3.147 \(\int \frac{x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=689 \[ \frac{3 i x \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i x \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (3,-e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (3,e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{6 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{5 \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tan ^{-1}\left (\sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{3 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x^2 \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}} \]

[Out]

(3*x)/(4*a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - (5*ArcTan[Sinh[c/2 + (I/4)*Pi + (d*x)/2]]*Cosh[c/2 + (I/4)*Pi
+ (d*x)/2])/(3*a^2*d^3*Sqrt[a + I*a*Sinh[c + d*x]]) + (((3*I)/8)*x^2*ArcTanh[E^((2*c - I*Pi)/4 + (d*x)/2)]*Cos
h[c/2 + (I/4)*Pi + (d*x)/2])/(a^2*d*Sqrt[a + I*a*Sinh[c + d*x]]) + (((3*I)/4)*x*Cosh[c/2 + (I/4)*Pi + (d*x)/2]
*PolyLog[2, -E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - (((3*I)/4)*x*Cosh[c/2 + (I
/4)*Pi + (d*x)/2]*PolyLog[2, E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - (((3*I)/2)
*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*PolyLog[3, -E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^3*Sqrt[a + I*a*Sinh[c + d*x]
]) + (((3*I)/2)*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*PolyLog[3, E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^3*Sqrt[a + I*a
*Sinh[c + d*x]]) + (x*Sech[c/2 + (I/4)*Pi + (d*x)/2]^2)/(6*a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - Tanh[c/2 + (
I/4)*Pi + (d*x)/2]/(6*a^2*d^3*Sqrt[a + I*a*Sinh[c + d*x]]) + (3*x^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(16*a^2*d*
Sqrt[a + I*a*Sinh[c + d*x]]) + (x^2*Sech[c/2 + (I/4)*Pi + (d*x)/2]^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(8*a^2*d*
Sqrt[a + I*a*Sinh[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.460092, antiderivative size = 689, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3319, 4186, 3768, 3770, 4182, 2531, 2282, 6589} \[ \frac{3 i x \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i x \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (3,-e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (3,e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{6 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{5 \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tan ^{-1}\left (\sinh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{3 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x^2 \cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{d x}{2}+\frac{1}{4} (2 c-i \pi )}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + I*a*Sinh[c + d*x])^(5/2),x]

[Out]

(3*x)/(4*a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - (5*ArcTan[Sinh[c/2 + (I/4)*Pi + (d*x)/2]]*Cosh[c/2 + (I/4)*Pi
+ (d*x)/2])/(3*a^2*d^3*Sqrt[a + I*a*Sinh[c + d*x]]) + (((3*I)/8)*x^2*ArcTanh[E^((2*c - I*Pi)/4 + (d*x)/2)]*Cos
h[c/2 + (I/4)*Pi + (d*x)/2])/(a^2*d*Sqrt[a + I*a*Sinh[c + d*x]]) + (((3*I)/4)*x*Cosh[c/2 + (I/4)*Pi + (d*x)/2]
*PolyLog[2, -E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - (((3*I)/4)*x*Cosh[c/2 + (I
/4)*Pi + (d*x)/2]*PolyLog[2, E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - (((3*I)/2)
*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*PolyLog[3, -E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^3*Sqrt[a + I*a*Sinh[c + d*x]
]) + (((3*I)/2)*Cosh[c/2 + (I/4)*Pi + (d*x)/2]*PolyLog[3, E^((2*c - I*Pi)/4 + (d*x)/2)])/(a^2*d^3*Sqrt[a + I*a
*Sinh[c + d*x]]) + (x*Sech[c/2 + (I/4)*Pi + (d*x)/2]^2)/(6*a^2*d^2*Sqrt[a + I*a*Sinh[c + d*x]]) - Tanh[c/2 + (
I/4)*Pi + (d*x)/2]/(6*a^2*d^3*Sqrt[a + I*a*Sinh[c + d*x]]) + (3*x^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(16*a^2*d*
Sqrt[a + I*a*Sinh[c + d*x]]) + (x^2*Sech[c/2 + (I/4)*Pi + (d*x)/2]^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(8*a^2*d*
Sqrt[a + I*a*Sinh[c + d*x]])

Rule 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (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 3768

