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

Optimal. Leaf size=241 \[ \frac{12 f^2 (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}-\frac{12 f^3 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}+\frac{6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}+\frac{6 i f^3 \sinh (c+d x)}{a d^4}-\frac{i (e+f x)^3 \cosh (c+d x)}{a d}-\frac{(e+f x)^3 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x)^3}{a d}+\frac{(e+f x)^4}{4 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.526142, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.355, Rules used = {5557, 3296, 2637, 32, 3318, 4184, 3716, 2190, 2531, 2282, 6589} \[ \frac{12 f^2 (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}-\frac{12 f^3 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}+\frac{6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}+\frac{6 i f^3 \sinh (c+d x)}{a d^4}-\frac{i (e+f x)^3 \cosh (c+d x)}{a d}-\frac{(e+f x)^3 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x)^3}{a d}+\frac{(e+f x)^4}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e + f*x)^3/(a*d)) + (e + f*x)^4/(4*a*f) - ((6*I)*f^2*(e + f*x)*Cosh[c + d*x])/(a*d^3) - (I*(e + f*x)^3*Cosh
[c + d*x])/(a*d) + (6*f*(e + f*x)^2*Log[1 + I*E^(c + d*x)])/(a*d^2) + (12*f^2*(e + f*x)*PolyLog[2, (-I)*E^(c +
d*x)])/(a*d^3) - (12*f^3*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^4) + ((6*I)*f^3*Sinh[c + d*x])/(a*d^4) + ((3*I)*f
*(e + f*x)^2*Sinh[c + d*x])/(a*d^2) - ((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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac{(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac{i \int (e+f x)^3 \sinh (c+d x) \, dx}{a}\\ &=-\frac{i (e+f x)^3 \cosh (c+d x)}{a d}+\frac{\int (e+f x)^3 \, dx}{a}+\frac{(3 i f) \int (e+f x)^2 \cosh (c+d x) \, dx}{a d}-\int \frac{(e+f x)^3}{a+i a \sinh (c+d x)} \, dx\\ &=\frac{(e+f x)^4}{4 a f}-\frac{i (e+f x)^3 \cosh (c+d x)}{a d}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\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{\left (6 i f^2\right ) \int (e+f x) \sinh (c+d x) \, dx}{a d^2}\\ &=\frac{(e+f x)^4}{4 a f}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^3 \cosh (c+d x)}{a d}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{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^3\right ) \int \cosh (c+d x) \, dx}{a d^3}\\ &=-\frac{(e+f x)^3}{a d}+\frac{(e+f x)^4}{4 a f}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^3 \cosh (c+d x)}{a d}+\frac{6 i f^3 \sinh (c+d x)}{a d^4}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{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{(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{(e+f x)^3}{a d}+\frac{(e+f x)^4}{4 a f}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^3 \cosh (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{6 i f^3 \sinh (c+d x)}{a d^4}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{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{\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{(e+f x)^3}{a d}+\frac{(e+f x)^4}{4 a f}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^3 \cosh (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{12 f^2 (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac{6 i f^3 \sinh (c+d x)}{a