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

Optimal. Leaf size=60 \[ \frac {3 x}{x^2+e^{2 i a}}+\frac {e^{2 i a}}{x \left (x^2+e^{2 i a}\right )}+2 e^{-i a} \tan ^{-1}\left (e^{-i a} x\right ) \]

[Out]

exp(2*I*a)/x/(exp(2*I*a)+x^2)+3*x/(exp(2*I*a)+x^2)+2*arctan(x/exp(I*a))/exp(I*a)

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Rubi [F]  time = 0.05, 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 = {} \[ \int \frac {\tan ^2(a+i \log (x))}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[Tan[a + I*Log[x]]^2/x^2,x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 72, normalized size = 1.20 \[ \frac {2 x (\cos (a)-i \sin (a))}{\left (x^2+1\right ) \cos (a)-i \left (x^2-1\right ) \sin (a)}+2 \cos (a) \tan ^{-1}(x (\cos (a)-i \sin (a)))-2 i \sin (a) \tan ^{-1}(x (\cos (a)-i \sin (a)))+\frac {1}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + I*Log[x]]^2/x^2,x]

[Out]

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

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fricas [A]  time = 0.58, size = 78, normalized size = 1.30 \[ \frac {3 \, x^{2} e^{\left (i \, a\right )} + {\left (i \, x^{3} + i \, x e^{\left (2 i \, a\right )}\right )} \log \left (x + i \, e^{\left (i \, a\right )}\right ) + {\left (-i \, x^{3} - i \, x e^{\left (2 i \, a\right )}\right )} \log \left (x - i \, e^{\left (i \, a\right )}\right ) + e^{\left (3 i \, a\right )}}{x^{3} e^{\left (i \, a\right )} + x e^{\left (3 i \, a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.56, size = 73, normalized size = 1.22 \[ 2 \, {\left (\arctan \left (x e^{\left (-i \, a\right )}\right ) e^{\left (-3 i \, a\right )} + \frac {x e^{\left (-2 i \, a\right )}}{x^{2} + e^{\left (2 i \, a\right )}}\right )} e^{\left (2 i \, a\right )} + \frac {5}{x {\left (\frac {e^{\left (2 i \, a\right )}}{x^{2}} + 1\right )}} + \frac {e^{\left (2 i \, a\right )}}{x^{3} {\left (\frac {e^{\left (2 i \, a\right )}}{x^{2}} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(arctan(x*e^(-I*a))*e^(-3*I*a) + x*e^(-2*I*a)/(x^2 + e^(2*I*a)))*e^(2*I*a) + 5/(x*(e^(2*I*a)/x^2 + 1)) + e^(
2*I*a)/(x^3*(e^(2*I*a)/x^2 + 1))

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maple [A]  time = 0.05, size = 38, normalized size = 0.63 \[ \frac {1}{x}+\frac {2}{x \left (1+\frac {{\mathrm e}^{2 i a}}{x^{2}}\right )}+2 \arctan \left (x \,{\mathrm e}^{-i a}\right ) {\mathrm e}^{-i a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 231, normalized size = 3.85 \[ \frac {6 \, x^{2} - {\left (x^{3} {\left (2 \, \cos \relax (a) - 2 i \, \sin \relax (a)\right )} + {\left ({\left (2 \, \cos \relax (a) - 2 i \, \sin \relax (a)\right )} \cos \left (2 \, a\right ) + 2 \, {\left (i \, \cos \relax (a) + \sin \relax (a)\right )} \sin \left (2 \, a\right )\right )} x\right )} \arctan \left (\frac {2 \, x \cos \relax (a)}{x^{2} + \cos \relax (a)^{2} - 2 \, x \sin \relax (a) + \sin \relax (a)^{2}}, \frac {x^{2} - \cos \relax (a)^{2} - \sin \relax (a)^{2}}{x^{2} + \cos \relax (a)^{2} - 2 \, x \sin \relax (a) + \sin \relax (a)^{2}}\right ) + {\left (x^{3} {\left (-i \, \cos \relax (a) - \sin \relax (a)\right )} + {\left ({\left (-i \, \cos \relax (a) - \sin \relax (a)\right )} \cos \left (2 \, a\right ) + {\left (\cos \relax (a) - i \, \sin \relax (a)\right )} \sin \left (2 \, a\right )\right )} x\right )} \log \left (\frac {x^{2} + \cos \relax (a)^{2} + 2 \, x \sin \relax (a) + \sin \relax (a)^{2}}{x^{2} + \cos \relax (a)^{2} - 2 \, x \sin \relax (a) + \sin \relax (a)^{2}}\right ) + 2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )}{2 \, x^{3} + x {\left (2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*x^2 - (x^3*(2*cos(a) - 2*I*sin(a)) + ((2*cos(a) - 2*I*sin(a))*cos(2*a) + 2*(I*cos(a) + sin(a))*sin(2*a))*x)
*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)) + (x^3*(-I*cos(a) - sin(a)) + ((-I*cos(a) - sin(a))*cos(2*a) + (cos(a) - I*sin(a))*sin(2
*a))*x)*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)) + 2*cos(2*a) +
2*I*sin(2*a))/(2*x^3 + x*(2*cos(2*a) + 2*I*sin(2*a)))

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mupad [B]  time = 2.20, size = 45, normalized size = 0.75 \[ \frac {2\,\mathrm {atan}\left (\frac {x}{\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}}}}\right )}{\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}}}}+\frac {3\,x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}}{x^3+{\mathrm {e}}^{a\,2{}\mathrm {i}}\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a + log(x)*1i)^2/x^2,x)

[Out]

(2*atan(x/exp(a*2i)^(1/2)))/exp(a*2i)^(1/2) + (exp(a*2i) + 3*x^2)/(x^3 + x*exp(a*2i))

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sympy [A]  time = 0.37, size = 54, normalized size = 0.90 \[ - \frac {- 3 x^{2} - e^{2 i a}}{x^{3} + x e^{2 i a}} - \left (i \log {\left (x - i e^{i a} \right )} - i \log {\left (x + i e^{i a} \right )}\right ) e^{- i a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*ln(x))**2/x**2,x)

[Out]

-(-3*x**2 - exp(2*I*a))/(x**3 + x*exp(2*I*a)) - (I*log(x - I*exp(I*a)) - I*log(x + I*exp(I*a)))*exp(-I*a)

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