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

Optimal. Leaf size=305 \[ -\frac{4 d^2 (c+d x) \text{PolyLog}\left (2,-i e^{e+f x}\right )}{a^2 f^3}+\frac{4 d^3 \text{PolyLog}\left (3,-i e^{e+f x}\right )}{a^2 f^4}-\frac{2 d^2 (c+d x) \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a^2 f^3}-\frac{2 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a^2 f^2}+\frac{d (c+d x)^2 \text{sech}^2\left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{2 a^2 f^2}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{3 a^2 f}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{6 a^2 f}+\frac{(c+d x)^3}{3 a^2 f}+\frac{4 d^3 \log \left (\cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )\right )}{a^2 f^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.397799, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3318, 4186, 4184, 3475, 3716, 2190, 2531, 2282, 6589} \[ -\frac{4 d^2 (c+d x) \text{PolyLog}\left (2,-i e^{e+f x}\right )}{a^2 f^3}+\frac{4 d^3 \text{PolyLog}\left (3,-i e^{e+f x}\right )}{a^2 f^4}-\frac{2 d^2 (c+d x) \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a^2 f^3}-\frac{2 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a^2 f^2}+\frac{d (c+d x)^2 \text{sech}^2\left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{2 a^2 f^2}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{3 a^2 f}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{6 a^2 f}+\frac{(c+d x)^3}{3 a^2 f}+\frac{4 d^3 \log \left (\cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )\right )}{a^2 f^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4186

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

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 3475

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

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

Mathematica [A]  time = 6.24652, size = 443, normalized size = 1.45 \[ \frac{\frac{2 d \left (-6 c d \left (1+i e^e\right ) \text{PolyLog}\left (2,i e^{-e-f x}\right )-6 d^2 \left (1+i e^e\right ) \left (x \text{PolyLog}\left (2,i e^{-e-f x}\right )+\frac{\text{PolyLog}\left (3,i e^{-e-f x}\right )}{f}\right )+\frac{3 \left (1+i e^e\right ) \left (2 d^2-c^2 f^2\right ) \left (f x-\log \left (-e^{e+f x}+i\right )\right )}{f}+3 c^2 f^2 x+6 c d \left (1+i e^e\right ) f x \log \left (1-i e^{-e-f x}\right )+3 c d f^2 x^2+3 d^2 \left (1+i e^e\right ) f x^2 \log \left (1-i e^{-e-f x}\right )+d^2 f^2 x^3-6 d^2 x\right )}{-1-i e^e}+\frac{(c+d x) \left (i \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2-6\right )\right ) \cosh \left (e+\frac{3 f x}{2}\right )+3 \sinh \left (\frac{f x}{2}\right ) \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2-4\right )\right )+3 i d f (c+d x) \sinh \left (e+\frac{f x}{2}\right )+3 d f (c+d x) \cosh \left (\frac{f x}{2}\right )+6 i d^2 \cosh \left (e+\frac{f x}{2}\right )\right )}{\left (\cosh \left (\frac{e}{2}\right )+i \sinh \left (\frac{e}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )^3}}{3 a^2 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.141, size = 723, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(6*I*d^3*x-3*I*f*d^3*x^2*exp(2*f*x+2*e)-I*f^2*d^3*x^3+3*f^2*d^3*x^3*exp(f*x+e)-3*f*d^3*x^2*exp(f*x+e)-3*f*
c^2*d*exp(f*x+e)-3*I*f^2*c*d^2*x^2+9*f^2*c*d^2*x^2*exp(f*x+e)+9*f^2*c^2*d*x*exp(f*x+e)-6*f*c*d^2*x*exp(f*x+e)-
3*I*f^2*c^2*d*x-6*I*c*d^2*exp(2*f*x+2*e)-I*f^2*c^3-6*I*f*c*d^2*x*exp(2*f*x+2*e)-6*I*d^3*x*exp(2*f*x+2*e)+6*I*c
*d^2-12*d^3*x*exp(f*x+e)-12*c*d^2*exp(f*x+e)+3*f^2*c^3*exp(f*x+e)-3*I*f*c^2*d*exp(2*f*x+2*e))/(exp(f*x+e)-I)^3
/f^3/a^2-4*d^2/f^2/a^2*ln(1+I*exp(f*x+e))*c*x-4*d^2/f^3/a^2*ln(1+I*exp(f*x+e))*c*e+4*d^2/f^2/a^2*c*e*x+4*d^2/f
^3/a^2*ln(exp(f*x+e)-I)*c*e-4*d^2/f^3/a^2*ln(exp(f*x+e))*c*e+4*d^3/f^4/a^2*ln(exp(f*x+e)-I)+4*d^3*polylog(3,-I
*exp(f*x+e))/a^2/f^4-4*d^3/f^4/a^2*ln(exp(f*x+e))-4*d^2/f^3/a^2*c*polylog(2,-I*exp(f*x+e))-2*d^3/f^3/a^2*e^2*x
+2*d^2/f/a^2*c*x^2+2*d^2/f^3/a^2*c*e^2-2*d^3/f^2/a^2*ln(1+I*exp(f*x+e))*x^2-4*d^3/f^3/a^2*polylog(2,-I*exp(f*x
+e))*x-2*d^3/f^4/a^2*ln(exp(f*x+e)-I)*e^2+2*d/f^2/a^2*ln(exp(f*x+e))*c^2-2*d/f^2/a^2*ln(exp(f*x+e)-I)*c^2+2*d^
3/f^4/a^2*ln(exp(f*x+e))*e^2+2/3*d^3/f/a^2*x^3+2*d^3/f^4/a^2*ln(1+I*exp(f*x+e))*e^2-4/3*d^3/f^4/a^2*e^3

