3.211 \(\int \frac {(e+f x)^3 \text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=419 \[ -\frac {12 f^3 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^4}-\frac {3 f^3 \text {Li}_3\left (e^{2 (c+d x)}\right )}{2 a d^4}+\frac {6 i f^3 \text {Li}_4\left (-e^{c+d x}\right )}{a d^4}-\frac {6 i f^3 \text {Li}_4\left (e^{c+d x}\right )}{a d^4}+\frac {12 f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {3 f^2 (e+f x) \text {Li}_2\left (e^{2 (c+d x)}\right )}{a d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (e^{c+d x}\right )}{a d^3}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {2 i (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {(e+f x)^3 \coth (c+d x)}{a d}-\frac {2 (e+f x)^3}{a d} \]

[Out]

-2*(f*x+e)^3/a/d-6*I*f^2*(f*x+e)*polylog(3,-exp(d*x+c))/a/d^3-(f*x+e)^3*coth(d*x+c)/a/d+6*f*(f*x+e)^2*ln(1+I*e
xp(d*x+c))/a/d^2+3*f*(f*x+e)^2*ln(1-exp(2*d*x+2*c))/a/d^2-3*I*f*(f*x+e)^2*polylog(2,exp(d*x+c))/a/d^2+12*f^2*(
f*x+e)*polylog(2,-I*exp(d*x+c))/a/d^3-6*I*f^3*polylog(4,exp(d*x+c))/a/d^4+3*f^2*(f*x+e)*polylog(2,exp(2*d*x+2*
c))/a/d^3+3*I*f*(f*x+e)^2*polylog(2,-exp(d*x+c))/a/d^2-12*f^3*polylog(3,-I*exp(d*x+c))/a/d^4+2*I*(f*x+e)^3*arc
tanh(exp(d*x+c))/a/d-3/2*f^3*polylog(3,exp(2*d*x+2*c))/a/d^4+6*I*f^3*polylog(4,-exp(d*x+c))/a/d^4+6*I*f^2*(f*x
+e)*polylog(3,exp(d*x+c))/a/d^3-(f*x+e)^3*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a/d

________________________________________________________________________________________

Rubi [A]  time = 0.81, antiderivative size = 419, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {5575, 4184, 3716, 2190, 2531, 2282, 6589, 4182, 6609, 3318} \[ \frac {12 f^2 (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {3 f^2 (e+f x) \text {PolyLog}\left (2,e^{2 (c+d x)}\right )}{a d^3}-\frac {6 i f^2 (e+f x) \text {PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}+\frac {6 i f^2 (e+f x) \text {PolyLog}\left (3,e^{c+d x}\right )}{a d^3}+\frac {3 i f (e+f x)^2 \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}-\frac {12 f^3 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac {3 f^3 \text {PolyLog}\left (3,e^{2 (c+d x)}\right )}{2 a d^4}+\frac {6 i f^3 \text {PolyLog}\left (4,-e^{c+d x}\right )}{a d^4}-\frac {6 i f^3 \text {PolyLog}\left (4,e^{c+d x}\right )}{a d^4}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 (c+d x)}\right )}{a d^2}+\frac {2 i (e+f x)^3 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {(e+f x)^3 \coth (c+d x)}{a d}-\frac {2 (e+f x)^3}{a d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Csch[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]

[Out]

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

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 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 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 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 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 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 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 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 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 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]

