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

Optimal. Leaf size=288 \[ \frac{i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{a f^2 \sqrt{a+i a \sinh (e+f x)}}-\frac{i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{1}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{x \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{2 a f \sqrt{a+i a \sinh (e+f x)}}+\frac{i x \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{a f \sqrt{a+i a \sinh (e+f x)}} \]

[Out]

1/(a*f^2*Sqrt[a + I*a*Sinh[e + f*x]]) + (I*x*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]]) + (I*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]]) - (I*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]]) + (x*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(2*a*f*Sqrt[a + I*a*Sinh[
e + f*x]])

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Rubi [A]  time = 0.169333, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3319, 4185, 4182, 2279, 2391} \[ \frac{i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{a f^2 \sqrt{a+i a \sinh (e+f x)}}-\frac{i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{1}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{x \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{2 a f \sqrt{a+i a \sinh (e+f x)}}+\frac{i x \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{a f \sqrt{a+i a \sinh (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(a*f^2*Sqrt[a + I*a*Sinh[e + f*x]]) + (I*x*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]]) + (I*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]]) - (I*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]]) + (x*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/(2*a*f*Sqrt[a + I*a*Sinh[
e + f*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 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{x}{(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 \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{1}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{x \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 \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{1}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{i x \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 \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 \log \left (1-e^{-i \left (\frac{i e}{2}+\frac{\pi }{4}\right )+\frac{f x}{2}}\right ) \, dx}{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 \log \left (1+e^{-i \left (\frac{i e}{2}+\frac{\pi }{4}\right )+\frac{f x}{2}}\right ) \, dx}{2 a f \sqrt{a+i a \sinh (e+f x)}}\\ &=\frac{1}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{i x \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 \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 ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{-i \left (\frac{i e}{2}+\frac{\pi }{4}\right )+\frac{f x}{2}}\right )}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{\sinh \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{-i \left (\frac{i e}{2}+\frac{\pi }{4}\right )+\frac{f x}{2}}\right )}{a f^2 \sqrt{a+i a \sinh (e+f x)}}\\ &=\frac{1}{a f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{i x \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{i \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{i \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 \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 = 0.852845, size = 332, normalized size = 1.15 \[ \frac{\left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right ) \left (\frac{i \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (-2 \text{PolyLog}\left (2,-\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}\right )+2 \text{PolyLog}\left (2,\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}\right )+\frac{1}{2} i (2 i e+2 i f x+\pi ) \left (\log \left (1-\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}\right )-\log \left (\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}+1\right )\right )+\pi \tan ^{-1}\left (\frac{\tanh \left (\frac{1}{4} (e+f x)\right )+i}{\sqrt{2}}\right )\right )}{\sqrt{2}}+2 f x \sinh \left (\frac{1}{2} (e+f x)\right )+(2+i f x) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )-\sqrt{2} e \tan ^{-1}\left (\frac{\tanh \left (\frac{1}{4} (e+f x)\right )+i}{\sqrt{2}}\right ) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )^2\right )}{2 f^2 (a+i a \sinh (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+ia\sinh \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left ({\left (-i \, f x - 2 i\right )} e^{\left (2 \, f x + 2 \, e\right )} +{\left (f x - 2\right )} e^{\left (f x + e\right )}\right )} \sqrt{i \, a e^{\left (2 \, f x + 2 \, e\right )} + 2 \, a e^{\left (f x + e\right )} - i \, a} e^{\left (-\frac{1}{2} \, f x - \frac{1}{2} \, e\right )} +{\left (a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, a^{2} f^{2} e^{\left (f x + e\right )} + i \, a^{2} f^{2}\right )}{\rm integral}\left (-\frac{i \, \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, f x + 2 \, e\right )} + 2 \, a e^{\left (f x + e\right )} - i \, a} x e^{\left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}}{2 \, a^{2} e^{\left (2 \, f x + 2 \, e\right )} - 4 i \, a^{2} e^{\left (f x + e\right )} - 2 \, a^{2}}, x\right )}{a^{2} f^{2} e^{\left (3 \, f x + 3 \, e\right )} - 3 i \, a^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, a^{2} f^{2} e^{\left (f x + e\right )} + i \, a^{2} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(1/2)*((-I*f*x - 2*I)*e^(2*f*x + 2*e) + (f*x - 2)*e^(f*x + e))*sqrt(I*a*e^(2*f*x + 2*e) + 2*a*e^(f*x + e)
- I*a)*e^(-1/2*f*x - 1/2*e) + (a^2*f^2*e^(3*f*x + 3*e) - 3*I*a^2*f^2*e^(2*f*x + 2*e) - 3*a^2*f^2*e^(f*x + e)
+ I*a^2*f^2)*integral(-I*sqrt(1/2)*sqrt(I*a*e^(2*f*x + 2*e) + 2*a*e^(f*x + e) - I*a)*x*e^(1/2*f*x + 1/2*e)/(2*
a^2*e^(2*f*x + 2*e) - 4*I*a^2*e^(f*x + e) - 2*a^2), x))/(a^2*f^2*e^(3*f*x + 3*e) - 3*I*a^2*f^2*e^(2*f*x + 2*e)
- 3*a^2*f^2*e^(f*x + e) + I*a^2*f^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a \left (i \sinh{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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