3.140 \(\int \frac{\tan (a+i \log (x))}{x^2} \, dx\)

Optimal. Leaf size=29 \[ 2 i e^{-i a} \tan ^{-1}\left (e^{-i a} x\right )+\frac{i}{x} \]

[Out]

I/x + ((2*I)*ArcTan[x/E^(I*a)])/E^(I*a)

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Rubi [F]  time = 0.0277522, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\tan (a+i \log (x))}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tan[a + I*Log[x]]/x^2,x]

[Out]

Defer[Int][Tan[a + I*Log[x]]/x^2, x]

Rubi steps

\begin{align*} \int \frac{\tan (a+i \log (x))}{x^2} \, dx &=\int \frac{\tan (a+i \log (x))}{x^2} \, dx\\ \end{align*}

Mathematica [A]  time = 0.0238917, size = 44, normalized size = 1.52 \[ 2 i \cos (a) \tan ^{-1}(x \cos (a)-i x \sin (a))+2 \sin (a) \tan ^{-1}(x \cos (a)-i x \sin (a))+\frac{i}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + I*Log[x]]/x^2,x]

[Out]

I/x + (2*I)*ArcTan[x*Cos[a] - I*x*Sin[a]]*Cos[a] + 2*ArcTan[x*Cos[a] - I*x*Sin[a]]*Sin[a]

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{\frac{\tan \left ( a+i\ln \left ( x \right ) \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a+I*ln(x))/x^2,x)

[Out]

int(tan(a+I*ln(x))/x^2,x)

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Maxima [B]  time = 1.63297, size = 171, normalized size = 5.9 \begin{align*} \frac{2 \, x{\left (-i \, \cos \left (a\right ) - \sin \left (a\right )\right )} \arctan \left (\frac{2 \, x \cos \left (a\right )}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}, \frac{x^{2} - \cos \left (a\right )^{2} - \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + x{\left (\cos \left (a\right ) - i \, \sin \left (a\right )\right )} \log \left (\frac{x^{2} + \cos \left (a\right )^{2} + 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + 2 i}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x^2,x, algorithm="maxima")

[Out]

1/2*(2*x*(-I*cos(a) - sin(a))*arctan2(2*x*cos(a)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2), (x^2 - cos(a)^2 - s
in(a)^2)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2)) + x*(cos(a) - I*sin(a))*log((x^2 + cos(a)^2 + 2*x*sin(a) +
sin(a)^2)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2)) + 2*I)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.456843, size = 27, normalized size = 0.93 \begin{align*} \left (\log{\left (x - i e^{i a} \right )} - \log{\left (x + i e^{i a} \right )}\right ) e^{- i a} + \frac{i}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*ln(x))/x**2,x)

[Out]

(log(x - I*exp(I*a)) - log(x + I*exp(I*a)))*exp(-I*a) + I/x

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Giac [A]  time = 1.17649, size = 38, normalized size = 1.31 \begin{align*} -\frac{2 \, \arctan \left (\frac{i \, x}{\sqrt{-e^{\left (2 i \, a\right )}}}\right )}{\sqrt{-e^{\left (2 i \, a\right )}}} + \frac{i}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x^2,x, algorithm="giac")

[Out]

-2*arctan(I*x/sqrt(-e^(2*I*a)))/sqrt(-e^(2*I*a)) + I/x