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

Optimal. Leaf size=287 \[ \frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}+\frac{2 f^2 \cosh (c+d x)}{a d^3}-\frac{i f^2 \sinh (c+d x) \cosh (c+d x)}{4 a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^2 \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.546854, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.419, Rules used = {5557, 3311, 32, 2635, 8, 3296, 2638, 3318, 4184, 3716, 2190, 2279, 2391} \[ \frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}+\frac{2 f^2 \cosh (c+d x)}{a d^3}-\frac{i f^2 \sinh (c+d x) \cosh (c+d x)}{4 a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^2 \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*f^2*x)/(a*d^2) - (I*(e + f*x)^2)/(a*d) + ((I/2)*(e + f*x)^3)/(a*f) + (2*f^2*Cosh[c + d*x])/(a*d^3) + ((
e + f*x)^2*Cosh[c + d*x])/(a*d) + ((4*I)*f*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a*d^2) + ((4*I)*f^2*PolyLog[2, (
-I)*E^(c + d*x)])/(a*d^3) - (2*f*(e + f*x)*Sinh[c + d*x])/(a*d^2) - ((I/4)*f^2*Cosh[c + d*x]*Sinh[c + d*x])/(a
*d^3) - ((I/2)*(e + f*x)^2*Cosh[c + d*x]*Sinh[c + d*x])/(a*d) + ((I/2)*f*(e + f*x)*Sinh[c + d*x]^2)/(a*d^2) -
(I*(e + f*x)^2*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 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2638

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

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
 e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [B]  time = 5.13595, size = 1661, normalized size = 5.79 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*I)*d^2*e^2*E^c*Cosh[(3*d*x)/2] + 6*d^2*e^2*E^(4*c)*Cosh[(3*d*x)/2] - (14*I)*d*e*E^c*f*Cosh[(3*d*x)/2] - 1
4*d*e*E^(4*c)*f*Cosh[(3*d*x)/2] - (15*I)*E^c*f^2*Cosh[(3*d*x)/2] + 15*E^(4*c)*f^2*Cosh[(3*d*x)/2] - (12*I)*d^2
*e*E^c*f*x*Cosh[(3*d*x)/2] + 12*d^2*e*E^(4*c)*f*x*Cosh[(3*d*x)/2] - (14*I)*d*E^c*f^2*x*Cosh[(3*d*x)/2] - 14*d*
E^(4*c)*f^2*x*Cosh[(3*d*x)/2] - (6*I)*d^2*E^c*f^2*x^2*Cosh[(3*d*x)/2] + 6*d^2*E^(4*c)*f^2*x^2*Cosh[(3*d*x)/2]
+ 2*d^2*e^2*Cosh[(5*d*x)/2] - (2*I)*d^2*e^2*E^(5*c)*Cosh[(5*d*x)/2] + 2*d*e*f*Cosh[(5*d*x)/2] + (2*I)*d*e*E^(5
*c)*f*Cosh[(5*d*x)/2] + f^2*Cosh[(5*d*x)/2] - I*E^(5*c)*f^2*Cosh[(5*d*x)/2] + 4*d^2*e*f*x*Cosh[(5*d*x)/2] - (4
*I)*d^2*e*E^(5*c)*f*x*Cosh[(5*d*x)/2] + 2*d*f^2*x*Cosh[(5*d*x)/2] + (2*I)*d*E^(5*c)*f^2*x*Cosh[(5*d*x)/2] + 2*
d^2*f^2*x^2*Cosh[(5*d*x)/2] - (2*I)*d^2*E^(5*c)*f^2*x^2*Cosh[(5*d*x)/2] + 8*E^(2*c)*Cosh[(d*x)/2]*(2*(1 - I*E^
c)*f^2 + 2*d*(1 + I*E^c)*f*(e + f*x) + d^2*(5 - I*E^c)*(e + f*x)^2 + d^3*(1 + I*E^c)*x*(3*e^2 + 3*e*f*x + f^2*