Rubi steps

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

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Mathematica [B]  time = 17.15, size = 1042, normalized size = 2.49 \[ -\frac {2 i \left (d^3 (e+f x)^3+3 d^2 \left (1+i e^c\right ) f \log \left (1-i e^{-c-d x}\right ) (e+f x)^2+6 i \left (i-e^c\right ) f^2 \left (d (e+f x) \text {Li}_2\left (i e^{-c-d x}\right )+f \text {Li}_3\left (i e^{-c-d x}\right )\right )\right )}{a d^4 \left (-i+e^c\right )}+\frac {i d^3 x^3 \log (\cosh (c+d x)-\sinh (c+d x)+1) f^3-i d^3 x^3 \log (-\cosh (c+d x)+\sinh (c+d x)+1) f^3+3 i \left (d^2 \text {Li}_2(\cosh (c+d x)-\sinh (c+d x)) x^2+2 (d x \text {Li}_3(\cosh (c+d x)-\sinh (c+d x))+\text {Li}_4(\cosh (c+d x)-\sinh (c+d x)))\right ) f^3-3 i \left (d^2 \text {Li}_2(\sinh (c+d x)-\cosh (c+d x)) x^2+2 (d x \text {Li}_3(\sinh (c+d x)-\cosh (c+d x))+\text {Li}_4(\sinh (c+d x)-\cosh (c+d x)))\right ) f^3+3 d^2 (i d e+f) x^2 \log (\cosh (c+d x)-\sinh (c+d x)+1) f^2+3 d^2 (f-i d e) x^2 \log (-\cosh (c+d x)+\sinh (c+d x)+1) f^2+6 i (d e+i f) (d x \text {Li}_2(\cosh (c+d x)-\sinh (c+d x))+\text {Li}_3(\cosh (c+d x)-\sinh (c+d x))) f^2-6 (i d e+f) (d x \text {Li}_2(\sinh (c+d x)-\cosh (c+d x))+\text {Li}_3(\sinh (c+d x)-\cosh (c+d x))) f^2+3 d^2 e (i d e+2 f) x \log (\cosh (c+d x)-\sinh (c+d x)+1) f+3 d^2 e (2 f-i d e) x \log (-\cosh (c+d x)+\sinh (c+d x)+1) f+3 i d e (d e+2 i f) \text {Li}_2(\cosh (c+d x)-\sinh (c+d x)) f-3 i d e (d e-2 i f) \text {Li}_2(\sinh (c+d x)-\cosh (c+d x)) f-d^3 (e+f x)^3 (\coth (c)-1)+i d^2 e^2 (d e+3 i f) (d x-\log (-\cosh (c+d x)-\sinh (c+d x)+1))-i d^2 e^2 (d e-3 i f) (d x-\log (\cosh (c+d x)+\sinh (c+d x)+1))}{a d^4}+\frac {\text {sech}\left (\frac {c}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\sinh \left (\frac {d x}{2}\right ) e^3-3 f x \sinh \left (\frac {d x}{2}\right ) e^2-3 f^2 x^2 \sinh \left (\frac {d x}{2}\right ) e-f^3 x^3 \sinh \left (\frac {d x}{2}\right )\right )}{2 a d}+\frac {\text {csch}\left (\frac {c}{2}\right ) \text {csch}\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\sinh \left (\frac {d x}{2}\right ) e^3+3 f x \sinh \left (\frac {d x}{2}\right ) e^2+3 f^2 x^2 \sinh \left (\frac {d x}{2}\right ) e+f^3 x^3 \sinh \left (\frac {d x}{2}\right )\right )}{2 a d}-\frac {2 \left (\sinh \left (\frac {d x}{2}\right ) e^3+3 f x \sinh \left (\frac {d x}{2}\right ) e^2+3 f^2 x^2 \sinh \left (\frac {d x}{2}\right ) e+f^3 x^3 \sinh \left (\frac {d x}{2}\right )\right )}{a d \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}\right )+i \sinh \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Csch[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]

[Out]

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

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fricas [C]  time = 0.98, size = 2577, normalized size = 6.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.37, size = 1535, normalized size = 3.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x)

[Out]

