\(\int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx\) [147]

Optimal result

Integrand size = 21, antiderivative size = 689 \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {3 x}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {5 \arctan \left (\sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right ) \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{3 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i x^2 \text {arctanh}\left (e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i x \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{2 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right )}{2 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {x \text {sech}^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{6 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {\tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{6 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 x^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{16 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {x^2 \text {sech}^2\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}} \]

[Out]

3/4*x/a^2/d^2/(a+I*a*sinh(d*x+c))^(1/2)-5/3*arctan(sinh(1/2*c+1/4*I*Pi+1/2*d*x))*cosh(1/2*c+1/4*I*Pi+1/2*d*x)/
a^2/d^3/(a+I*a*sinh(d*x+c))^(1/2)-3/8*I*x^2*arctanh(exp(1/2*c+3/4*I*Pi+1/2*d*x))*cosh(1/2*c+1/4*I*Pi+1/2*d*x)/
a^2/d/(a+I*a*sinh(d*x+c))^(1/2)+3/4*I*x*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*polylog(2,exp(1/2*c+3/4*I*Pi+1/2*d*x))/a^
2/d^2/(a+I*a*sinh(d*x+c))^(1/2)-3/4*I*x*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*polylog(2,-exp(1/2*c+3/4*I*Pi+1/2*d*x))/a
^2/d^2/(a+I*a*sinh(d*x+c))^(1/2)-3/2*I*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*polylog(3,exp(1/2*c+3/4*I*Pi+1/2*d*x))/a^2
/d^3/(a+I*a*sinh(d*x+c))^(1/2)+3/2*I*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*polylog(3,-exp(1/2*c+3/4*I*Pi+1/2*d*x))/a^2/
d^3/(a+I*a*sinh(d*x+c))^(1/2)+1/6*x*sech(1/2*c+1/4*I*Pi+1/2*d*x)^2/a^2/d^2/(a+I*a*sinh(d*x+c))^(1/2)-1/6*tanh(
1/2*c+1/4*I*Pi+1/2*d*x)/a^2/d^3/(a+I*a*sinh(d*x+c))^(1/2)+3/16*x^2*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a^2/d/(a+I*a*s
inh(d*x+c))^(1/2)+1/8*x^2*sech(1/2*c+1/4*I*Pi+1/2*d*x)^2*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a^2/d/(a+I*a*sinh(d*x+c)
)^(1/2)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 689, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3400, 4271, 3853, 3855, 4267, 2611, 2320, 6724} \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=-\frac {5 \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \arctan \left (\sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{3 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i x^2 \text {arctanh}\left (e^{\frac {d x}{2}+\frac {1}{4} (2 c-i \pi )}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i \operatorname {PolyLog}\left (3,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}-\frac {\tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{6 a^2 d^3 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 i x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}-\frac {3 i x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 x}{4 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {x \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{6 a^2 d^2 \sqrt {a+i a \sinh (c+d x)}}+\frac {3 x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{16 a^2 d \sqrt {a+i a \sinh (c+d x)}}+\frac {x^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{8 a^2 d \sqrt {a+i a \sinh (c+d x)}} \]

[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 2320

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 2611

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 3400

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 3853

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 3855

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

Rule 4267

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 4271

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]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 6724

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*} \text {integral}& = \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 \arctan \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 \text {arctanh}\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 \arctan \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 \text {arctanh}\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 ) \operatorname {PolyLog}\left (2,-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 ) \operatorname {PolyLog}\left (2,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 \operatorname {PolyLog}\left (2,-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 \operatorname {PolyLog}\left (2,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 \arctan \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 \text {arctanh}\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 ) \operatorname {PolyLog}\left (2,-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 ) \operatorname {PolyLog}\left (2,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 ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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 ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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 \arctan \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 \text {arctanh}\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 ) \operatorname {PolyLog}\left (2,-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 ) \operatorname {PolyLog}\left (2,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 ) \operatorname {PolyLog}\left (3,-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 ) \operatorname {PolyLog}\left (3,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] (verified)

Time = 1.48 (sec) , antiderivative size = 482, normalized size of antiderivative = 0.70 \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 d x (4+3 i d x) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )+\left (-8 i+36 d x+9 i d^2 x^2\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^3-\left (\frac {1}{2}-\frac {i}{2}\right ) (-1)^{3/4} \left (-160 \text {arctanh}\left ((-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+18 c^2 \text {arctanh}\left ((-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+9 c^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-9 d^2 x^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-9 c^2 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+9 d^2 x^2 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+36 d x \operatorname {PolyLog}\left (2,-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-36 d x \operatorname {PolyLog}\left (2,(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )-72 \operatorname {PolyLog}\left (3,-(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )+72 \operatorname {PolyLog}\left (3,(-1)^{3/4} e^{\frac {1}{2} (c+d x)}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^4+24 d^2 x^2 \sinh \left (\frac {1}{2} (c+d x)\right )+2 \left (-8+9 d^2 x^2\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{48 d^3 (a+i a \sinh (c+d x))^{5/2}} \]

[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))

Maple [F]

\[\int \frac {x^{2}}{\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]

[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)

Fricas [F]

\[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

1/24*(24*(a^3*d^3*e^(4*d*x + 4*c) - 4*I*a^3*d^3*e^(3*d*x + 3*c) - 6*a^3*d^3*e^(2*d*x + 2*c) + 4*I*a^3*d^3*e^(d
*x + c) + a^3*d^3)*integral(1/48*(-9*I*d^2*x^2 + 80*I)*sqrt(1/2*I*a*e^(-d*x - c))*e^(d*x + c)/(a^3*d^2*e^(d*x
+ c) - I*a^3*d^2), x) + ((-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) + (-33*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(1/2*I
*a*e^(-d*x - c)))/(a^3*d^3*e^(4*d*x + 4*c) - 4*I*a^3*d^3*e^(3*d*x + 3*c) - 6*a^3*d^3*e^(2*d*x + 2*c) + 4*I*a^3
*d^3*e^(d*x + c) + a^3*d^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {x^{2}}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int \frac {x^2}{{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[In]

int(x^2/(a + a*sinh(c + d*x)*1i)^(5/2),x)

[Out]

int(x^2/(a + a*sinh(c + d*x)*1i)^(5/2), x)