3.205 \(\int \frac{(e+f x)^3 \text{csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=313 \[ \frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,e^{c+d x}\right )}{a d^3}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}-\frac{12 i f^3 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac{6 f^3 \text{PolyLog}\left (4,-e^{c+d x}\right )}{a d^4}+\frac{6 f^3 \text{PolyLog}\left (4,e^{c+d x}\right )}{a d^4}+\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^3}{a d} \]

[Out]

((-I)*(e + f*x)^3)/(a*d) - (2*(e + f*x)^3*ArcTanh[E^(c + d*x)])/(a*d) + ((6*I)*f*(e + f*x)^2*Log[1 + I*E^(c +
d*x)])/(a*d^2) - (3*f*(e + f*x)^2*PolyLog[2, -E^(c + d*x)])/(a*d^2) + ((12*I)*f^2*(e + f*x)*PolyLog[2, (-I)*E^
(c + d*x)])/(a*d^3) + (3*f*(e + f*x)^2*PolyLog[2, E^(c + d*x)])/(a*d^2) + (6*f^2*(e + f*x)*PolyLog[3, -E^(c +
d*x)])/(a*d^3) - ((12*I)*f^3*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^4) - (6*f^2*(e + f*x)*PolyLog[3, E^(c + d*x)])
/(a*d^3) - (6*f^3*PolyLog[4, -E^(c + d*x)])/(a*d^4) + (6*f^3*PolyLog[4, E^(c + d*x)])/(a*d^4) - (I*(e + f*x)^3
*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.482013, antiderivative size = 313, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {5575, 4182, 2531, 6609, 2282, 6589, 3318, 4184, 3716, 2190} \[ \frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,e^{c+d x}\right )}{a d^3}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,e^{c+d x}\right )}{a d^2}-\frac{12 i f^3 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac{6 f^3 \text{PolyLog}\left (4,-e^{c+d x}\right )}{a d^4}+\frac{6 f^3 \text{PolyLog}\left (4,e^{c+d x}\right )}{a d^4}+\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^3}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(e + f*x)^3)/(a*d) - (2*(e + f*x)^3*ArcTanh[E^(c + d*x)])/(a*d) + ((6*I)*f*(e + f*x)^2*Log[1 + I*E^(c +
d*x)])/(a*d^2) - (3*f*(e + f*x)^2*PolyLog[2, -E^(c + d*x)])/(a*d^2) + ((12*I)*f^2*(e + f*x)*PolyLog[2, (-I)*E^
(c + d*x)])/(a*d^3) + (3*f*(e + f*x)^2*PolyLog[2, E^(c + d*x)])/(a*d^2) + (6*f^2*(e + f*x)*PolyLog[3, -E^(c +
d*x)])/(a*d^3) - ((12*I)*f^3*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^4) - (6*f^2*(e + f*x)*PolyLog[3, E^(c + d*x)])
/(a*d^3) - (6*f^3*PolyLog[4, -E^(c + d*x)])/(a*d^4) + (6*f^3*PolyLog[4, E^(c + d*x)])/(a*d^4) - (I*(e + f*x)^3
*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

Rule 5575

Int[(Csch[(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*Csch[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Csch[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 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, 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 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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]

