3.141 \(\int \frac {\tan (a+i \log (x))}{x^3} \, dx\)

Optimal. Leaf size=35 \[ \frac {i}{2 x^2}-i e^{-2 i a} \log \left (1+\frac {e^{2 i a}}{x^2}\right ) \]

[Out]

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

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Rubi [F]  time = 0.03, 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 = {} \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tan[a + I*Log[x]]/x^3,x]

[Out]

Defer[Int][Tan[a + I*Log[x]]/x^3, x]

Rubi steps

\begin {align*} \int \frac {\tan (a+i \log (x))}{x^3} \, dx &=\int \frac {\tan (a+i \log (x))}{x^3} \, dx\\ \end {align*}

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Mathematica [B]  time = 0.04, size = 132, normalized size = 3.77 \[ \cos (2 a) \left (-\tan ^{-1}\left (\frac {\left (x^2+1\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right )\right )+i \sin (2 a) \tan ^{-1}\left (\frac {\left (x^2+1\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right )-\frac {1}{2} i \cos (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )-\frac {1}{2} \sin (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+2 \sin (2 a) \log (x)+2 i \cos (2 a) \log (x)+\frac {i}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + I*Log[x]]/x^3,x]

[Out]

(I/2)/x^2 - ArcTan[((1 + x^2)*Cos[a])/(Sin[a] - x^2*Sin[a])]*Cos[2*a] + (2*I)*Cos[2*a]*Log[x] - (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] + 2*Log[x]*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 = 37, normalized size = 1.06 \[ \frac {{\left (-2 i \, x^{2} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) + 4 i \, x^{2} \log \relax (x) + i \, e^{\left (2 i \, a\right )}\right )} e^{\left (-2 i \, a\right )}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x^3,x, algorithm="fricas")

[Out]

1/2*(-2*I*x^2*log(x^2 + e^(2*I*a)) + 4*I*x^2*log(x) + I*e^(2*I*a))*e^(-2*I*a)/x^2

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giac [A]  time = 0.53, size = 33, normalized size = 0.94 \[ -i \, e^{\left (-2 i \, a\right )} \log \left (-i \, x^{2} - i \, e^{\left (2 i \, a\right )}\right ) + 2 i \, e^{\left (-2 i \, a\right )} \log \relax (x) + \frac {i}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 36, normalized size = 1.03 \[ \frac {i}{2 x^{2}}+2 i {\mathrm e}^{-2 i a} \ln \relax (x )-i {\mathrm e}^{-2 i a} \ln \left ({\mathrm e}^{2 i a}+x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a+I*ln(x))/x^3,x)

[Out]

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

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maxima [B]  time = 0.36, size = 96, normalized size = 2.74 \[ -\frac {x^{2} {\left (i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \log \left (x^{4} + 2 \, x^{2} \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right ) - {\left ({\left (2 \, \cos \left (2 \, a\right ) - 2 i \, \sin \left (2 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + 4 \, {\left (i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \log \relax (x)\right )} x^{2} - i}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.29, size = 35, normalized size = 1.00 \[ -{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\ln \left (x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}+{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\ln \relax (x)\,2{}\mathrm {i}+\frac {1{}\mathrm {i}}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a + log(x)*1i)/x^3,x)

[Out]

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

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sympy [A]  time = 0.35, size = 39, normalized size = 1.11 \[ 2 i e^{- 2 i a} \log {\relax (x )} - i e^{- 2 i a} \log {\left (x^{2} + e^{2 i a} \right )} + \frac {i}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*ln(x))/x**3,x)

[Out]

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

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