d^4}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{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{\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{(e+f x)^3}{a d}+\frac{(e+f x)^4}{4 a f}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^3 \cosh (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{12 f^2 (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac{6 i f^3 \sinh (c+d x)}{a d^4}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{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{\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{(e+f x)^3}{a d}+\frac{(e+f x)^4}{4 a f}-\frac{6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^3 \cosh (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{12 f^2 (e+f x) \text{Li}_2\left (-i 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^3 \sinh (c+d x)}{a d^4}+\frac{3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac{(e+f x)^3 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}\\ \end{align*}

Mathematica [B]  time = 6.5947, size = 857, normalized size = 3.56 \[ \frac{\frac{i f^3 x^4 \sinh \left (c+\frac{d x}{2}\right ) d^4+4 i e f^2 x^3 \sinh \left (c+\frac{d x}{2}\right ) d^4+6 i e^2 f x^2 \sinh \left (c+\frac{d x}{2}\right ) d^4+4 i e^3 x \sinh \left (c+\frac{d x}{2}\right ) d^4-10 e^3 \sinh \left (\frac{d x}{2}\right ) d^3-10 f^3 x^3 \sinh \left (\frac{d x}{2}\right ) d^3-30 e f^2 x^2 \sinh \left (\frac{d x}{2}\right ) d^3-30 e^2 f x \sinh \left (\frac{d x}{2}\right ) d^3+2 e^3 \sinh \left (2 c+\frac{3 d x}{2}\right ) d^3+2 f^3 x^3 \sinh \left (2 c+\frac{3 d x}{2}\right ) d^3+6 e f^2 x^2 \sinh \left (2 c+\frac{3 d x}{2}\right ) d^3+6 e^2 f x \sinh \left (2 c+\frac{3 d x}{2}\right ) d^3-6 f^3 x^2 \cosh \left (2 c+\frac{3 d x}{2}\right ) d^2-6 e^2 f \cosh \left (2 c+\frac{3 d x}{2}\right ) d^2-12 e f^2 x \cosh \left (2 c+\frac{3 d x}{2}\right ) d^2+6 i f^3 x^2 \sinh \left (c+\frac{d x}{2}\right ) d^2+6 i e^2 f \sinh \left (c+\frac{d x}{2}\right ) d^2+12 i e f^2 x \sinh \left (c+\frac{d x}{2}\right ) d^2+6 i f^3 x^2 \sinh \left (c+\frac{3 d x}{2}\right ) d^2+6 i e^2 f \sinh \left (c+\frac{3 d x}{2}\right ) d^2+12 i e f^2 x \sinh \left (c+\frac{3 d x}{2}\right ) d^2-2 i (e+f x) \left (6 f^2+d^2 (e+f x)^2\right ) \cosh \left (c+\frac{d x}{2}\right ) d-2 i (e+f x) \left (6 f^2+d^2 (e+f x)^2\right ) \cosh \left (c+\frac{3 d x}{2}\right ) d-12 e f^2 \sinh \left (\frac{d x}{2}\right ) d-12 f^3 x \sinh \left (\frac{d x}{2}\right ) d+12 e f^2 \sinh \left (2 c+\frac{3 d x}{2}\right ) d+12 f^3 x \sinh \left (2 c+\frac{3 d x}{2}\right ) d+\left (x \left (4 e^3+6 f x e^2+4 f^2 x^2 e+f^3 x^3\right ) d^4+6 f (e+f x)^2 d^2+12 f^3\right ) \cosh \left (\frac{d x}{2}\right )-12 f^3 \cosh \left (2 c+\frac{3 d x}{2}\right )+12 i f^3 \sinh \left (c+\frac{d x}{2}\right )+12 i f^3 \sinh \left (c+\frac{3 d x}{2}\right )}{\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 )}-\frac{8 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{PolyLog}\left (2,i e^{-c-d x}\right )+f \text{PolyLog}\left (3,i e^{-c-d x}\right )\right )\right )}{-i+e^c}}{4 a d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-8*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)])))/(-I + E^c) + ((12*f^3 + 6*d^2*f*(e + f*x