Rubi steps

\begin{align*} \int \frac{(e+f x)^3 \text{csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\left (i \int \frac{(e+f x)^3}{a+i a \sinh (c+d x)} \, dx\right )+\frac{\int (e+f x)^3 \text{csch}(c+d x) \, dx}{a}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{i \int (e+f x)^3 \csc ^2\left (\frac{1}{2} \left (i c+\frac{\pi }{2}\right )+\frac{i d x}{2}\right ) \, dx}{2 a}-\frac{(3 f) \int (e+f x)^2 \log \left (1-e^{c+d x}\right ) \, dx}{a d}+\frac{(3 f) \int (e+f x)^2 \log \left (1+e^{c+d x}\right ) \, dx}{a d}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{3 f (e+f x)^2 \text{Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac{3 f (e+f x)^2 \text{Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{(3 i f) \int (e+f x)^2 \coth \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}+\frac{\left (6 f^2\right ) \int (e+f x) \text{Li}_2\left (-e^{c+d x}\right ) \, dx}{a d^2}-\frac{\left (6 f^2\right ) \int (e+f x) \text{Li}_2\left (e^{c+d x}\right ) \, dx}{a d^2}\\ &=-\frac{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac{3 f (e+f x)^2 \text{Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac{3 f (e+f x)^2 \text{Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(6 f) \int \frac{e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )} (e+f x)^2}{1+i e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}} \, dx}{a d}-\frac{\left (6 f^3\right ) \int \text{Li}_3\left (-e^{c+d x}\right ) \, dx}{a d^3}+\frac{\left (6 f^3\right ) \int \text{Li}_3\left (e^{c+d x}\right ) \, dx}{a d^3}\\ &=-\frac{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{3 f (e+f x)^2 \text{Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac{3 f (e+f x)^2 \text{Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{\left (12 i f^2\right ) \int (e+f x) \log \left (1+i e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a d^2}-\frac{\left (6 f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{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(x)}{x} \, dx,x,e^{c+d x}\right )}{a d^4}\\ &=-\frac{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{3 f (e+f x)^2 \text{Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac{12 i f^2 (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac{3 f (e+f x)^2 \text{Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac{6 f^3 \text{Li}_4\left (-e^{c+d x}\right )}{a d^4}+\frac{6 f^3 \text{Li}_4\left (e^{c+d x}\right )}{a d^4}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{\left (12 i f^3\right ) \int \text{Li}_2\left (-i e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a d^3}\\ &=-\frac{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{3 f (e+f x)^2 \text{Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac{12 i f^2 (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac{3 f (e+f x)^2 \text{Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac{6 f^3 \text{Li}_4\left (-e^{c+d x}\right )}{a d^4}+\frac{6 f^3 \text{Li}_4\left (e^{c+d x}\right )}{a d^4}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{\left (12 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}\right )}{a d^4}\\ &=-\frac{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{3 f (e+f x)^2 \text{Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac{12 i f^2 (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac{3 f (e+f x)^2 \text{Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac{12 i f^3 \text{Li}_3\left (-i e^{c+d x}\right )}{a d^4}-\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac{6 f^3 \text{Li}_4\left (-e^{c+d x}\right )}{a d^4}+\frac{6 f^3 \text{Li}_4\left (e^{c+d x}\right )}{a d^4}-\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}\\ \end{align*}

Mathematica [A]  time = 5.59883, size = 363, normalized size = 1.16 \[ \frac{-3 f \left (d^2 (e+f x)^2 \text{PolyLog}(2,-\sinh (c+d x)-\cosh (c+d x))-2 d f (e+f x) \text{PolyLog}(3,-\sinh (c+d x)-\cosh (c+d x))+2 f^2 \text{PolyLog}(4,-\sinh (c+d x)-\cosh (c+d x))\right )+3 f \left (d^2 (e+f x)^2 \text{PolyLog}(2,\sinh (c+d x)+\cosh (c+d x))-2 d f (e+f x) \text{PolyLog}(3,\sinh (c+d x)+\cosh (c+d x))+2 f^2 \text{PolyLog}(4,\sinh (c+d x)+\cosh (c+d x))\right )-12 i f^2 \left (d (e+f x) \text{PolyLog}\left (2,i e^{-c-d x}\right )+f \text{PolyLog}\left (3,i e^{-c-d x}\right )\right )+\frac{2 d^3 (e+f x)^3}{e^c-i}+6 i d^2 f (e+f x)^2 \log \left (1-i e^{-c-d x}\right )-\frac{2 i d^3 \sinh \left (\frac{d x}{2}\right ) (e+f x)^3}{\left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}-2 d^3 (e+f x)^3 \tanh ^{-1}(\sinh (c+d x)+\cosh (c+d x))}{a d^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*d^3*(e + f*x)^3)/(-I + E^c) - 2*d^3*(e + f*x)^3*ArcTanh[Cosh[c + d*x] + Sinh[c + d*x]] + (6*I)*d^2*f*(e +
f*x)^2*Log[1 - I*E^(-c - d*x)] - (12*I)*f^2*(d*(e + f*x)*PolyLog[2, I*E^(-c - d*x)] + f*PolyLog[3, I*E^(-c - d
*x)]) - 3*f*(d^2*(e + f*x)^2*PolyLog[2, -Cosh[c + d*x] - Sinh[c + d*x]] - 2*d*f*(e + f*x)*PolyLog[3, -Cosh[c +
d*x] - Sinh[c + d*x]] + 2*f^2*PolyLog[4, -Cosh[c + d*x] - Sinh[c + d*x]]) + 3*f*(d^2*(e + f*x)^2*PolyLog[2, C
osh[c + d*x] + Sinh[c + d*x]] - 2*d*f*(e + f*x)*PolyLog[3, Cosh[c + d*x] + Sinh[c + d*x]] + 2*f^2*PolyLog[4, C
osh[c + d*x] + Sinh[c + d*x]]) - ((2*I)*d^3*(e + f*x)^3*Sinh[(d*x)/2])/((Cosh[c/2] + I*Sinh[c/2])*(Cosh[(c + d
*x)/2] + I*Sinh[(c + d*x)/2])))/(a*d^4)