6/a/d^2*e*f^2*ln(exp(d*x+c)+1)*x+I/a/d*f^3*ln(exp(d*x+c)+1)*x^3+I/a/d^4*f^3*c^3*ln(exp(d*x+c)-1)+6*I*f^3*polyl
og(4,-exp(d*x+c))/a/d^4-6*I*f^3*polylog(4,exp(d*x+c))/a/d^4+24/a/d^3*f^2*e*c*ln(exp(d*x+c))+12/a/d^2*f^2*e*ln(
1+I*exp(d*x+c))*x+12/a/d^3*f^2*e*ln(1+I*exp(d*x+c))*c-24/a/d^2*f^2*e*c*x-12/a/d^3*f^2*e*c*ln(exp(d*x+c)-I)-12*
f^3*polylog(3,-I*exp(d*x+c))/a/d^4-6*f^3*polylog(3,-exp(d*x+c))/a/d^4-6*f^3*polylog(3,exp(d*x+c))/a/d^4-2*I*(f
^3*x^3*exp(2*d*x+2*c)+3*e*f^2*x^2*exp(2*d*x+2*c)+3*e^2*f*x*exp(2*d*x+2*c)-2*x^3*f^3-I*exp(d*x+c)*f^3*x^3+e^3*e
xp(2*d*x+2*c)-6*e*f^2*x^2-3*I*exp(d*x+c)*e*f^2*x^2-6*e^2*f*x-3*I*exp(d*x+c)*e^2*f*x-2*e^3-I*exp(d*x+c)*e^3)/(e
xp(2*d*x+2*c)-1)/(exp(d*x+c)-I)/d/a-4/a/d*f^3*x^3+8/a/d^4*f^3*c^3+3*I/a/d*e*f^2*ln(exp(d*x+c)+1)*x^2+6*I/a/d^2
*e*f^2*polylog(2,-exp(d*x+c))*x-3*I/a/d*e*f^2*ln(1-exp(d*x+c))*x^2+3*I/a/d^3*e*f^2*ln(1-exp(d*x+c))*c^2-6*I/a/
d^2*e*f^2*polylog(2,exp(d*x+c))*x+3*I/a/d*ln(exp(d*x+c)+1)*e^2*f*x-3*I/a/d*ln(1-exp(d*x+c))*e^2*f*x-3*I/a/d^3*
e*f^2*c^2*ln(exp(d*x+c)-1)-3*I/a/d^2*ln(1-exp(d*x+c))*c*e^2*f+3*I/a/d^2*e^2*f*c*ln(exp(d*x+c)-1)-6/a/d^3*e*f^2
*c*ln(exp(d*x+c)-1)+6/a/d^2*e*f^2*ln(1-exp(d*x+c))*x+6/a/d^3*e*f^2*ln(1-exp(d*x+c))*c+3*I/a/d^2*e^2*f*polylog(
2,-exp(d*x+c))-3*I/a/d^2*e^2*f*polylog(2,exp(d*x+c))-I/a/d^4*f^3*c^3*ln(1-exp(d*x+c))+3*I/a/d^2*f^3*polylog(2,
-exp(d*x+c))*x^2-6*I/a/d^3*f^3*polylog(3,-exp(d*x+c))*x-I/a/d*f^3*ln(1-exp(d*x+c))*x^3-3*I/a/d^2*f^3*polylog(2
,exp(d*x+c))*x^2+6*I/a/d^3*f^3*polylog(3,exp(d*x+c))*x-6*I/a/d^3*e*f^2*polylog(3,-exp(d*x+c))+6*I/a/d^3*e*f^2*
polylog(3,exp(d*x+c))+12/a/d^3*f^3*polylog(2,-I*exp(d*x+c))*x+6/a/d^4*f^3*c^2*ln(exp(d*x+c)-I)-12/a/d^2*f*ln(e
xp(d*x+c))*e^2+12/a/d^3*f^3*c^2*x+6/a/d^2*f*ln(exp(d*x+c)-I)*e^2-12/a/d^3*f^2*e*c^2-12/a/d*f^2*e*x^2+12/a/d^3*
f^2*e*polylog(2,-I*exp(d*x+c))-12/a/d^4*f^3*c^2*ln(exp(d*x+c))+6/a/d^2*f^3*ln(1+I*exp(d*x+c))*x^2-6/a/d^4*f^3*
ln(1+I*exp(d*x+c))*c^2+3/a/d^4*f^3*c^2*ln(exp(d*x+c)-1)+6/a/d^3*e*f^2*polylog(2,-exp(d*x+c))+6/a/d^3*e*f^2*pol
ylog(2,exp(d*x+c))+3/a/d^2*e^2*f*ln(exp(d*x+c)+1)+3/a/d^2*e^2*f*ln(exp(d*x+c)-1)+3/a/d^2*f^3*ln(exp(d*x+c)+1)*
x^2+3/a/d^2*f^3*ln(1-exp(d*x+c))*x^2-3/a/d^4*f^3*ln(1-exp(d*x+c))*c^2+6/a/d^3*f^3*polylog(2,exp(d*x+c))*x+6/a/
d^3*f^3*polylog(2,-exp(d*x+c))*x+I/a/d*e^3*ln(exp(d*x+c)+1)-I/a/d*e^3*ln(exp(d*x+c)-1)

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maxima [B]  time = 1.07, size = 938, normalized size = 2.24 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e+f\,x\right )}^3}{{\mathrm {sinh}\left (c+d\,x\right )}^2\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^3/(sinh(c + d*x)^2*(a + a*sinh(c + d*x)*1i)),x)

[Out]

int((e + f*x)^3/(sinh(c + d*x)^2*(a + a*sinh(c + d*x)*1i)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*csch(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)

[Out]

Timed out

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