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]

-x^2/2 + (2*E^((4*I)*a))/(E^((2*I)*a) + x^2) + 2*E^((2*I)*a)*Log[E^((2*I)*a) + x^2]

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Rubi [F]  time = 0.0318918, 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 = {} \[ \int x \tan ^2(a+i \log (x)) \, dx \]

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*}

Mathematica [B]  time = 0.113743, size = 135, normalized size = 2.65 \[ \cos (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+\frac{2 \cos (3 a)+2 i \sin (3 a)}{\left (x^2+1\right ) \cos (a)-i \left (x^2-1\right ) \sin (a)}+i \sin (2 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )+2 i \cos (2 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cot (a)}{x^2-1}\right )-2 \sin (2 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cot (a)}{x^2-1}\right )-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Tan[a + I*Log[x]]^2,x]

[Out]

-x^2/2 + (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*Arc
Tan[((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)*S
in[3*a])/((1 + x^2)*Cos[a] - I*(-1 + x^2)*Sin[a])

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Maple [A]  time = 0.072, size = 50, normalized size = 1. \begin{align*} -{\frac{5\,{x}^{2}}{2}}+2\,{\frac{{x}^{2}}{ \left ({{\rm e}^{i \left ( a+i\ln \left ( x \right ) \right ) }} \right ) ^{2}+1}}+2\, \left ({{\rm e}^{ia}} \right ) ^{2}\ln \left ( \left ({{\rm e}^{ia}} \right ) ^{2}+{x}^{2} \right ) \end{align*}

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

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Maxima [B]  time = 1.02712, size = 261, normalized size = 5.12 \begin{align*} -\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 )} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, x^{2} +{\left (e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1\right )}{\rm integral}\left (-\frac{x e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 5 \, x}{e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1}, x\right )}{e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^2 + (e^(2*I*a - 2*log(x)) + 1)*integral(-(x*e^(2*I*a - 2*log(x)) + 5*x)/(e^(2*I*a - 2*log(x)) + 1), x))/(
e^(2*I*a - 2*log(x)) + 1)

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Sympy [A]  time = 0.742233, size = 42, normalized size = 0.82 \begin{align*} - \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}} \end{align*}

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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Giac [B]  time = 1.28563, size = 298, normalized size = 5.84 \begin{align*} -\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}} \end{align*}

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)