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

Optimal. Leaf size=33 \[ \frac{i x^2}{2}-i e^{2 i a} \log \left (x^2+e^{2 i a}\right ) \]

[Out]

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

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Rubi [F]  time = 0.017491, 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 = {} \[ \int x \tan (a+i \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x*Tan[a + I*Log[x]],x]

[Out]

Defer[Int][x*Tan[a + I*Log[x]], x]

Rubi steps

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

Mathematica [B]  time = 0.0224594, size = 114, normalized size = 3.45 \[ -\frac{1}{2} i \cos (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+\frac{1}{2} \sin (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )-\cos (2 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right )-i \sin (2 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right )+\frac{i x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Tan[a + I*Log[x]],x]

[Out]

(I/2)*x^2 - ArcTan[((1 + x^2)*Cos[a])/(Sin[a] - x^2*Sin[a])]*Cos[2*a] - (I/2)*Cos[2*a]*Log[1 + x^4 + 2*x^2*Cos
[2*a]] - I*ArcTan[((1 + x^2)*Cos[a])/(Sin[a] - x^2*Sin[a])]*Sin[2*a] + (Log[1 + x^4 + 2*x^2*Cos[2*a]]*Sin[2*a]
)/2

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Maple [A]  time = 0.062, size = 37, normalized size = 1.1 \begin{align*} -{\frac{i}{2}}{x}^{2}-i \left ( -{x}^{2}+ \left ({{\rm e}^{ia}} \right ) ^{2}\ln \left ( \left ({{\rm e}^{ia}} \right ) ^{2}+{x}^{2} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*x^2-I*(-x^2+exp(I*a)^2*ln(exp(I*a)^2+x^2))

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Maxima [B]  time = 1.05837, size = 99, normalized size = 3. \begin{align*} \frac{1}{2} i \, x^{2} + \frac{1}{2} \,{\left (2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + \frac{1}{2} \,{\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \log \left (x^{4} + 2 \, x^{2} \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*x^2 + 1/2*(2*cos(2*a) + 2*I*sin(2*a))*arctan2(sin(2*a), x^2 + cos(2*a)) + 1/2*(-I*cos(2*a) + sin(2*a))*l
og(x^4 + 2*x^2*cos(2*a) + cos(2*a)^2 + sin(2*a)^2)

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

[Out]

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

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Sympy [A]  time = 0.518396, size = 26, normalized size = 0.79 \begin{align*} \frac{i x^{2}}{2} - i e^{2 i a} \log{\left (x^{2} + e^{2 i a} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x**2/2 - I*exp(2*I*a)*log(x**2 + exp(2*I*a))

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Giac [A]  time = 1.19309, size = 34, normalized size = 1.03 \begin{align*} \frac{1}{2} i \, x^{2} - i \, e^{\left (2 i \, a\right )} \log \left (-i \, x^{2} - i \, e^{\left (2 i \, a\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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