x^2) + 8*d*(1 + I*E^c)*f*(e + f*x)*Log[1 - I*E^(-c - d*x)]) - 40*d^2*e^2*E^(2*c)*Sinh[(d*x)/2] - (8*I)*d^2*e^2
*E^(3*c)*Sinh[(d*x)/2] - 16*d*e*E^(2*c)*f*Sinh[(d*x)/2] + (16*I)*d*e*E^(3*c)*f*Sinh[(d*x)/2] - 16*E^(2*c)*f^2*
Sinh[(d*x)/2] - (16*I)*E^(3*c)*f^2*Sinh[(d*x)/2] - 24*d^3*e^2*E^(2*c)*x*Sinh[(d*x)/2] + (24*I)*d^3*e^2*E^(3*c)
*x*Sinh[(d*x)/2] - 80*d^2*e*E^(2*c)*f*x*Sinh[(d*x)/2] - (16*I)*d^2*e*E^(3*c)*f*x*Sinh[(d*x)/2] - 16*d*E^(2*c)*
f^2*x*Sinh[(d*x)/2] + (16*I)*d*E^(3*c)*f^2*x*Sinh[(d*x)/2] - 24*d^3*e*E^(2*c)*f*x^2*Sinh[(d*x)/2] + (24*I)*d^3
*e*E^(3*c)*f*x^2*Sinh[(d*x)/2] - 40*d^2*E^(2*c)*f^2*x^2*Sinh[(d*x)/2] - (8*I)*d^2*E^(3*c)*f^2*x^2*Sinh[(d*x)/2
] - 8*d^3*E^(2*c)*f^2*x^3*Sinh[(d*x)/2] + (8*I)*d^3*E^(3*c)*f^2*x^3*Sinh[(d*x)/2] - 64*d*e*E^(2*c)*f*Log[1 - I
*E^(-c - d*x)]*Sinh[(d*x)/2] + (64*I)*d*e*E^(3*c)*f*Log[1 - I*E^(-c - d*x)]*Sinh[(d*x)/2] - 64*d*E^(2*c)*f^2*x
*Log[1 - I*E^(-c - d*x)]*Sinh[(d*x)/2] + (64*I)*d*E^(3*c)*f^2*x*Log[1 - I*E^(-c - d*x)]*Sinh[(d*x)/2] + 64*E^(
2*c)*f^2*PolyLog[2, I*E^(-c - d*x)]*((-1 - I*E^c)*Cosh[(d*x)/2] + (1 - I*E^c)*Sinh[(d*x)/2]) + (6*I)*d^2*e^2*E
^c*Sinh[(3*d*x)/2] + 6*d^2*e^2*E^(4*c)*Sinh[(3*d*x)/2] + (14*I)*d*e*E^c*f*Sinh[(3*d*x)/2] - 14*d*e*E^(4*c)*f*S
inh[(3*d*x)/2] + (15*I)*E^c*f^2*Sinh[(3*d*x)/2] + 15*E^(4*c)*f^2*Sinh[(3*d*x)/2] + (12*I)*d^2*e*E^c*f*x*Sinh[(
3*d*x)/2] + 12*d^2*e*E^(4*c)*f*x*Sinh[(3*d*x)/2] + (14*I)*d*E^c*f^2*x*Sinh[(3*d*x)/2] - 14*d*E^(4*c)*f^2*x*Sin
h[(3*d*x)/2] + (6*I)*d^2*E^c*f^2*x^2*Sinh[(3*d*x)/2] + 6*d^2*E^(4*c)*f^2*x^2*Sinh[(3*d*x)/2] - 2*d^2*e^2*Sinh[
(5*d*x)/2] - (2*I)*d^2*e^2*E^(5*c)*Sinh[(5*d*x)/2] - 2*d*e*f*Sinh[(5*d*x)/2] + (2*I)*d*e*E^(5*c)*f*Sinh[(5*d*x
)/2] - f^2*Sinh[(5*d*x)/2] - I*E^(5*c)*f^2*Sinh[(5*d*x)/2] - 4*d^2*e*f*x*Sinh[(5*d*x)/2] - (4*I)*d^2*e*E^(5*c)
*f*x*Sinh[(5*d*x)/2] - 2*d*f^2*x*Sinh[(5*d*x)/2] + (2*I)*d*E^(5*c)*f^2*x*Sinh[(5*d*x)/2] - 2*d^2*f^2*x^2*Sinh[
(5*d*x)/2] - (2*I)*d^2*E^(5*c)*f^2*x^2*Sinh[(5*d*x)/2])/(16*a*d^3*E^(2*c)*((-I + E^c)*Cosh[(d*x)/2] + (I + E^c
)*Sinh[(d*x)/2]))

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Maple [A]  time = 0.128, size = 508, normalized size = 1.8 \begin{align*}{\frac{-2\,i{f}^{2}{c}^{2}}{a{d}^{3}}}+{\frac{{\frac{i}{16}} \left ( 2\,{f}^{2}{x}^{2}{d}^{2}+4\,{d}^{2}efx+2\,{d}^{2}{e}^{2}+2\,d{f}^{2}x+2\,efd+{f}^{2} \right ){{\rm e}^{-2\,dx-2\,c}}}{a{d}^{3}}}-{\frac{2\,i{f}^{2}{x}^{2}}{da}}-{\frac{4\,i\ln \left ({{\rm e}^{dx+c}} \right ) ef}{a{d}^{2}}}+{\frac{ \left ({f}^{2}{x}^{2}{d}^{2}+2\,{d}^{2}efx+{d}^{2}{e}^{2}-2\,d{f}^{2}x-2\,efd+2\,{f}^{2} \right ){{\rm e}^{dx+c}}}{2\,a{d}^{3}}}+{\frac{ \left ({f}^{2}{x}^{2}{d}^{2}+2\,{d}^{2}efx+{d}^{2}{e}^{2}+2\,d{f}^{2}x+2\,efd+2\,{f}^{2} \right ){{\rm e}^{-dx-c}}}{2\,a{d}^{3}}}+{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{3}}}+2\,{\frac{{x}^{2}{f}^{2}+2\,efx+{e}^{2}}{da \left ({{\rm e}^{dx+c}}-i \right ) }}-{\frac{4\,i{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}-{\frac{{\frac{i}{16}} \left ( 2\,{f}^{2}{x}^{2}{d}^{2}+4\,{d}^{2}efx+2\,{d}^{2}{e}^{2}-2\,d{f}^{2}x-2\,efd+{f}^{2} \right ){{\rm e}^{2\,dx+2\,c}}}{a{d}^{3}}}+{\frac{4\,i{f}^{2}c\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{{\frac{3\,i}{2}}{e}^{2}x}{a}}-{\frac{4\,i{f}^{2}cx}{a{d}^{2}}}+{\frac{{\frac{i}{2}}{x}^{3}{f}^{2}}{a}}+{\frac{4\,i{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{{\frac{3\,i}{2}}ef{x}^{2}}{a}}+{\frac{4\,i\ln \left ({{\rm e}^{dx+c}}-i \right ) ef}{a{d}^{2}}}+{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/a/d^3*f^2*c^2+1/16*I*(2*d^2*f^2*x^2+4*d^2*e*f*x+2*d^2*e^2+2*d*f^2*x+2*d*e*f+f^2)/a/d^3*exp(-2*d*x-2*c)-2*
I/a/d*f^2*x^2-4*I/a/d^2*ln(exp(d*x+c))*e*f+1/2*(d^2*f^2*x^2+2*d^2*e*f*x+d^2*e^2-2*d*f^2*x-2*d*e*f+2*f^2)/a/d^3
*exp(d*x+c)+1/2*(d^2*f^2*x^2+2*d^2*e*f*x+d^2*e^2+2*d*f^2*x+2*d*e*f+2*f^2)/a/d^3*exp(-d*x-c)+4*I/a/d^3*f^2*ln(1
+I*exp(d*x+c))*c+2*(f^2*x^2+2*e*f*x+e^2)/d/a/(exp(d*x+c)-I)-4*I/a/d^3*f^2*c*ln(exp(d*x+c)-I)-1/16*I*(2*d^2*f^2
*x^2+4*d^2*e*f*x+2*d^2*e^2-2*d*f^2*x-2*d*e*f+f^2)/a/d^3*exp(2*d*x+2*c)+4*I/a/d^3*f^2*c*ln(exp(d*x+c))+3/2*I/a*
e^2*x-4*I/a/d^2*f^2*c*x+1/2*I/a*x^3*f^2+4*I*f^2*polylog(2,-I*exp(d*x+c))/a/d^3+3/2*I/a*e*f*x^2+4*I/a/d^2*ln(ex
p(d*x+c)-I)*e*f+4*I/a/d^2*f^2*ln(1+I*exp(d*x+c))*x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.67821, size = 1419, normalized size = 4.94 \begin{align*} \frac{2 \, d^{2} f^{2} x^{2} + 2 \, d^{2} e^{2} + 2 \, d e f + f^{2} + 2 \,{\left (2 \, d^{2} e f + d f^{2}\right )} x +{\left (64 i \, f^{2} e^{\left (3 \, d x + 3 \, c\right )} + 64 \, f^{2} e^{\left (2 \, d x + 2 \, c\right )}\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) +{\left (-2 i \, d^{2} f^{2} x^{2} - 2 i \, d^{2} e^{2} + 2 i \, d e f - i \, f^{2} +{\left (-4 i \, d^{2} e f + 2 i \, d f^{2}\right )} x\right )} e^{\left (5 \, d x + 5 \, c\right )} +{\left (6 \, d^{2} f^{2} x^{2} + 6 \, d^{2} e^{2} - 14 \, d e f + 15 \, f^{2} + 2 \,{\left (6 \, d^{2} e f - 7 \, d f^{2}\right )} x\right )} e^{\left (4 \, d x + 4 \, c\right )} +{\left (8 i \, d^{3} f^{2} x^{3} - 8 i \, d^{2} e^{2} +{\left (-64 i \, c + 16 i\right )} d e f +{\left (32 i \, c^{2} - 16 i\right )} f^{2} +{\left (24 i \, d^{3} e f - 40 i \, d^{2} f^{2}\right )} x^{2} +{\left (24 i \, d^{3} e^{2} - 80 i \, d^{2} e f + 16 i \, d f^{2}\right )} x\right )} e^{\left (3 \, d x + 3 \, c\right )} + 8 \,{\left (d^{3} f^{2} x^{3} + 5 \, d^{2} e^{2} - 2 \,{\left (4 \, c - 1\right )} d e f + 2 \,{\left (2 \, c^{2} + 1\right )} f^{2} +{\left (3 \, d^{3} e f + d^{2} f^{2}\right )} x^{2} +{\left (3 \, d^{3} e^{2} + 2 \, d^{2} e f + 2 \, d f^{2}\right )} x\right )} e^{\left (2 \, d x + 2 \, c\right )} +{\left (-6 i \, d^{2} f^{2} x^{2} - 6 i \, d^{2} e^{2} - 14 i \, d e f - 15 i \, f^{2} +{\left (-12 i \, d^{2} e f - 14 i \, d f^{2}\right )} x\right )} e^{\left (d x + c\right )} +{\left ({\left (64 i \, d e f - 64 i \, c f^{2}\right )} e^{\left (3 \, d x + 3 \, c\right )} + 64 \,{\left (d e f - c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left ({\left (64 i \, d f^{2} x + 64 i \, c f^{2}\right )} e^{\left (3 \, d x + 3 \, c\right )} + 64 \,{\left (d f^{2} x + c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{16 \, a d^{3} e^{\left (3 \, d x + 3 \, c\right )} - 16 i \, a d^{3} e^{\left (2 \, d x + 2 \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d^2*f^2*x^2 + 2*d^2*e^2 + 2*d*e*f + f^2 + 2*(2*d^2*e*f + d*f^2)*x + (64*I*f^2*e^(3*d*x + 3*c) + 64*f^2*e^(2
*d*x + 2*c))*dilog(-I*e^(d*x + c)) + (-2*I*d^2*f^2*x^2 - 2*I*d^2*e^2 + 2*I*d*e*f - I*f^2 + (-4*I*d^2*e*f + 2*I
*d*f^2)*x)*e^(5*d*x + 5*c) + (6*d^2*f^2*x^2 + 6*d^2*e^2 - 14*d*e*f + 15*f^2 + 2*(6*d^2*e*f - 7*d*f^2)*x)*e^(4*
d*x + 4*c) + (8*I*d^3*f^2*x^3 - 8*I*d^2*e^2 + (-64*I*c + 16*I)*d*e*f + (32*I*c^2 - 16*I)*f^2 + (24*I*d^3*e*f -
 40*I*d^2*f^2)*x^2 + (24*I*d^3*e^2 - 80*I*d^2*e*f + 16*I*d*f^2)*x)*e^(3*d*x + 3*c) + 8*(d^3*f^2*x^3 + 5*d^2*e^
2 - 2*(4*c - 1)*d*e*f + 2*(2*c^2 + 1)*f^2 + (3*d^3*e*f + d^2*f^2)*x^2 + (3*d^3*e^2 + 2*d^2*e*f + 2*d*f^2)*x)*e
^(2*d*x + 2*c) + (-6*I*d^2*f^2*x^2 - 6*I*d^2*e^2 - 14*I*d*e*f - 15*I*f^2 + (-12*I*d^2*e*f - 14*I*d*f^2)*x)*e^(
d*x + c) + ((64*I*d*e*f - 64*I*c*f^2)*e^(3*d*x + 3*c) + 64*(d*e*f - c*f^2)*e^(2*d*x + 2*c))*log(e^(d*x + c) -
I) + ((64*I*d*f^2*x + 64*I*c*f^2)*e^(3*d*x + 3*c) + 64*(d*f^2*x + c*f^2)*e^(2*d*x + 2*c))*log(I*e^(d*x + c) +
1))/(16*a*d^3*e^(3*d*x + 3*c) - 16*I*a*d^3*e^(2*d*x + 2*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)**2*sinh(d*x+c)**3/(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 )}^{2} \sinh \left (d x + c\right )^{3}}{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)^2*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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