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

Optimal. Leaf size=163 \[ -\frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{12 i f^3 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}+\frac{i (e+f x)^3}{a d}-\frac{i (e+f x)^4}{4 a f} \]

[Out]

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

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Rubi [A]  time = 0.356154, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {5557, 32, 3318, 4184, 3716, 2190, 2531, 2282, 6589} \[ -\frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{12 i f^3 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac{6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i (e+f x)^3 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}+\frac{i (e+f x)^3}{a d}-\frac{i (e+f x)^4}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5557

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sinh[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sinh[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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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]

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 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]

Rubi steps

\begin{align*} \int \frac{(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac{(e+f x)^3}{a+i a \sinh (c+d x)} \, dx-\frac{i \int (e+f x)^3 \, dx}{a}\\ &=-\frac{i (e+f x)^4}{4 a f}+\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{i (e+f x)^4}{4 a f}+\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{i (e+f x)^3}{a d}-\frac{i (e+f x)^4}{4 a f}+\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{i (e+f x)^3}{a d}-\frac{i (e+f x)^4}{4 a f}-\frac{6 i f (e+f x)^2 \log \left (1+i 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{\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{i (e+f x)^3}{a d}-\frac{i (e+f x)^4}{4 a f}-\frac{6 i f (e+f x)^2 \log \left (1+i 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{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{i (e+f x)^4}{4 a f}-\frac{6 i f (e+f x)^2 \log \left (1+i 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{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{i (e+f x)^4}{4 a f}-\frac{6 i f (e+f x)^2 \log \left (1+i 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{12 i f^3 \text{Li}_3\left (-i 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 = 3.27641, size = 232, normalized size = 1.42 \[ \frac{-\frac{8 \left (6 i \left (-e^c+i\right ) f^2 \left (d (e+f x) \text{PolyLog}\left (2,i e^{-c-d x}\right )+f \text{PolyLog}\left (3,i e^{-c-d x}\right )\right )+3 \left (1+i e^c\right ) d^2 f (e+f x)^2 \log \left (1-i e^{-c-d x}\right )+d^3 (e+f x)^3\right )}{\left (e^c-i\right ) d^4}+\frac{8 i \sinh \left (\frac{d x}{2}\right ) (e+f x)^3}{d \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}-i x \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3) - (8*(d^3*(e + f*x)^3 + 3*d^2*(1 + I*E^c)*f*(e + f*x)^2*Lo
g[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)])))/(d^4*(-I + E^c)) + ((8*I)*(e + f*x)^3*Sinh[(d*x)/2])/(d*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[(c + d*x)/2] +
I*Sinh[(c + d*x)/2])))/(4*a)

