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

Optimal. Leaf size=451 \[ \frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^3}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2}-\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^4}-\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^4}-\frac{3 f^2 (e+f x) \text{PolyLog}\left (3,e^{2 (c+d x)}\right )}{2 a d^3}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{2 a d^2}+\frac{3 f^3 \text{PolyLog}\left (4,e^{2 (c+d x)}\right )}{4 a d^4}-\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d}-\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d}+\frac{(e+f x)^3 \log \left (1-e^{2 (c+d x)}\right )}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.769999, antiderivative size = 451, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5569, 3716, 2190, 2531, 6609, 2282, 6589, 5561} \[ \frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^3}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^2}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^2}-\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{a d^4}-\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{a d^4}-\frac{3 f^2 (e+f x) \text{PolyLog}\left (3,e^{2 (c+d x)}\right )}{2 a d^3}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,e^{2 (c+d x)}\right )}{2 a d^2}+\frac{3 f^3 \text{PolyLog}\left (4,e^{2 (c+d x)}\right )}{4 a d^4}-\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{a d}-\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{a d}+\frac{(e+f x)^3 \log \left (1-e^{2 (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5569

Int[(Coth[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Coth[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Cosh[c + d*x]*Coth[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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 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 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 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]

Rubi steps

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

Mathematica [B]  time = 19.0461, size = 1924, normalized size = 4.27 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(2*c)*((e + f*x)^4/(E^(2*c)*f) - (2*(1 - E^(-2*c))*(e + f*x)^3*Log[1 - E^(-c - d*x)])/d - (2*(1 - E^(-2*c)
)*(e + f*x)^3*Log[1 + E^(-c - d*x)])/d + (6*(-1 + E^(2*c))*f*(d^2*(e + f*x)^2*PolyLog[2, -E^(-c - d*x)] + 2*f*
(d*(e + f*x)*PolyLog[3, -E^(-c - d*x)] + f*PolyLog[4, -E^(-c - d*x)])))/(d^4*E^(2*c)) + (6*(-1 + E^(2*c))*f*(d
^2*(e + f*x)^2*PolyLog[2, E^(-c - d*x)] + 2*f*(d*(e + f*x)*PolyLog[3, E^(-c - d*x)] + f*PolyLog[4, E^(-c - d*x
)])))/(d^4*E^(2*c))))/(2*a*(-1 + E^(2*c))) + (4*e^3*E^(2*c)*x + 6*e^2*E^(2*c)*f*x^2 + 4*e*E^(2*c)*f^2*x^3 + E^
(2*c)*f^3*x^4 + (4*a*Sqrt[-(a^2 + b^2)^2]*e^3*E^(2*c)*ArcTan[(a + b*E^(c + d*x))/Sqrt[-a^2 - b^2]])/((a^2 + b^
2)^(3/2)*d) + (4*a*Sqrt[-(a^2 + b^2)^2]*e^3*E^(2*c)*ArcTanh[(a + b*E^(c + d*x))/Sqrt[a^2 + b^2]])/((-a^2 - b^2
)^(3/2)*d) + (2*e^3*Log[b - 2*a*E^(c + d*x) - b*E^(2*(c + d*x))])/d - (2*e^3*E^(2*c)*Log[2*a*E^(c + d*x) + b*(
-1 + E^(2*(c + d*x)))])/d + (6*e^2*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*
e^2*E^(2*c)*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (6*e*f^2*x^2*Log[1 + (b*E^
(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e*E^(2*c)*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c -
Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (2*f^3*x^3*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d
- (2*E^(2*c)*f^3*x^3*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (6*e^2*f*x*Log[1 + (
b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e^2*E^(2*c)*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c
+ Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (6*e*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])
])/d - (6*e*E^(2*c)*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (2*f^3*x^3*Log
[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (2*E^(2*c)*f^3*x^3*Log[1 + (b*E^(2*c + d*x))/
(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*(-1 + E^(2*c))*f*(e + f*x)^2*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^
c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^2 - (6*(-1 + E^(2*c))*f*(e + f*x)^2*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c
+ Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^2 - (12*e*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*
c)]))])/d^3 + (12*e*E^(2*c)*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 - (1
2*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (12*E^(2*c)*f^3*x*PolyLog[
3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 - (12*e*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(
a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (12*e*E^(2*c)*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^
2 + b^2)*E^(2*c)]))])/d^3 - (12*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^
3 + (12*E^(2*c)*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (12*f^3*Poly
Log[4, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^4 - (12*E^(2*c)*f^3*PolyLog[4, -((b*E^(2*c
+ d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^4 + (12*f^3*PolyLog[4, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^
2 + b^2)*E^(2*c)]))])/d^4 - (12*E^(2*c)*f^3*PolyLog[4, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])
)])/d^4)/(2*a*(-1 + E^(2*c)))

