2*I/d*f/(2*a^2+2*b^2)*ln(1-I*exp(d*x+c))*b*x+2*I/d^2*f/(2*a^2+2*b^
2)*ln(1-I*exp(d*x+c))*b*c-2*I/d*f/(2*a^2+2*b^2)*ln(1+I*exp(d*x+c))*b*x-2*I/d^2*f/(2*a^2+2*b^2)*ln(1+I*exp(d*x+
c))*b*c-2/d/(a^2+b^2)^(1/2)*b^2*e/(2*a^2+2*b^2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+2/d^2/(a^2+b
^2)^(1/2)*b^2*f*c/(2*a^2+2*b^2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+2/d^2*f/(2*a^2+2*b^2)*dilog(
1-I*exp(d*x+c))*a-2/d^2*f/(2*a^2+2*b^2)*dilog((b*exp(d*x+c)+(a^2+b^2)^(1/2)+a)/(a+(a^2+b^2)^(1/2)))*a-2/d^2*f/
(2*a^2+2*b^2)*dilog((-b*exp(d*x+c)+(a^2+b^2)^(1/2)-a)/(-a+(a^2+b^2)^(1/2)))*a+2/d*e/(2*a^2+2*b^2)*a*ln(1+exp(2
*d*x+2*c))+4/d*e/(2*a^2+2*b^2)*b*arctan(exp(d*x+c))-2/d*e/(2*a^2+2*b^2)*a*ln(b*exp(2*d*x+2*c)+2*a*exp(d*x+c)-b
)+2/d*e/(2*a^2+2*b^2)*(a^2+b^2)^(1/2)*arctanh(1/2*(2*b*exp(d*x+c)+2*a)/(a^2+b^2)^(1/2))+2/d^2*f/(2*a^2+2*b^2)*
dilog(1+I*exp(d*x+c))*a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -e{\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 )} + f \int \frac{2 \, 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)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-e*(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)) + f*integrate(2*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 [A]  time = 2.46716, size = 1544, normalized size = 3.67 \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)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(a*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) + a*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) - (a*f + I*b*f)*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) - (a*f - I*b*f)*dilog(-I*cosh(d*x +
c) - I*sinh(d*x + c)) + (a*d*e - a*c*f)*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*e - a*c*f)*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*
f*x + a*c*f)*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*f*x + a*c*f)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*s
qrt((a^2 + b^2)/b^2) - b)/b) - (a*d*e + I*b*d*e - a*c*f - I*b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) + I) - (a
*d*e - I*b*d*e - a*c*f + I*b*c*f)*log(cosh(d*x + c) + sinh(d*x + c) - I) - (a*d*f*x - I*b*d*f*x + a*c*f - I*b*
c*f)*log(I*cosh(d*x + c) + I*sinh(d*x + c) + 1) - (a*d*f*x + I*b*d*f*x + a*c*f + I*b*c*f)*log(-I*cosh(d*x + c)
- I*sinh(d*x + c) + 1))/((a^2 + b^2)*d^2)

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

[Out]

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

[Out]

Timed out