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Maple [B]  time = 0.103, size = 501, normalized size = 3.1 \begin{align*}{\frac{12\,i{f}^{3}{\it polylog} \left ( 3,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{4}}}-{\frac{6\,i{f}^{3}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ){x}^{2}}{a{d}^{2}}}-{\frac{12\,i{f}^{3}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) x}{a{d}^{3}}}+{\frac{6\,ie{f}^{2}{c}^{2}}{a{d}^{3}}}-2\,{\frac{{x}^{3}{f}^{3}+3\,e{f}^{2}{x}^{2}+3\,{e}^{2}fx+{e}^{3}}{da \left ({{\rm e}^{dx+c}}-i \right ) }}-{\frac{6\,i\ln \left ({{\rm e}^{dx+c}}-i \right ){e}^{2}f}{a{d}^{2}}}+{\frac{12\,ie{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}-{\frac{4\,i{f}^{3}{c}^{3}}{a{d}^{4}}}-{\frac{12\,ie{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{6\,i{f}^{3}{c}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) }{a{d}^{4}}}+{\frac{12\,ie{f}^{2}cx}{a{d}^{2}}}-{\frac{6\,i{f}^{3}{c}^{2}x}{a{d}^{3}}}-{\frac{{\frac{i}{4}}{x}^{4}{f}^{3}}{a}}-{\frac{12\,ie{f}^{2}c\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}-{\frac{i{e}^{3}x}{a}}-{\frac{6\,i{f}^{3}{c}^{2}\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{4}}}+{\frac{2\,i{f}^{3}{x}^{3}}{da}}-{\frac{12\,ie{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{3}}}-{\frac{ie{f}^{2}{x}^{3}}{a}}-{\frac{12\,ie{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}+{\frac{6\,i{f}^{3}{c}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{4}}}+{\frac{6\,ie{f}^{2}{x}^{2}}{da}}-{\frac{{\frac{3\,i}{2}}{e}^{2}f{x}^{2}}{a}}+{\frac{6\,i\ln \left ({{\rm e}^{dx+c}} \right ){e}^{2}f}{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.7308, size = 428, normalized size = 2.63 \begin{align*} \frac{3}{2} \, e^{2} f{\left (\frac{-i \, d x^{2} +{\left (d x^{2} e^{c} - 4 \, x e^{c}\right )} e^{\left (d x\right )}}{i \, a d e^{\left (d x + c\right )} + a d} - \frac{4 i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} - e^{3}{\left (\frac{i \,{\left (d x + c\right )}}{a d} + \frac{2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} - \frac{d f^{3} x^{4} + 24 \, e f^{2} x^{2} + 4 \,{\left (d e f^{2} + 2 \, f^{3}\right )} x^{3} +{\left (i \, d f^{3} x^{4} e^{c} + 4 i \, d e f^{2} x^{3} e^{c}\right )} e^{\left (d x\right )}}{4 \,{\left (a d e^{\left (d x + c\right )} - i \, a d\right )}} - \frac{12 i \,{\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) +{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e f^{2}}{a d^{3}} - \frac{6 i \,{\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \,{\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} + \frac{2 i \, d^{3} f^{3} x^{3} + 6 i \, d^{3} e f^{2} x^{2}}{a d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*e^2*f*((-I*d*x^2 + (d*x^2*e^c - 4*x*e^c)*e^(d*x))/(I*a*d*e^(d*x + c) + a*d) - 4*I*log((e^(d*x + c) - I)*e^
(-c))/(a*d^2)) - e^3*(I*(d*x + c)/(a*d) + 2/((a*e^(-d*x - c) + I*a)*d)) - 1/4*(d*f^3*x^4 + 24*e*f^2*x^2 + 4*(d
*e*f^2 + 2*f^3)*x^3 + (I*d*f^3*x^4*e^c + 4*I*d*e*f^2*x^3*e^c)*e^(d*x))/(a*d*e^(d*x + c) - I*a*d) - 12*I*(d*x*l
og(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*e*f^2/(a*d^3) - 6*I*(d^2*x^2*log(I*e^(d*x + c) + 1) + 2*d*x*dil
og(-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 = 2.48288, size = 1096, normalized size = 6.72 \begin{align*} -\frac{d^{4} f^{3} x^{4} + 4 \, d^{4} e f^{2} x^{3} + 6 \, d^{4} e^{2} f x^{2} + 4 \, d^{4} e^{3} x + 8 \, d^{3} e^{3} - 24 \, c d^{2} e^{2} f + 24 \, c^{2} d e f^{2} - 8 \, c^{3} f^{3} +{\left (48 \, d f^{3} x + 48 \, d e f^{2} -{\left (-48 i \, d f^{3} x - 48 i \, d e f^{2}\right )} e^{\left (d x + c\right )}\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) -{\left (-i \, d^{4} f^{3} x^{4} + 24 i \, c d^{2} e^{2} f - 24 i \, c^{2} d e f^{2} + 8 i \, c^{3} f^{3} +{\left (-4 i \, d^{4} e f^{2} + 8 i \, d^{3} f^{3}\right )} x^{3} +{\left (-6 i \, d^{4} e^{2} f + 24 i \, d^{3} e f^{2}\right )} x^{2} +{\left (-4 i \, d^{4} e^{3} + 24 i \, d^{3} e^{2} f\right )} x\right )} e^{\left (d x + c\right )} +{\left (24 \, d^{2} e^{2} f - 48 \, c d e f^{2} + 24 \, c^{2} f^{3} -{\left (-24 i \, d^{2} e^{2} f + 48 i \, c d e f^{2} - 24 i \, c^{2} f^{3}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (24 \, d^{2} f^{3} x^{2} + 48 \, d^{2} e f^{2} x + 48 \, c d e f^{2} - 24 \, c^{2} f^{3} -{\left (-24 i \, d^{2} f^{3} x^{2} - 48 i \, d^{2} e f^{2} x - 48 i \, c d e f^{2} + 24 i \, c^{2} f^{3}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) -{\left (48 i \, f^{3} e^{\left (d x + c\right )} + 48 \, f^{3}\right )}{\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right )}{4 \, a d^{4} e^{\left (d x + c\right )} - 4 i \, a d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d^4*f^3*x^4 + 4*d^4*e*f^2*x^3 + 6*d^4*e^2*f*x^2 + 4*d^4*e^3*x + 8*d^3*e^3 - 24*c*d^2*e^2*f + 24*c^2*d*e*f^2
- 8*c^3*f^3 + (48*d*f^3*x + 48*d*e*f^2 - (-48*I*d*f^3*x - 48*I*d*e*f^2)*e^(d*x + c))*dilog(-I*e^(d*x + c)) - (
-I*d^4*f^3*x^4 + 24*I*c*d^2*e^2*f - 24*I*c^2*d*e*f^2 + 8*I*c^3*f^3 + (-4*I*d^4*e*f^2 + 8*I*d^3*f^3)*x^3 + (-6*
I*d^4*e^2*f + 24*I*d^3*e*f^2)*x^2 + (-4*I*d^4*e^3 + 24*I*d^3*e^2*f)*x)*e^(d*x + c) + (24*d^2*e^2*f - 48*c*d*e*
f^2 + 24*c^2*f^3 - (-24*I*d^2*e^2*f + 48*I*c*d*e*f^2 - 24*I*c^2*f^3)*e^(d*x + c))*log(e^(d*x + c) - I) + (24*d
^2*f^3*x^2 + 48*d^2*e*f^2*x + 48*c*d*e*f^2 - 24*c^2*f^3 - (-24*I*d^2*f^3*x^2 - 48*I*d^2*e*f^2*x - 48*I*c*d*e*f
^2 + 24*I*c^2*f^3)*e^(d*x + c))*log(I*e^(d*x + c) + 1) - (48*I*f^3*e^(d*x + c) + 48*f^3)*polylog(3, -I*e^(d*x
+ c)))/(4*a*d^4*e^(d*x + c) - 4*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*sinh(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} \sinh \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*sinh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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