3.145 \(\int x \tan ^2(a+i \log (x)) \, dx\)

Optimal. Leaf size=51 \[ \frac {2 e^{4 i a}}{x^2+e^{2 i a}}+2 e^{2 i a} \log \left (x^2+e^{2 i a}\right )-\frac {x^2}{2} \]

[Out]

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

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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 = {} \[ \int x \tan ^2(a+i \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x*Tan[a + I*Log[x]]^2,x]

[Out]

Defer[Int][x*Tan[a + I*Log[x]]^2, x]

Rubi steps

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

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Mathematica [B]  time = 0.12, size = 135, normalized size = 2.65 \[ \frac {2 \cos (3 a)+2 i \sin (3 a)}{\left (x^2+1\right ) \cos (a)-i \left (x^2-1\right ) \sin (a)}+2 i \cos (2 a) \tan ^{-1}\left (\frac {\left (x^2+1\right ) \cot (a)}{x^2-1}\right )-2 \sin (2 a) \tan ^{-1}\left (\frac {\left (x^2+1\right ) \cot (a)}{x^2-1}\right )+\cos (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+i \sin (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )-\frac {x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Tan[a + I*Log[x]]^2,x]

[Out]

-1/2*x^2 + (2*I)*ArcTan[((1 + x^2)*Cot[a])/(-1 + x^2)]*Cos[2*a] + Cos[2*a]*Log[1 + x^4 + 2*x^2*Cos[2*a]] - 2*A
rcTan[((1 + x^2)*Cot[a])/(-1 + x^2)]*Sin[2*a] + I*Log[1 + x^4 + 2*x^2*Cos[2*a]]*Sin[2*a] + (2*Cos[3*a] + (2*I)
*Sin[3*a])/((1 + x^2)*Cos[a] - I*(-1 + x^2)*Sin[a])

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fricas [A]  time = 0.48, size = 54, normalized size = 1.06 \[ -\frac {x^{4} + x^{2} e^{\left (2 i \, a\right )} - 4 \, {\left (x^{2} e^{\left (2 i \, a\right )} + e^{\left (4 i \, a\right )}\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) - 4 \, e^{\left (4 i \, a\right )}}{2 \, {\left (x^{2} + e^{\left (2 i \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.57, size = 221, normalized size = 4.33 \[ -\frac {x^{4}}{2 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} + \frac {2 \, x^{2} e^{\left (2 i \, a\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right )}{x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}} - \frac {5 \, x^{2} e^{\left (2 i \, a\right )}}{2 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} + \frac {4 \, e^{\left (4 i \, a\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right )}{x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}} - \frac {3 \, e^{\left (4 i \, a\right )}}{2 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} + \frac {2 \, e^{\left (6 i \, a\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right )}{{\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )} x^{2}} + \frac {e^{\left (6 i \, a\right )}}{2 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^4/(x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a)) + 2*x^2*e^(2*I*a)*log(x^2 + e^(2*I*a))/(x^2 + e^(4*I*a)/x^2 + 2*e
^(2*I*a)) - 5/2*x^2*e^(2*I*a)/(x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a)) + 4*e^(4*I*a)*log(x^2 + e^(2*I*a))/(x^2 + e^
(4*I*a)/x^2 + 2*e^(2*I*a)) - 3/2*e^(4*I*a)/(x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a)) + 2*e^(6*I*a)*log(x^2 + e^(2*I*
a))/((x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a))*x^2) + 1/2*e^(6*I*a)/((x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a))*x^2)

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maple [A]  time = 0.05, size = 42, normalized size = 0.82 \[ -\frac {5 x^{2}}{2}+\frac {2 x^{2}}{1+\frac {{\mathrm e}^{2 i a}}{x^{2}}}+2 \,{\mathrm e}^{2 i a} \ln \left ({\mathrm e}^{2 i a}+x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*x^2+2*x^2/(1+exp(2*I*a)/x^2)+2*exp(2*I*a)*ln(exp(2*I*a)+x^2)

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maxima [B]  time = 0.35, size = 193, normalized size = 3.78 \[ -\frac {x^{4} + {\left (4 \, {\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + \cos \left (2 \, a\right ) + i \, \sin \left (2 \, a\right )\right )} x^{2} - {\left (4 i \, \cos \left (2 \, a\right )^{2} - 8 \, \cos \left (2 \, a\right ) \sin \left (2 \, a\right ) - 4 i \, \sin \left (2 \, a\right )^{2}\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) - {\left (x^{2} {\left (2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )\right )} + 2 \, \cos \left (2 \, a\right )^{2} + 4 i \, \cos \left (2 \, a\right ) \sin \left (2 \, a\right ) - 2 \, \sin \left (2 \, a\right )^{2}\right )} \log \left (x^{4} + 2 \, x^{2} \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right ) - 4 \, \cos \left (4 \, a\right ) - 4 i \, \sin \left (4 \, a\right )}{2 \, x^{2} + 2 \, \cos \left (2 \, a\right ) + 2 i \, \sin \left (2 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^4 + (4*(-I*cos(2*a) + sin(2*a))*arctan2(sin(2*a), x^2 + cos(2*a)) + cos(2*a) + I*sin(2*a))*x^2 - (4*I*cos(
2*a)^2 - 8*cos(2*a)*sin(2*a) - 4*I*sin(2*a)^2)*arctan2(sin(2*a), x^2 + cos(2*a)) - (x^2*(2*cos(2*a) + 2*I*sin(
2*a)) + 2*cos(2*a)^2 + 4*I*cos(2*a)*sin(2*a) - 2*sin(2*a)^2)*log(x^4 + 2*x^2*cos(2*a) + cos(2*a)^2 + sin(2*a)^
2) - 4*cos(4*a) - 4*I*sin(4*a))/(2*x^2 + 2*cos(2*a) + 2*I*sin(2*a))

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mupad [B]  time = 2.21, size = 41, normalized size = 0.80 \[ \frac {2\,{\mathrm {e}}^{a\,4{}\mathrm {i}}}{x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}}+2\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,\ln \left (x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )-\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tan(a + log(x)*1i)^2,x)

[Out]

(2*exp(a*4i))/(exp(a*2i) + x^2) + 2*exp(a*2i)*log(exp(a*2i) + x^2) - x^2/2

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sympy [A]  time = 0.29, size = 42, normalized size = 0.82 \[ - \frac {x^{2}}{2} + 2 e^{2 i a} \log {\left (x^{2} + e^{2 i a} \right )} + \frac {2 e^{4 i a}}{x^{2} + e^{2 i a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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