________________________________________________________________________________________

Maple [B]  time = 0.268, size = 1034, normalized size = 3.3 \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)^3*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

2*(f^3*x^3+3*e*f^2*x^2+3*e^2*f*x+e^3)/d/a/(exp(d*x+c)-I)+6*I/d^2/a*f^3*ln(1+I*exp(d*x+c))*x^2-6*I/d^2/a*ln(exp
(d*x+c))*e^2*f-6*I/d/a*e*f^2*x^2+6*I/d^2/a*ln(exp(d*x+c)-I)*e^2*f+6*I/d^3/a*f^3*c^2*x+6*I/d^4/a*f^3*c^2*ln(exp
(d*x+c)-I)+12*I/d^3/a*e*f^2*polylog(2,-I*exp(d*x+c))-6*I/d^4/a*f^3*c^2*ln(exp(d*x+c))-6*I/d^3/a*e*f^2*c^2+12*I
/d^3/a*f^3*polylog(2,-I*exp(d*x+c))*x-6*I/d^4/a*f^3*c^2*ln(1+I*exp(d*x+c))-12*I/d^2/a*e*f^2*c*x+12*I/d^3/a*e*f
^2*ln(1+I*exp(d*x+c))*c+12*I/d^2/a*e*f^2*ln(1+I*exp(d*x+c))*x+12*I/d^3/a*e*f^2*c*ln(exp(d*x+c))-12*I*f^3*polyl
og(3,-I*exp(d*x+c))/a/d^4-6*f^3*polylog(4,-exp(d*x+c))/a/d^4+6*f^3*polylog(4,exp(d*x+c))/a/d^4-2*I/d/a*f^3*x^3
+4*I/d^4/a*f^3*c^3+3/d^3/a*e*f^2*c^2*ln(exp(d*x+c)-1)-12*I/d^3/a*e*f^2*c*ln(exp(d*x+c)-I)-3/d^2/a*e^2*f*c*ln(e
xp(d*x+c)-1)+3/d^2/a*ln(1-exp(d*x+c))*c*e^2*f+3/d/a*ln(1-exp(d*x+c))*e^2*f*x-3/d/a*ln(exp(d*x+c)+1)*e^2*f*x-3/
d^3/a*e*f^2*c^2*ln(1-exp(d*x+c))+3/d/a*e*f^2*ln(1-exp(d*x+c))*x^2+6/d^2/a*e*f^2*polylog(2,exp(d*x+c))*x-3/d/a*
e*f^2*ln(exp(d*x+c)+1)*x^2-6/d^2/a*e*f^2*polylog(2,-exp(d*x+c))*x+1/d/a*e^3*ln(exp(d*x+c)-1)-1/d/a*e^3*ln(exp(
d*x+c)+1)+3/d^2/a*f^3*polylog(2,exp(d*x+c))*x^2-6/d^3/a*f^3*polylog(3,exp(d*x+c))*x+3/d^2/a*e^2*f*polylog(2,ex
p(d*x+c))-3/d^2/a*e^2*f*polylog(2,-exp(d*x+c))-6/d^3/a*e*f^2*polylog(3,exp(d*x+c))+6/d^3/a*e*f^2*polylog(3,-ex
p(d*x+c))-1/d^4/a*f^3*c^3*ln(exp(d*x+c)-1)-1/d/a*f^3*ln(exp(d*x+c)+1)*x^3-3/d^2/a*f^3*polylog(2,-exp(d*x+c))*x
^2+6/d^3/a*f^3*polylog(3,-exp(d*x+c))*x+1/d/a*f^3*ln(1-exp(d*x+c))*x^3+1/d^4/a*f^3*ln(1-exp(d*x+c))*c^3