)^2 + d^4*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3))*Cosh[(d*x)/2] - (2*I)*d*(e + f*x)*(6*f^2 + d^2*(e + f
*x)^2)*Cosh[c + (d*x)/2] - (2*I)*d*(e + f*x)*(6*f^2 + d^2*(e + f*x)^2)*Cosh[c + (3*d*x)/2] - 6*d^2*e^2*f*Cosh[
2*c + (3*d*x)/2] - 12*f^3*Cosh[2*c + (3*d*x)/2] - 12*d^2*e*f^2*x*Cosh[2*c + (3*d*x)/2] - 6*d^2*f^3*x^2*Cosh[2*
c + (3*d*x)/2] - 10*d^3*e^3*Sinh[(d*x)/2] - 12*d*e*f^2*Sinh[(d*x)/2] - 30*d^3*e^2*f*x*Sinh[(d*x)/2] - 12*d*f^3
*x*Sinh[(d*x)/2] - 30*d^3*e*f^2*x^2*Sinh[(d*x)/2] - 10*d^3*f^3*x^3*Sinh[(d*x)/2] + (6*I)*d^2*e^2*f*Sinh[c + (d
*x)/2] + (12*I)*f^3*Sinh[c + (d*x)/2] + (4*I)*d^4*e^3*x*Sinh[c + (d*x)/2] + (12*I)*d^2*e*f^2*x*Sinh[c + (d*x)/
2] + (6*I)*d^4*e^2*f*x^2*Sinh[c + (d*x)/2] + (6*I)*d^2*f^3*x^2*Sinh[c + (d*x)/2] + (4*I)*d^4*e*f^2*x^3*Sinh[c
+ (d*x)/2] + I*d^4*f^3*x^4*Sinh[c + (d*x)/2] + (6*I)*d^2*e^2*f*Sinh[c + (3*d*x)/2] + (12*I)*f^3*Sinh[c + (3*d*
x)/2] + (12*I)*d^2*e*f^2*x*Sinh[c + (3*d*x)/2] + (6*I)*d^2*f^3*x^2*Sinh[c + (3*d*x)/2] + 2*d^3*e^3*Sinh[2*c +
(3*d*x)/2] + 12*d*e*f^2*Sinh[2*c + (3*d*x)/2] + 6*d^3*e^2*f*x*Sinh[2*c + (3*d*x)/2] + 12*d*f^3*x*Sinh[2*c + (3
*d*x)/2] + 6*d^3*e*f^2*x^2*Sinh[2*c + (3*d*x)/2] + 2*d^3*f^3*x^3*Sinh[2*c + (3*d*x)/2])/((Cosh[c/2] + I*Sinh[c
/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])))/(4*a*d^4)

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Maple [B]  time = 0.183, size = 688, normalized size = 2.9 \begin{align*} 12\,{\frac{e{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}+12\,{\frac{e{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{3}}}+12\,{\frac{e{f}^{2}c\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}-12\,{\frac{e{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}-12\,{\frac{e{f}^{2}cx}{a{d}^{2}}}+4\,{\frac{{c}^{3}{f}^{3}}{a{d}^{4}}}-2\,{\frac{{x}^{3}{f}^{3}}{da}}-{\frac{{\frac{i}{2}} \left ({f}^{3}{x}^{3}{d}^{3}+3\,{d}^{3}e{f}^{2}{x}^{2}+3\,{d}^{3}{e}^{2}fx-3\,{d}^{2}{f}^{3}{x}^{2}+{d}^{3}{e}^{3}-6\,{d}^{2}e{f}^{2}x-3\,{e}^{2}f{d}^{2}+6\,d{f}^{3}x+6\,e{f}^{2}d-6\,{f}^{3} \right ){{\rm e}^{dx+c}}}{a{d}^{4}}}-{\frac{2\,i \left ({x}^{3}{f}^{3}+3\,e{f}^{2}{x}^{2}+3\,{e}^{2}fx+{e}^{3} \right ) }{da \left ({{\rm e}^{dx+c}}-i \right ) }}-12\,{\frac{{f}^{3}{\it polylog} \left ( 3,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{4}}}+6\,{\frac{{f}^{3}{c}^{2}x}{a{d}^{3}}}-6\,{\frac{f\ln \left ({{\rm e}^{dx+c}} \right ){e}^{2}}{a{d}^{2}}}+6\,{\frac{{f}^{3}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ){x}^{2}}{a{d}^{2}}}-6\,{\frac{{f}^{3}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ){c}^{2}}{a{d}^{4}}}-6\,{\frac{e{c}^{2}{f}^{2}}{a{d}^{3}}}+6\,{\frac{f\ln \left ({{\rm e}^{dx+c}}-i \right ){e}^{2}}{a{d}^{2}}}+6\,{\frac{{f}^{3}{c}^{2}\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{4}}}-6\,{\frac{e{f}^{2}{x}^{2}}{da}}-6\,{\frac{{f}^{3}{c}^{2}\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{4}}}+12\,{\frac{{f}^{3}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) x}{a{d}^{3}}}+12\,{\frac{e{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{e{f}^{2}{x}^{3}}{a}}+{\frac{3\,{e}^{2}f{x}^{2}}{2\,a}}-{\frac{{\frac{i}{2}} \left ({f}^{3}{x}^{3}{d}^{3}+3\,{d}^{3}e{f}^{2}{x}^{2}+3\,{d}^{3}{e}^{2}fx+3\,{d}^{2}{f}^{3}{x}^{2}+{d}^{3}{e}^{3}+6\,{d}^{2}e{f}^{2}x+3\,{e}^{2}f{d}^{2}+6\,d{f}^{3}x+6\,e{f}^{2}d+6\,{f}^{3} \right ){{\rm e}^{-dx-c}}}{a{d}^{4}}}+{\frac{{x}^{4}{f}^{3}}{4\,a}}+{\frac{{e}^{3}x}{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.98274, size = 906, normalized size = 3.76 \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*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-3/4*e^2*f*(4*x*e^(d*x + c)/(a*d*e^(d*x + c) - I*a*d) - (-2*I*d^2*x^2*e^c - 2*I*d*x*e^c - (2*I*d*x*e^(3*c) - 2