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Maxima [B]  time = 1.99941, size = 856, normalized size = 2.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*d*(3*(2*f*x*e^(3*f*x + 3*e) + (-6*I*f*x*e^(2*e) - 2*I*e^(2*e))*e^(2*f*x) - 2*e^(f*x + e))/(3*a^2*f^2*e^(3*
f*x + 3*e) - 9*I*a^2*f^2*e^(2*f*x + 2*e) - 9*a^2*f^2*e^(f*x + e) + 3*I*a^2*f^2) - 2*log(-I*(I*e^(f*x + e) + 1)
*e^(-e))/(a^2*f^2)) + c^3*(6*e^(-f*x - e)/((9*a^2*e^(-f*x - e) - 9*I*a^2*e^(-2*f*x - 2*e) - 3*a^2*e^(-3*f*x -
3*e) + 3*I*a^2)*f) + 2*I/((9*a^2*e^(-f*x - e) - 9*I*a^2*e^(-2*f*x - 2*e) - 3*a^2*e^(-3*f*x - 3*e) + 3*I*a^2)*f
)) + (-2*I*d^3*f^2*x^3 - 6*I*c*d^2*f^2*x^2 + 12*I*d^3*x + 12*I*c*d^2 - (6*I*d^3*f*x^2*e^(2*e) + 12*I*c*d^2*e^(
2*e) + (12*I*c*d^2*f*e^(2*e) + 12*I*d^3*e^(2*e))*x)*e^(2*f*x) + 6*(d^3*f^2*x^3*e^e - 4*c*d^2*e^e + (3*c*d^2*f^
2*e^e - d^3*f*e^e)*x^2 - 2*(c*d^2*f*e^e + 2*d^3*e^e)*x)*e^(f*x))/(3*a^2*f^3*e^(3*f*x + 3*e) - 9*I*a^2*f^3*e^(2
*f*x + 2*e) - 9*a^2*f^3*e^(f*x + e) + 3*I*a^2*f^3) - 4*(f*x*log(I*e^(f*x + e) + 1) + dilog(-I*e^(f*x + e)))*c*
d^2/(a^2*f^3) - 4*d^3*x/(a^2*f^3) - 2*(f^2*x^2*log(I*e^(f*x + e) + 1) + 2*f*x*dilog(-I*e^(f*x + e)) - 2*polylo
g(3, -I*e^(f*x + e)))*d^3/(a^2*f^4) + 4*d^3*log(e^(f*x + e) - I)/(a^2*f^4) + 2/3*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^
2)/(a^2*f^4)

