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

Optimal. Leaf size=207 \[ \frac{4 i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{f^2 \sqrt{a+i a \sinh (e+f x)}}-\frac{4 i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{4 i x \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{f \sqrt{a+i a \sinh (e+f x)}} \]

[Out]

((4*I)*x*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]])
+ ((4*I)*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]]) - ((4*I)*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*Si
nh[e + f*x]])

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Rubi [A]  time = 0.104719, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3319, 4182, 2279, 2391} \[ \frac{4 i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,-e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{f^2 \sqrt{a+i a \sinh (e+f x)}}-\frac{4 i \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \text{PolyLog}\left (2,e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{f^2 \sqrt{a+i a \sinh (e+f x)}}+\frac{4 i x \cosh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right ) \tanh ^{-1}\left (e^{\frac{f x}{2}+\frac{1}{4} (2 e-i \pi )}\right )}{f \sqrt{a+i a \sinh (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + I*a*Sinh[e + f*x]],x]

[Out]

((4*I)*x*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]])
+ ((4*I)*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]]) - ((4*I)*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*Si
nh[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 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}{\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 \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 \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 (2 \sinh \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right )\right ) \int \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 (2 \sinh \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right )\right ) \int \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 \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 ) \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 )}{f^2 \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{\log (1+x)}{x} \, dx,x,e^{-i \left (\frac{i e}{2}+\frac{\pi }{4}\right )+\frac{f x}{2}}\right )}{f^2 \sqrt{a+i a \sinh (e+f x)}}\\ &=\frac{4 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 )}{f \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}_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{4 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 )}{f^2 \sqrt{a+i a \sinh (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.505013, size = 221, normalized size = 1.07 \[ \frac{\sqrt{2} \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right ) \left (-2 i \left (\text{PolyLog}\left (2,-\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}\right )-\text{PolyLog}\left (2,\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}\right )\right )-\frac{1}{2} (2 i e+2 i f x+\pi ) \left (\log \left (1-\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}\right )-\log \left (\sqrt [4]{-1} e^{-\frac{e}{2}-\frac{f x}{2}}+1\right )\right )-2 e \tan ^{-1}\left (\frac{\tanh \left (\frac{1}{4} (e+f x)\right )+i}{\sqrt{2}}\right )+i \pi \tan ^{-1}\left (\frac{\tanh \left (\frac{1}{4} (e+f x)\right )+i}{\sqrt{2}}\right )\right )}{f^2 \sqrt{a+i a \sinh (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{x{\frac{1}{\sqrt{a+ia\sinh \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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