*I*e^(3*c))*e^(2*d*x) + 2*(d^2*x^2*e^(2*c) - 3*d*x*e^(2*c) + e^(2*c))*e^(d*x) - 2*(d*x + 1)*e^(-d*x) - 2*I*e^c
)/(a*d^2*e^(d*x + 2*c) - I*a*d^2*e^c) - 8*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) + 1/4*e^3*(4*(d*x + c)/(a*d)
+ 2*(-5*I*e^(-d*x - c) + 1)/((I*a*e^(-d*x - c) + a*e^(-2*d*x - 2*c))*d) - 2*I*e^(-d*x - c)/(a*d)) + 1/4*(-I*d^
4*f^3*x^4 - 12*I*d*e*f^2 - (4*I*d^4*e*f^2 + 10*I*d^3*f^3)*x^3 - 12*I*f^3 - (30*I*d^3*e*f^2 + 6*I*d^2*f^3)*x^2
- (12*I*d^2*e*f^2 + 12*I*d*f^3)*x - (2*I*d^3*f^3*x^3*e^(2*c) + (6*I*d^3*e*f^2 - 6*I*d^2*f^3)*x^2*e^(2*c) + (-1
2*I*d^2*e*f^2 + 12*I*d*f^3)*x*e^(2*c) + (12*I*d*e*f^2 - 12*I*f^3)*e^(2*c))*e^(2*d*x) + (d^4*f^3*x^4*e^c + 2*(2
*d^4*e*f^2 - d^3*f^3)*x^3*e^c - 6*(d^3*e*f^2 - d^2*f^3)*x^2*e^c + 12*(d^2*e*f^2 - d*f^3)*x*e^c - 12*(d*e*f^2 -
f^3)*e^c)*e^(d*x))/(a*d^4*e^(d*x + c) - I*a*d^4) + 12*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*e*
f^2/(a*d^3) + 6*(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*(d^3*f^3*x^3 + 3*d^3*e*f^2*x^2)/(a*d^4)

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Fricas [C]  time = 2.6279, size = 1904, normalized size = 7.9 \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*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(2*d^3*f^3*x^3 + 2*d^3*e^3 + 6*d^2*e^2*f + 12*d*e*f^2 + 12*f^3 + 6*(d^3*e*f^2 + d^2*f^3)*x^2 + 6*(d^3*e^2*f +
2*d^2*e*f^2 + 2*d*f^3)*x - (48*(d*f^3*x + d*e*f^2)*e^(2*d*x + 2*c) + (-48*I*d*f^3*x - 48*I*d*e*f^2)*e^(d*x +
c))*dilog(-I*e^(d*x + c)) - (-2*I*d^3*f^3*x^3 - 2*I*d^3*e^3 + 6*I*d^2*e^2*f - 12*I*d*e*f^2 + 12*I*f^3 + (-6*I*
d^3*e*f^2 + 6*I*d^2*f^3)*x^2 + (-6*I*d^3*e^2*f + 12*I*d^2*e*f^2 - 12*I*d*f^3)*x)*e^(3*d*x + 3*c) - (d^4*f^3*x^
4 - 2*d^3*e^3 - 6*(4*c - 1)*d^2*e^2*f + 12*(2*c^2 - 1)*d*e*f^2 - 4*(2*c^3 - 3)*f^3 + 2*(2*d^4*e*f^2 - 5*d^3*f^
3)*x^3 + 6*(d^4*e^2*f - 5*d^3*e*f^2 + d^2*f^3)*x^2 + 2*(2*d^4*e^3 - 15*d^3*e^2*f + 6*d^2*e*f^2 - 6*d*f^3)*x)*e
^(2*d*x + 2*c) - (-I*d^4*f^3*x^4 - 10*I*d^3*e^3 + (24*I*c - 6*I)*d^2*e^2*f + (-24*I*c^2 - 12*I)*d*e*f^2 + (8*I
*c^3 - 12*I)*f^3 + (-4*I*d^4*e*f^2 - 2*I*d^3*f^3)*x^3 + (-6*I*d^4*e^2*f - 6*I*d^3*e*f^2 - 6*I*d^2*f^3)*x^2 + (
-4*I*d^4*e^3 - 6*I*d^3*e^2*f - 12*I*d^2*e*f^2 - 12*I*d*f^3)*x)*e^(d*x + c) - (24*(d^2*e^2*f - 2*c*d*e*f^2 + c^
2*f^3)*e^(2*d*x + 2*c) + (-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 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*e^(2*d*x + 2*c) + (-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*(f^3*e^(2*d*x + 2*c) - I*f^3*e^
(d*x + c))*polylog(3, -I*e^(d*x + c)))/(4*a*d^4*e^(2*d*x + 2*c) - 4*I*a*d^4*e^(d*x + c))

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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)**2/(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 )^{2}}{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)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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