3.2.36 \(\int \frac {x^3}{\sqrt {a+i a \sinh (e+f x)}} \, dx\) [136]

Optimal. Leaf size=493 \[ \frac {4 i x^3 \tanh ^{-1}\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {12 i x^2 \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {PolyLog}\left (2,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {12 i x^2 \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {PolyLog}\left (2,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {48 i x \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {PolyLog}\left (3,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {48 i x \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {PolyLog}\left (3,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {96 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {PolyLog}\left (4,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^4 \sqrt {a+i a \sinh (e+f x)}}-\frac {96 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {PolyLog}\left (4,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^4 \sqrt {a+i a \sinh (e+f x)}} \]

[Out]

-4*I*x^3*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)/f/(a+I*a*sinh(f*x+e))^(1/2)+12*I*x^
2*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))/f^2/(a+I*a*sinh(f*x+e))^(1/2)-12*I*x^2*c
osh(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))/f^2/(a+I*a*sinh(f*x+e))^(1/2)-48*I*x*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))/f^3/(a+I*a*sinh(f*x+e))^(1/2)+48*I*x*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))/f^3/(a+I*a*sinh(f*x+e))^(1/2)+96*I*cosh(1/2*e+1/4*I
*Pi+1/2*f*x)*polylog(4,exp(1/2*e+3/4*I*Pi+1/2*f*x))/f^4/(a+I*a*sinh(f*x+e))^(1/2)-96*I*cosh(1/2*e+1/4*I*Pi+1/2
*f*x)*polylog(4,-exp(1/2*e+3/4*I*Pi+1/2*f*x))/f^4/(a+I*a*sinh(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.18, antiderivative size = 493, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3400, 4267, 2611, 6744, 2320, 6724} \begin {gather*} \frac {96 i \text {Li}_4\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 )}{f^4 \sqrt {a+i a \sinh (e+f x)}}-\frac {96 i \text {Li}_4\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 )}{f^4 \sqrt {a+i a \sinh (e+f x)}}-\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}+\frac {4 i x^3 \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 )}{f \sqrt {a+i a \sinh (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

((4*I)*x^3*ArcTanh[E^((2*e - I*Pi)/4 + (f*x)/2)]*Cosh[e/2 + (I/4)*Pi + (f*x)/2])/(f*Sqrt[a + I*a*Sinh[e + f*x]
]) + ((12*I)*x^2*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[2, -E^((2*e - I*Pi)/4 + (f*x)/2)])/(f^2*Sqrt[a + I*a*S
inh[e + f*x]]) - ((12*I)*x^2*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[2, E^((2*e - I*Pi)/4 + (f*x)/2)])/(f^2*Sqr
t[a + I*a*Sinh[e + f*x]]) - ((48*I)*x*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[3, -E^((2*e - I*Pi)/4 + (f*x)/2)]
)/(f^3*Sqrt[a + I*a*Sinh[e + f*x]]) + ((48*I)*x*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[3, E^((2*e - I*Pi)/4 +
(f*x)/2)])/(f^3*Sqrt[a + I*a*Sinh[e + f*x]]) + ((96*I)*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[4, -E^((2*e - I*
Pi)/4 + (f*x)/2)])/(f^4*Sqrt[a + I*a*Sinh[e + f*x]]) - ((96*I)*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[4, E^((2
*e - I*Pi)/4 + (f*x)/2)])/(f^4*Sqrt[a + I*a*Sinh[e + f*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 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 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]

Rule 6744

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

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {a+i a \sinh (e+f x)}} \, dx &=\frac {\sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \int x^3 \text {csch}\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \, dx}{\sqrt {a+i a \sinh (e+f x)}}\\ &=\frac {4 i x^3 \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 )}{f \sqrt {a+i a \sinh (e+f x)}}-\frac {\left (6 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \int x^2 \log \left (1-e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right ) \, dx}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {\left (6 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \int x^2 \log \left (1+e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right ) \, dx}{f \sqrt {a+i a \sinh (e+f x)}}\\ &=\frac {4 i x^3 \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 )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}+\frac {\left (24 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \int x \text {Li}_2\left (-e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right ) \, dx}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {\left (24 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \int x \text {Li}_2\left (e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right ) \, dx}{f^2 \sqrt {a+i a \sinh (e+f x)}}\\ &=\frac {4 i x^3 \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 )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}-\frac {\left (48 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \int \text {Li}_3\left (-e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right ) \, dx}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {\left (48 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \int \text {Li}_3\left (e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right ) \, dx}{f^3 \sqrt {a+i a \sinh (e+f x)}}\\ &=\frac {4 i x^3 \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 )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}-\frac {\left (96 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right )}{f^4 \sqrt {a+i a \sinh (e+f x)}}+\frac {\left (96 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{-i \left (\frac {i e}{2}+\frac {\pi }{4}\right )+\frac {f x}{2}}\right )}{f^4 \sqrt {a+i a \sinh (e+f x)}}\\ &=\frac {4 i x^3 \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 )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {12 i x^2 \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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {48 i x \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {96 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {Li}_4\left (-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^4 \sqrt {a+i a \sinh (e+f x)}}-\frac {96 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \text {Li}_4\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^4 \sqrt {a+i a \sinh (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.82, size = 331, normalized size = 0.67 \begin {gather*} \frac {(1-i) (-1)^{3/4} \left (2 i e^3 \text {ArcTan}\left (\sqrt [4]{-1} e^{\frac {1}{2} (e+f x)}\right )+e^3 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+f^3 x^3 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-e^3 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-f^3 x^3 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-6 f^2 x^2 \text {PolyLog}\left (2,-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+6 f^2 x^2 \text {PolyLog}\left (2,(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+24 f x \text {PolyLog}\left (3,-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-24 f x \text {PolyLog}\left (3,(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-48 \text {PolyLog}\left (4,-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+48 \text {PolyLog}\left (4,(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )\right ) \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )}{f^4 \sqrt {a+i a \sinh (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

((1 - I)*(-1)^(3/4)*((2*I)*e^3*ArcTan[(-1)^(1/4)*E^((e + f*x)/2)] + e^3*Log[1 - (-1)^(3/4)*E^((e + f*x)/2)] +
f^3*x^3*Log[1 - (-1)^(3/4)*E^((e + f*x)/2)] - e^3*Log[1 + (-1)^(3/4)*E^((e + f*x)/2)] - f^3*x^3*Log[1 + (-1)^(
3/4)*E^((e + f*x)/2)] - 6*f^2*x^2*PolyLog[2, -((-1)^(3/4)*E^((e + f*x)/2))] + 6*f^2*x^2*PolyLog[2, (-1)^(3/4)*
E^((e + f*x)/2)] + 24*f*x*PolyLog[3, -((-1)^(3/4)*E^((e + f*x)/2))] - 24*f*x*PolyLog[3, (-1)^(3/4)*E^((e + f*x
)/2)] - 48*PolyLog[4, -((-1)^(3/4)*E^((e + f*x)/2))] + 48*PolyLog[4, (-1)^(3/4)*E^((e + f*x)/2)])*(Cosh[(e + f
*x)/2] + I*Sinh[(e + f*x)/2]))/(f^4*Sqrt[a + I*a*Sinh[e + f*x]])

________________________________________________________________________________________

Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \frac {x^{3}}{\sqrt {a +i a \sinh \left (f x +e \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-2*I*sqrt(1/2*I*a*e^(-f*x - e))*x^3*e^(f*x + e)/(a*e^(f*x + e) - I*a), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a+I*a*sinh(f*x+e))**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3}{\sqrt {a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________