3.142 \(\int \frac {x^2}{(a+i a \sinh (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=506 \[ -\frac {4 i \text {Li}_3\left (-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{a f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {4 i \text {Li}_3\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{a f^3 \sqrt {a+i a \sinh (e+f x)}}-\frac {4 \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tan ^{-1}\left (\sinh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )\right )}{a f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {2 i x \text {Li}_2\left (-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{a f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {2 i x \text {Li}_2\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{a f^2 \sqrt {a+i a \sinh (e+f x)}}+\frac {2 x}{a f^2 \sqrt {a+i a \sinh (e+f x)}}+\frac {x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{2 a f \sqrt {a+i a \sinh (e+f x)}}+\frac {i x^2 \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{a f \sqrt {a+i a \sinh (e+f x)}} \]

[Out]

2*x/a/f^2/(a+I*a*sinh(f*x+e))^(1/2)-4*arctan(sinh(1/2*e+1/4*I*Pi+1/2*f*x))*cosh(1/2*e+1/4*I*Pi+1/2*f*x)/a/f^3/
(a+I*a*sinh(f*x+e))^(1/2)-I*x^2*arctanh(exp(1/2*e+3/4*I*Pi+1/2*f*x))*cosh(1/2*e+1/4*I*Pi+1/2*f*x)/a/f/(a+I*a*s
inh(f*x+e))^(1/2)+2*I*x*cosh(1/2*e+1/4*I*Pi+1/2*f*x)*polylog(2,exp(1/2*e+3/4*I*Pi+1/2*f*x))/a/f^2/(a+I*a*sinh(
f*x+e))^(1/2)-2*I*x*cosh(1/2*e+1/4*I*Pi+1/2*f*x)*polylog(2,-exp(1/2*e+3/4*I*Pi+1/2*f*x))/a/f^2/(a+I*a*sinh(f*x
+e))^(1/2)-4*I*cosh(1/2*e+1/4*I*Pi+1/2*f*x)*polylog(3,exp(1/2*e+3/4*I*Pi+1/2*f*x))/a/f^3/(a+I*a*sinh(f*x+e))^(
1/2)+4*I*cosh(1/2*e+1/4*I*Pi+1/2*f*x)*polylog(3,-exp(1/2*e+3/4*I*Pi+1/2*f*x))/a/f^3/(a+I*a*sinh(f*x+e))^(1/2)+
1/2*x^2*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/a/f/(a+I*a*sinh(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 506, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3319, 4186, 3770, 4182, 2531, 2282, 6589} \[ \frac {2 i x \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (2,-e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{a f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {2 i x \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (2,e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{a f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {4 i \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (3,-e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{a f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {4 i \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (3,e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{a f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {2 x}{a f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {4 \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tan ^{-1}\left (\sinh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )\right )}{a f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {x^2 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{2 a f \sqrt {a+i a \sinh (e+f x)}}+\frac {i x^2 \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{a f \sqrt {a+i a \sinh (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/(a*f^2*Sqrt[a + I*a*Sinh[e + f*x]]) - (4*ArcTan[Sinh[e/2 + (I/4)*Pi + (f*x)/2]]*Cosh[e/2 + (I/4)*Pi + (f
*x)/2])/(a*f^3*Sqrt[a + I*a*Sinh[e + f*x]]) + (I*x^2*ArcTanh[E^((2*e - I*Pi)/4 + (f*x)/2)]*Cosh[e/2 + (I/4)*Pi
+ (f*x)/2])/(a*f*Sqrt[a + I*a*Sinh[e + f*x]]) + ((2*I)*x*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[2, -E^((2*e -
I*Pi)/4 + (f*x)/2)])/(a*f^2*Sqrt[a + I*a*Sinh[e + f*x]]) - ((2*I)*x*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[2,
E^((2*e - I*Pi)/4 + (f*x)/2)])/(a*f^2*Sqrt[a + I*a*Sinh[e + f*x]]) - ((4*I)*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*Po
lyLog[3, -E^((2*e - I*Pi)/4 + (f*x)/2)])/(a*f^3*Sqrt[a + I*a*Sinh[e + f*x]]) + ((4*I)*Cosh[e/2 + (I/4)*Pi + (f
*x)/2]*PolyLog[3, E^((2*e - I*Pi)/4 + (f*x)/2)])/(a*f^3*Sqrt[a + I*a*Sinh[e + f*x]]) + (x^2*Tanh[e/2 + (I/4)*P
i + (f*x)/2])/(2*a*f*Sqrt[a + I*a*Sinh[e + f*x]])