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

Rule 3770

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

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 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{x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx &=\frac{\sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \int x^2 \text{csch}^5\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{4 a^2 \sqrt{a+i a \sinh (c+d x)}}\\ &=\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}-\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \int x^2 \text{csch}^3\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{16 a^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{\sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \int \text{csch}^3\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}\\ &=\frac{3 x}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \int x^2 \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{32 a^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \int \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{12 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \int \text{csch}\left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}\\ &=\frac{3 x}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{5 \tan ^{-1}\left (\sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{3 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x^2 \tanh ^{-1}\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}-\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \int x \log \left (1-e^{-i \left (\frac{i c}{2}+\frac{\pi }{4}\right )+\frac{d x}{2}}\right ) \, dx}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \int x \log \left (1+e^{-i \left (\frac{i c}{2}+\frac{\pi }{4}\right )+\frac{d x}{2}}\right ) \, dx}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}\\ &=\frac{3 x}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{5 \tan ^{-1}\left (\sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{3 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x^2 \tanh ^{-1}\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_2\left (-e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i x \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_2\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \int \text{Li}_2\left (-e^{-i \left (\frac{i c}{2}+\frac{\pi }{4}\right )+\frac{d x}{2}}\right ) \, dx}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \int \text{Li}_2\left (e^{-i \left (\frac{i c}{2}+\frac{\pi }{4}\right )+\frac{d x}{2}}\right ) \, dx}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}\\ &=\frac{3 x}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{5 \tan ^{-1}\left (\sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{3 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x^2 \tanh ^{-1}\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_2\left (-e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i x \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_2\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}+\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{-i \left (\frac{i c}{2}+\frac{\pi }{4}\right )+\frac{d x}{2}}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}-\frac{\left (3 \sinh \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{-i \left (\frac{i c}{2}+\frac{\pi }{4}\right )+\frac{d x}{2}}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}\\ &=\frac{3 x}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{5 \tan ^{-1}\left (\sinh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{3 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x^2 \tanh ^{-1}\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right ) \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i x \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_2\left (-e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i x \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_2\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{4 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{3 i \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_3\left (-e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 i \cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \text{Li}_3\left (e^{\frac{1}{4} (2 c-i \pi )+\frac{d x}{2}}\right )}{2 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{x \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^2 \sqrt{a+i a \sinh (c+d x)}}-\frac{\tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{6 a^2 d^3 \sqrt{a+i a \sinh (c+d x)}}+\frac{3 x^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{16 a^2 d \sqrt{a+i a \sinh (c+d x)}}+\frac{x^2 \text{sech}^2\left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right ) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{8 a^2 d \sqrt{a+i a \sinh (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.28438, size = 482, normalized size = 0.7 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\left (-\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^4 \left (36 d x \text{PolyLog}\left (2,-(-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )-36 d x \text{PolyLog}\left (2,(-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )-72 \text{PolyLog}\left (3,-(-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )+72 \text{PolyLog}\left (3,(-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )+9 c^2 \log \left (1-(-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )-9 c^2 \log \left ((-1)^{3/4} e^{\frac{1}{2} (c+d x)}+1\right )+18 c^2 \tanh ^{-1}\left ((-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )-9 d^2 x^2 \log \left (1-(-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )+9 d^2 x^2 \log \left ((-1)^{3/4} e^{\frac{1}{2} (c+d x)}+1\right )-160 \tanh ^{-1}\left ((-1)^{3/4} e^{\frac{1}{2} (c+d x)}\right )\right )+24 d^2 x^2 \sinh \left (\frac{1}{2} (c+d x)\right )+\left (9 i d^2 x^2+36 d x-8 i\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^3+2 \left (9 d^2 x^2-8\right ) \sinh \left (\frac{1}{2} (c+d x)\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2+4 d x (4+3 i d x) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )\right )}{48 d^3 (a+i a \sinh (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + I*a*Sinh[c + d*x])^(5/2),x]

[Out]

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(4*d*x*(4 + (3*I)*d*x)*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) +
(-8*I + 36*d*x + (9*I)*d^2*x^2)*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^3 - (1/2 - I/2)*(-1)^(3/4)*(-160*Arc
Tanh[(-1)^(3/4)*E^((c + d*x)/2)] + 18*c^2*ArcTanh[(-1)^(3/4)*E^((c + d*x)/2)] + 9*c^2*Log[1 - (-1)^(3/4)*E^((c
+ d*x)/2)] - 9*d^2*x^2*Log[1 - (-1)^(3/4)*E^((c + d*x)/2)] - 9*c^2*Log[1 + (-1)^(3/4)*E^((c + d*x)/2)] + 9*d^
2*x^2*Log[1 + (-1)^(3/4)*E^((c + d*x)/2)] + 36*d*x*PolyLog[2, -((-1)^(3/4)*E^((c + d*x)/2))] - 36*d*x*PolyLog[
2, (-1)^(3/4)*E^((c + d*x)/2)] - 72*PolyLog[3, -((-1)^(3/4)*E^((c + d*x)/2))] + 72*PolyLog[3, (-1)^(3/4)*E^((c
+ d*x)/2)])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^4 + 24*d^2*x^2*Sinh[(c + d*x)/2] + 2*(-8 + 9*d^2*x^2)*(
Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2*Sinh[(c + d*x)/2]))/(48*d^3*(a + I*a*Sinh[c + d*x])^(5/2))