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Fricas [C]  time = 2.60388, size = 2147, normalized size = 7.04 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*I*d^3*e^3 + 6*I*c^2*d*e*f^2 - 2*I*c^3*f^3 - 12*I*d^3*e + (-6*I*c*d^2*e^2 + 12*I*c*d^2)*f + (-12*I*d^3*f*x -
12*I*c*d^2*f - 12*(d^3*f*x + c*d^2*f)*e^(3*f*x + 3*e) + (36*I*d^3*f*x + 36*I*c*d^2*f)*e^(2*f*x + 2*e) + 36*(d
^3*f*x + c*d^2*f)*e^(f*x + e))*dilog(-I*e^(f*x + e)) + 2*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + d^3*e^3 - 3*c*d^2*e^
2*f + 3*c^2*d*e*f^2 - 6*d^3*e + 3*(c^2*d*f^3 - 2*d^3*f)*x)*e^(3*f*x + 3*e) + (-6*I*d^3*f^3*x^3 - 6*I*d^3*e^3 +
36*I*d^3*e + (-18*I*c^2*d*e - 6*I*c^2*d)*f^2 + (-18*I*c*d^2*f^3 - 6*I*d^3*f^2)*x^2 + (18*I*c*d^2*e^2 - 12*I*c
*d^2)*f + (-18*I*c^2*d*f^3 - 12*I*c*d^2*f^2 + 24*I*d^3*f)*x)*e^(2*f*x + 2*e) - 6*(d^3*f^2*x^2 + d^3*e^3 - c^3*
f^3 - 6*d^3*e + (3*c^2*d*e + c^2*d)*f^2 - (3*c*d^2*e^2 - 4*c*d^2)*f + 2*(c*d^2*f^2 - d^3*f)*x)*e^(f*x + e) + (
-6*I*d^3*e^2 + 12*I*c*d^2*e*f - 6*I*c^2*d*f^2 + 12*I*d^3 - 6*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 - 2*d^3)*e^(3*
f*x + 3*e) + (18*I*d^3*e^2 - 36*I*c*d^2*e*f + 18*I*c^2*d*f^2 - 36*I*d^3)*e^(2*f*x + 2*e) + 18*(d^3*e^2 - 2*c*d
^2*e*f + c^2*d*f^2 - 2*d^3)*e^(f*x + e))*log(e^(f*x + e) - I) + (-6*I*d^3*f^2*x^2 - 12*I*c*d^2*f^2*x + 6*I*d^3
*e^2 - 12*I*c*d^2*e*f - 6*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*e^(3*f*x + 3*e) + (18*I*d^3*f^
2*x^2 + 36*I*c*d^2*f^2*x - 18*I*d^3*e^2 + 36*I*c*d^2*e*f)*e^(2*f*x + 2*e) + 18*(d^3*f^2*x^2 + 2*c*d^2*f^2*x -
d^3*e^2 + 2*c*d^2*e*f)*e^(f*x + e))*log(I*e^(f*x + e) + 1) + (12*d^3*e^(3*f*x + 3*e) - 36*I*d^3*e^(2*f*x + 2*e
) - 36*d^3*e^(f*x + e) + 12*I*d^3)*polylog(3, -I*e^(f*x + e)))/(3*a^2*f^4*e^(3*f*x + 3*e) - 9*I*a^2*f^4*e^(2*f
*x + 2*e) - 9*a^2*f^4*e^(f*x + e) + 3*I*a^2*f^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((d*x+c)**3/(a+I*a*sinh(f*x+e))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{3}}{{\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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