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Maple [F]  time = 0.495, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3}{\rm coth} \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*coth(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

int((f*x+e)^3*coth(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{\log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a d} - \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d}\right )} + \frac{3 \,{\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} e^{2} f}{a d^{2}} + \frac{3 \,{\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} e^{2} f}{a d^{2}} + \frac{3 \,{\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} e f^{2}}{a d^{3}} + \frac{3 \,{\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (d x + c\right )})\right )} e f^{2}}{a d^{3}} + \frac{{\left (d^{3} x^{3} \log \left (e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2}{\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 6 \, d x{\rm Li}_{3}(-e^{\left (d x + c\right )}) + 6 \,{\rm Li}_{4}(-e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} + \frac{{\left (d^{3} x^{3} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2}{\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 6 \, d x{\rm Li}_{3}(e^{\left (d x + c\right )}) + 6 \,{\rm Li}_{4}(e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} - \frac{d^{4} f^{3} x^{4} + 4 \, d^{4} e f^{2} x^{3} + 6 \, d^{4} e^{2} f x^{2}}{2 \, a d^{4}} + \int -\frac{2 \,{\left (b f^{3} x^{3} + 3 \, b e f^{2} x^{2} + 3 \, b e^{2} f x -{\left (a f^{3} x^{3} e^{c} + 3 \, a e f^{2} x^{2} e^{c} + 3 \, a e^{2} f x e^{c}\right )} e^{\left (d x\right )}\right )}}{a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} e^{\left (d x + c\right )} - a b}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^3*(log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/(a*d) - log(e^(-d*x - c) + 1)/(a*d) - log(e^(-d*x - c) -
1)/(a*d)) + 3*(d*x*log(e^(d*x + c) + 1) + dilog(-e^(d*x + c)))*e^2*f/(a*d^2) + 3*(d*x*log(-e^(d*x + c) + 1) +
dilog(e^(d*x + c)))*e^2*f/(a*d^2) + 3*(d^2*x^2*log(e^(d*x + c) + 1) + 2*d*x*dilog(-e^(d*x + c)) - 2*polylog(3
, -e^(d*x + c)))*e*f^2/(a*d^3) + 3*(d^2*x^2*log(-e^(d*x + c) + 1) + 2*d*x*dilog(e^(d*x + c)) - 2*polylog(3, e^
(d*x + c)))*e*f^2/(a*d^3) + (d^3*x^3*log(e^(d*x + c) + 1) + 3*d^2*x^2*dilog(-e^(d*x + c)) - 6*d*x*polylog(3, -
e^(d*x + c)) + 6*polylog(4, -e^(d*x + c)))*f^3/(a*d^4) + (d^3*x^3*log(-e^(d*x + c) + 1) + 3*d^2*x^2*dilog(e^(d
*x + c)) - 6*d*x*polylog(3, e^(d*x + c)) + 6*polylog(4, e^(d*x + c)))*f^3/(a*d^4) - 1/2*(d^4*f^3*x^4 + 4*d^4*e
*f^2*x^3 + 6*d^4*e^2*f*x^2)/(a*d^4) + integrate(-2*(b*f^3*x^3 + 3*b*e*f^2*x^2 + 3*b*e^2*f*x - (a*f^3*x^3*e^c +
3*a*e*f^2*x^2*e^c + 3*a*e^2*f*x*e^c)*e^(d*x))/(a*b*e^(2*d*x + 2*c) + 2*a^2*e^(d*x + c) - a*b), x)

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Fricas [C]  time = 2.85818, size = 2996, normalized size = 6.64 \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*coth(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(6*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*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*f^3*polylog(4, cosh(d*x + c) + sinh(d*x + c)) - 6*f^3*polylog(4, -cosh(d*x + c) - sinh(d*x
+ c)) + 3*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + 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*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + 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*(
d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*dilog(cosh(d*x + c) + sinh(d*x + c)) - 3*(d^2*f^3*x^2 + 2*d^2*e*f^2*x
+ d^2*e^2*f)*dilog(-cosh(d*x + c) - sinh(d*x + c)) + (d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - 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) + (d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*
e*f^2 - 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) + (d^3*f^3*x^3 +
3*d^3*e*f^2*x^2 + 3*d^3*e^2*f*x + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + 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) + (d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3
*d^3*e^2*f*x + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + 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) - (d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*d^3*e^2*f*x + d^3*e
^3)*log(cosh(d*x + c) + sinh(d*x + c) + 1) - (d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*log(cosh(d*x
+ c) + sinh(d*x + c) - 1) - (d^3*f^3*x^3 + 3*d^3*e*f^2*x^2 + 3*d^3*e^2*f*x + 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + c
^3*f^3)*log(-cosh(d*x + c) - sinh(d*x + c) + 1) - 6*(d*f^3*x + 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*(d*f^3*x + 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*(d*f^3*x +
d*e*f^2)*polylog(3, cosh(d*x + c) + sinh(d*x + c)) + 6*(d*f^3*x + d*e*f^2)*polylog(3, -cosh(d*x + c) - sinh(d
*x + c)))/(a*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} \coth{\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*coth(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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