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Maxima [B]  time = 1.96947, size = 783, normalized size = 2.5 \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*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-e^3*(log(e^(-d*x - c) + 1)/(a*d) - log(e^(-d*x - c) - 1)/(a*d) - 2/((a*e^(-d*x - c) + I*a)*d)) - 6*I*e^2*f*x/
(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) + dilo
g(e^(d*x + c)))*e^2*f/(a*d^2) + 6*I*e^2*f*log(I*e^(d*x + c) + 1)/(a*d^2) + 2*(f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*
x)/(a*d*e^(d*x + c) - I*a*d) - 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) + 12*I*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*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) + 6*I*(d^2*x^2*log(I*e^(d*x + c) + 1) + 2*d*x*dilog(-I*e^(d*x + c)) -
2*polylog(3, -I*e^(d*x + c)))*f^3/(a*d^4) - (2*I*d^3*f^3*x^3 + 6*I*d^3*e*f^2*x^2)/(a*d^4)

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Fricas [C]  time = 2.82403, size = 2367, normalized size = 7.56 \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*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(2*d^3*e^3 - 6*c*d^2*e^2*f + 6*c^2*d*e*f^2 - 2*c^3*f^3 + (12*d*f^3*x + 12*d*e*f^2 + (12*I*d*f^3*x + 12*I*d*e*f
^2)*e^(d*x + c))*dilog(-I*e^(d*x + c)) + (3*I*d^2*f^3*x^2 + 6*I*d^2*e*f^2*x + 3*I*d^2*e^2*f - 3*(d^2*f^3*x^2 +
2*d^2*e*f^2*x + d^2*e^2*f)*e^(d*x + c))*dilog(-e^(d*x + c)) + (-3*I*d^2*f^3*x^2 - 6*I*d^2*e*f^2*x - 3*I*d^2*e
^2*f + 3*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*e^(d*x + c))*dilog(e^(d*x + c)) + (-2*I*d^3*f^3*x^3 - 6*I*d
^3*e*f^2*x^2 - 6*I*d^3*e^2*f*x - 6*I*c*d^2*e^2*f + 6*I*c^2*d*e*f^2 - 2*I*c^3*f^3)*e^(d*x + c) + (I*d^3*f^3*x^3
+ 3*I*d^3*e*f^2*x^2 + 3*I*d^3*e^2*f*x + I*d^3*e^3 - (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)
*e^(d*x + c))*log(e^(d*x + c) + 1) + (6*d^2*e^2*f - 12*c*d*e*f^2 + 6*c^2*f^3 + (6*I*d^2*e^2*f - 12*I*c*d*e*f^2
+ 6*I*c^2*f^3)*e^(d*x + c))*log(e^(d*x + c) - I) + (-I*d^3*e^3 + 3*I*c*d^2*e^2*f - 3*I*c^2*d*e*f^2 + I*c^3*f^
3 + (d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*e^(d*x + c))*log(e^(d*x + c) - 1) + (6*d^2*f^3*x^2 + 1
2*d^2*e*f^2*x + 12*c*d*e*f^2 - 6*c^2*f^3 + (6*I*d^2*f^3*x^2 + 12*I*d^2*e*f^2*x + 12*I*c*d*e*f^2 - 6*I*c^2*f^3)
*e^(d*x + c))*log(I*e^(d*x + c) + 1) + (-I*d^3*f^3*x^3 - 3*I*d^3*e*f^2*x^2 - 3*I*d^3*e^2*f*x - 3*I*c*d^2*e^2*f
+ 3*I*c^2*d*e*f^2 - I*c^3*f^3 + (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)*e^(d*x + c))*log(-e^(d*x + c) + 1) - 6*(f^3*e^(d*x + c) - I*f^3)*polylog(4, -e^(d*x + c)) + 6*(f^
3*e^(d*x + c) - I*f^3)*polylog(4, e^(d*x + c)) + (-12*I*f^3*e^(d*x + c) - 12*f^3)*polylog(3, -I*e^(d*x + c)) +
(-6*I*d*f^3*x - 6*I*d*e*f^2 + 6*(d*f^3*x + d*e*f^2)*e^(d*x + c))*polylog(3, -e^(d*x + c)) + (6*I*d*f^3*x + 6*
I*d*e*f^2 - 6*(d*f^3*x + d*e*f^2)*e^(d*x + c))*polylog(3, e^(d*x + c)))/(a*d^4*e^(d*x + c) - I*a*d^4)

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Sympy [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*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{3} \operatorname{csch}\left (d x + c\right )}{i \, a \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*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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