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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( a+ia\sinh \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+I*a*sinh(d*x+c))^(5/2),x)

[Out]

int(x^2/(a+I*a*sinh(d*x+c))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(x^2/(I*a*sinh(d*x + c) + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left ({\left (-9 i \, d^{2} x^{2} - 36 i \, d x + 8 i\right )} e^{\left (4 \, d x + 4 \, c\right )} -{\left (33 \, d^{2} x^{2} + 140 \, d x - 8\right )} e^{\left (3 \, d x + 3 \, c\right )} +{\left (-33 i \, d^{2} x^{2} + 140 i \, d x + 8 i\right )} e^{\left (2 \, d x + 2 \, c\right )} -{\left (9 \, d^{2} x^{2} - 36 \, d x - 8\right )} e^{\left (d x + c\right )}\right )} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )} +{\left (24 \, a^{3} d^{3} e^{\left (5 \, d x + 5 \, c\right )} - 120 i \, a^{3} d^{3} e^{\left (4 \, d x + 4 \, c\right )} - 240 \, a^{3} d^{3} e^{\left (3 \, d x + 3 \, c\right )} + 240 i \, a^{3} d^{3} e^{\left (2 \, d x + 2 \, c\right )} + 120 \, a^{3} d^{3} e^{\left (d x + c\right )} - 24 i \, a^{3} d^{3}\right )}{\rm integral}\left (\frac{\sqrt{\frac{1}{2}}{\left (-9 i \, d^{2} x^{2} + 80 i\right )} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{48 \, a^{3} d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 96 i \, a^{3} d^{2} e^{\left (d x + c\right )} - 48 \, a^{3} d^{2}}, x\right )}{24 \, a^{3} d^{3} e^{\left (5 \, d x + 5 \, c\right )} - 120 i \, a^{3} d^{3} e^{\left (4 \, d x + 4 \, c\right )} - 240 \, a^{3} d^{3} e^{\left (3 \, d x + 3 \, c\right )} + 240 i \, a^{3} d^{3} e^{\left (2 \, d x + 2 \, c\right )} + 120 \, a^{3} d^{3} e^{\left (d x + c\right )} - 24 i \, a^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

(sqrt(1/2)*((-9*I*d^2*x^2 - 36*I*d*x + 8*I)*e^(4*d*x + 4*c) - (33*d^2*x^2 + 140*d*x - 8)*e^(3*d*x + 3*c) + (-3
3*I*d^2*x^2 + 140*I*d*x + 8*I)*e^(2*d*x + 2*c) - (9*d^2*x^2 - 36*d*x - 8)*e^(d*x + c))*sqrt(I*a*e^(2*d*x + 2*c
) + 2*a*e^(d*x + c) - I*a)*e^(-1/2*d*x - 1/2*c) + (24*a^3*d^3*e^(5*d*x + 5*c) - 120*I*a^3*d^3*e^(4*d*x + 4*c)
- 240*a^3*d^3*e^(3*d*x + 3*c) + 240*I*a^3*d^3*e^(2*d*x + 2*c) + 120*a^3*d^3*e^(d*x + c) - 24*I*a^3*d^3)*integr
al(sqrt(1/2)*(-9*I*d^2*x^2 + 80*I)*sqrt(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - I*a)*e^(1/2*d*x + 1/2*c)/(48*a
^3*d^2*e^(2*d*x + 2*c) - 96*I*a^3*d^2*e^(d*x + c) - 48*a^3*d^2), x))/(24*a^3*d^3*e^(5*d*x + 5*c) - 120*I*a^3*d
^3*e^(4*d*x + 4*c) - 240*a^3*d^3*e^(3*d*x + 3*c) + 240*I*a^3*d^3*e^(2*d*x + 2*c) + 120*a^3*d^3*e^(d*x + c) - 2
4*I*a^3*d^3)

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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(x**2/(a+I*a*sinh(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(x^2/(I*a*sinh(d*x + c) + a)^(5/2), x)