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

________________________________________________________________________________________

Mathematica [A]  time = 1.68, size = 384, normalized size = 0.76 \[ \frac {\left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right ) \left (-\left (\frac {1}{2}-\frac {i}{2}\right ) (-1)^{3/4} \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (e^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-e^2 \log \left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}+1\right )+2 e^2 \tanh ^{-1}\left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-f^2 x^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+f^2 x^2 \log \left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}+1\right )+4 f x \text {Li}_2\left (-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-4 f x \text {Li}_2\left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-8 \text {Li}_3\left (-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+8 \text {Li}_3\left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-16 \tanh ^{-1}\left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )\right )+2 f^2 x^2 \sinh \left (\frac {1}{2} (e+f x)\right )+f x (4+i f x) \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )\right )}{2 f^3 (a+i a \sinh (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2])*(f*x*(4 + I*f*x)*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2]) - (1/2 -
I/2)*(-1)^(3/4)*(-16*ArcTanh[(-1)^(3/4)*E^((e + f*x)/2)] + 2*e^2*ArcTanh[(-1)^(3/4)*E^((e + f*x)/2)] + e^2*Lo
g[1 - (-1)^(3/4)*E^((e + f*x)/2)] - f^2*x^2*Log[1 - (-1)^(3/4)*E^((e + f*x)/2)] - e^2*Log[1 + (-1)^(3/4)*E^((e
+ f*x)/2)] + f^2*x^2*Log[1 + (-1)^(3/4)*E^((e + f*x)/2)] + 4*f*x*PolyLog[2, -((-1)^(3/4)*E^((e + f*x)/2))] -
4*f*x*PolyLog[2, (-1)^(3/4)*E^((e + f*x)/2)] - 8*PolyLog[3, -((-1)^(3/4)*E^((e + f*x)/2))] + 8*PolyLog[3, (-1)
^(3/4)*E^((e + f*x)/2)])*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2])^2 + 2*f^2*x^2*Sinh[(e + f*x)/2]))/(2*f^3*(a
+ I*a*Sinh[e + f*x])^(3/2))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \frac {{\left (a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 2 i \, a^{2} f^{2} e^{\left (f x + e\right )} - a^{2} f^{2}\right )} {\rm integral}\left (\frac {{\left (-i \, f^{2} x^{2} + 8 i\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-f x - e\right )}} e^{\left (f x + e\right )}}{2 \, {\left (a^{2} f^{2} e^{\left (f x + e\right )} - i \, a^{2} f^{2}\right )}}, x\right ) + {\left ({\left (-i \, f x^{2} - 4 i \, x\right )} e^{\left (2 \, f x + 2 \, e\right )} + {\left (f x^{2} - 4 \, x\right )} e^{\left (f x + e\right )}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-f x - e\right )}}}{a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 2 i \, a^{2} f^{2} e^{\left (f x + e\right )} - a^{2} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*f^2*e^(2*f*x + 2*e) - 2*I*a^2*f^2*e^(f*x + e) - a^2*f^2)*integral(1/2*(-I*f^2*x^2 + 8*I)*sqrt(1/2*I*a*e^
(-f*x - e))*e^(f*x + e)/(a^2*f^2*e^(f*x + e) - I*a^2*f^2), x) + ((-I*f*x^2 - 4*I*x)*e^(2*f*x + 2*e) + (f*x^2 -
4*x)*e^(f*x + e))*sqrt(1/2*I*a*e^(-f*x - e)))/(a^2*f^2*e^(2*f*x + 2*e) - 2*I*a^2*f^2*e^(f*x + e) - a^2*f^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a +i a \sinh \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + a*sinh(e + f*x)*1i)^(3/2),x)

[Out]

int(x^2/(a + a*sinh(e + f*x)*1i)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (i a \left (\sinh {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(I*a*(sinh(e + f*x) - I))**(3/2), x)

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