3.137 \(\int \frac {x^2}{\sqrt {a+i a \sinh (e+f x)}} \, dx\)

Optimal. Leaf size=349 \[ -\frac {16 i \text {Li}_3\left (-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {16 i \text {Li}_3\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}+\frac {4 i x^2 \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{f \sqrt {a+i a \sinh (e+f x)}} \]

[Out]

-4*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)/f/(a+I*a*sinh(f*x+e))^(1/2)+8*I*x*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)-8*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))/f^2/(a+I*a*sinh(f*x+e))^(1/2)-16*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))/f^3/(a+I*a*sinh(f*x+e))^(1/2)+16*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))/f^3/(a+I*a*sinh(f*x+e))^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3319, 4182, 2531, 2282, 6589} \[ \frac {8 i x \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (2,-e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {8 i x \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (2,e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {16 i \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (3,-e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {16 i \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \text {PolyLog}\left (3,e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {4 i x^2 \cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{f \sqrt {a+i a \sinh (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

((4*I)*x^2*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]
]) + ((8*I)*x*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*Sinh
[e + f*x]]) - ((8*I)*x*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*Sinh[e + f*x]]) - ((16*I)*Cosh[e/2 + (I/4)*Pi + (f*x)/2]*PolyLog[3, -E^((2*e - I*Pi)/4 + (f*x)/2)])/(f^3*S
qrt[a + I*a*Sinh[e + f*x]]) + ((16*I)*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]])

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 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 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}{\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^2 \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^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 )}{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 ) \int x \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 (4 \sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )\right ) \int x \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^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 )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {8 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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {8 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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}+\frac {\left (8 \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}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {\left (8 \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}{f^2 \sqrt {a+i a \sinh (e+f x)}}\\ &=\frac {4 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 )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {8 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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {8 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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}+\frac {\left (16 \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}-\frac {\left (16 \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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}\\ &=\frac {4 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 )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {8 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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {8 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 )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {16 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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {16 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 )}{f^3 \sqrt {a+i a \sinh (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 276, normalized size = 0.79 \[ \frac {(1+i) (-1)^{3/4} \left (\sinh \left (\frac {1}{2} (e+f x)\right )-i \cosh \left (\frac {1}{2} (e+f x)\right )\right ) \left (-e^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+e^2 \log \left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}+1\right )-2 i e^2 \tan ^{-1}\left (\sqrt [4]{-1} e^{\frac {1}{2} (e+f x)}\right )+f^2 x^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-f^2 x^2 \log \left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}+1\right )-4 f x \text {Li}_2\left (-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+4 f x \text {Li}_2\left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+8 \text {Li}_3\left (-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-8 \text {Li}_3\left ((-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

((1 + I)*(-1)^(3/4)*((-2*I)*e^2*ArcTan[(-1)^(1/4)*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)] + 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)])*((-I)*Cosh
[(e + f*x)/2] + Sinh[(e + f*x)/2]))/(f^3*Sqrt[a + I*a*Sinh[e + f*x]])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {2 i \, \sqrt {\frac {1}{2} i \, a e^{\left (-f x - e\right )}} x^{2} e^{\left (f x + e\right )}}{a e^{\left (f x + e\right )} - i \, a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a +i a \sinh \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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