3.138 \(\int \tan (a+i \log (x)) \, dx\)

Optimal. Leaf size=27 \[ i x-2 i e^{i a} \tan ^{-1}\left (e^{-i a} x\right ) \]

[Out]

I*x - (2*I)*E^(I*a)*ArcTan[x/E^(I*a)]

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Rubi [F]  time = 0.0074411, 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 = {} \[ \int \tan (a+i \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tan[a + I*Log[x]],x]

[Out]

Defer[Int][Tan[a + I*Log[x]], x]

Rubi steps

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

Mathematica [A]  time = 0.0071506, size = 42, normalized size = 1.56 \[ -2 i \cos (a) \tan ^{-1}(x \cos (a)-i x \sin (a))+2 \sin (a) \tan ^{-1}(x \cos (a)-i x \sin (a))+i x \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + I*Log[x]],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 = 0.058, size = 0, normalized size = 0. \begin{align*} \int \tan \left ( a+i\ln \left ( x \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a+I*ln(x)),x)

[Out]

int(tan(a+I*ln(x)),x)

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Maxima [B]  time = 1.70817, size = 165, normalized size = 6.11 \begin{align*}{\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 ) - \frac{1}{2} \,{\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 ) + i \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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 - sin(a)^2)/(
x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2)) - 1/2*(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)) + 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}{e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-I*e^(2*I*a - 2*log(x)) + I)/(e^(2*I*a - 2*log(x)) + 1), x)

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Sympy [A]  time = 0.50162, size = 27, normalized size = 1. \begin{align*} i x + \left (- \log{\left (x - i e^{i a} \right )} + \log{\left (x + i e^{i a} \right )}\right ) e^{i a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14532, size = 41, normalized size = 1.52 \begin{align*} \frac{2 \, \arctan \left (\frac{i \, x}{\sqrt{-e^{\left (2 i \, a\right )}}}\right ) e^{\left (2 i \, a\right )}}{\sqrt{-e^{\left (2 i \, a\right )}}} + i \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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