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]

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

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Rubi [F]  time = 0.02, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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*}

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Mathematica [B]  time = 0.02, size = 114, normalized size = 3.45 \[ -\cos (2 a) \tan ^{-1}\left (\frac {\left (x^2+1\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right )-i \sin (2 a) \tan ^{-1}\left (\frac {\left (x^2+1\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right )-\frac {1}{2} i \cos (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+\frac {1}{2} \sin (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+\frac {i x^2}{2} \]

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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fricas [A]  time = 0.46, size = 21, normalized size = 0.64 \[ \frac {1}{2} i \, x^{2} - i \, e^{\left (2 i \, a\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 25, normalized size = 0.76 \[ \frac {1}{2} i \, x^{2} - i \, e^{\left (2 i \, a\right )} \log \left (-i \, x^{2} - i \, e^{\left (2 i \, a\right )}\right ) \]

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))

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maple [A]  time = 0.05, size = 26, normalized size = 0.79 \[ \frac {i x^{2}}{2}-i {\mathrm e}^{2 i a} \ln \left ({\mathrm e}^{2 i a}+x^{2}\right ) \]

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*exp(2*I*a)*ln(exp(2*I*a)+x^2)

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maxima [B]  time = 0.34, size = 73, normalized size = 2.21 \[ \frac {1}{2} i \, x^{2} + \frac {1}{2} \, {\left (2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + \frac {1}{2} \, {\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \log \left (x^{4} + 2 \, x^{2} \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right ) \]

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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mupad [B]  time = 2.19, size = 25, normalized size = 0.76 \[ -{\mathrm {e}}^{a\,2{}\mathrm {i}}\,\ln \left (x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}+\frac {x^2\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tan(a + log(x)*1i),x)

[Out]

(x^2*1i)/2 - exp(a*2i)*log(exp(a*2i) + x^2)*1i

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sympy [A]  time = 0.19, size = 26, normalized size = 0.79 \[ \frac {i x^{2}}{2} - i e^{2 i a} \log {\left (x^{2} + e^{2 i a} \right )} \]

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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