3.149 \(\int \frac{\tan ^2(a+i \log (x))}{x^3} \, dx\)

Optimal. Leaf size=55 \[ -\frac{2 e^{-2 i a}}{1+\frac{e^{2 i a}}{x^2}}-2 e^{-2 i a} \log \left (1+\frac{e^{2 i a}}{x^2}\right )+\frac{1}{2 x^2} \]

[Out]

-2/(E^((2*I)*a)*(1 + E^((2*I)*a)/x^2)) + 1/(2*x^2) - (2*Log[1 + E^((2*I)*a)/x^2])/E^((2*I)*a)

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Rubi [F]  time = 0.052695, 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 = {} \[ \int \frac{\tan ^2(a+i \log (x))}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tan[a + I*Log[x]]^2/x^3,x]

[Out]

Defer[Int][Tan[a + I*Log[x]]^2/x^3, x]

Rubi steps

\begin{align*} \int \frac{\tan ^2(a+i \log (x))}{x^3} \, dx &=\int \frac{\tan ^2(a+i \log (x))}{x^3} \, dx\\ \end{align*}

Mathematica [B]  time = 0.170479, size = 150, normalized size = 2.73 \[ -\cos (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+\frac{2 \cos (a)-2 i \sin (a)}{\left (x^2+1\right ) \cos (a)-i \left (x^2-1\right ) \sin (a)}+i \sin (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )-2 i \cos (2 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cot (a)}{x^2-1}\right )-2 \sin (2 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cot (a)}{x^2-1}\right )-4 i \sin (2 a) \log (x)+4 \cos (2 a) \log (x)+\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + I*Log[x]]^2/x^3,x]

[Out]

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

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Maple [A]  time = 0.065, size = 61, normalized size = 1.1 \begin{align*}{\frac{1}{2\,{x}^{2}}}+2\,{\frac{1}{{x}^{2} \left ( \left ({{\rm e}^{i \left ( a+i\ln \left ( x \right ) \right ) }} \right ) ^{2}+1 \right ) }}+4\,{\frac{\ln \left ( x \right ) }{ \left ({{\rm e}^{ia}} \right ) ^{2}}}-2\,{\frac{\ln \left ( \left ({{\rm e}^{ia}} \right ) ^{2}+{x}^{2} \right ) }{ \left ({{\rm e}^{ia}} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a+I*ln(x))^2/x^3,x)

[Out]

1/2/x^2+2/x^2/(exp(I*(a+I*ln(x)))^2+1)+4/exp(I*a)^2*ln(x)-2/exp(I*a)^2*ln(exp(I*a)^2+x^2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))^2/x^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (x^{2} e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + x^{2}\right )}{\rm integral}\left (-\frac{e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} - 3}{x^{3} e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + x^{3}}, x\right ) + 2}{x^{2} e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))^2/x^3,x, algorithm="fricas")

[Out]

((x^2*e^(2*I*a - 2*log(x)) + x^2)*integral(-(e^(2*I*a - 2*log(x)) - 3)/(x^3*e^(2*I*a - 2*log(x)) + x^3), x) +
2)/(x^2*e^(2*I*a - 2*log(x)) + x^2)

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Sympy [A]  time = 1.0463, size = 60, normalized size = 1.09 \begin{align*} \frac{5 x^{2} + e^{2 i a}}{2 x^{4} + 2 x^{2} e^{2 i a}} + 4 e^{- 2 i a} \log{\left (x \right )} - 2 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(tan(a+I*ln(x))**2/x**3,x)

[Out]

(5*x**2 + exp(2*I*a))/(2*x**4 + 2*x**2*exp(2*I*a)) + 4*exp(-2*I*a)*log(x) - 2*exp(-2*I*a)*log(x**2 + exp(2*I*a
))

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Giac [B]  time = 1.21621, size = 240, normalized size = 4.36 \begin{align*} -\frac{2 \, \log \left (-x^{2} - e^{\left (2 i \, a\right )}\right )}{\frac{e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}} + \frac{4 \, \log \left (x\right )}{\frac{e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}} - \frac{2}{\frac{e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}} - \frac{2 \, e^{\left (2 i \, a\right )} \log \left (-x^{2} - e^{\left (2 i \, a\right )}\right )}{x^{2}{\left (\frac{e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} + \frac{4 \, e^{\left (2 i \, a\right )} \log \left (x\right )}{x^{2}{\left (\frac{e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} + \frac{e^{\left (2 i \, a\right )}}{2 \, x^{2}{\left (\frac{e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} + \frac{e^{\left (4 i \, a\right )}}{2 \, x^{4}{\left (\frac{e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))^2/x^3,x, algorithm="giac")

[Out]

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