3.143 \(\int x^3 \tan ^2(a+i \log (x)) \, dx\)

Optimal. Leaf size=63 \[ 2 e^{2 i a} x^2-\frac{2 e^{6 i a}}{x^2+e^{2 i a}}-4 e^{4 i a} \log \left (x^2+e^{2 i a}\right )-\frac{x^4}{4} \]

[Out]

2*E^((2*I)*a)*x^2 - x^4/4 - (2*E^((6*I)*a))/(E^((2*I)*a) + x^2) - 4*E^((4*I)*a)*Log[E^((2*I)*a) + x^2]

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Rubi [F]  time = 0.0699981, 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 = {} \[ \int x^3 \tan ^2(a+i \log (x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[x^3*Tan[a + I*Log[x]]^2,x]

[Out]

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

Rubi steps

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

Mathematica [B]  time = 0.167011, size = 155, normalized size = 2.46 \[ 2 i x^2 \sin (2 a)+2 x^2 \cos (2 a)-2 \cos (4 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )-\frac{2 (\cos (5 a)+i \sin (5 a))}{\left (x^2+1\right ) \cos (a)-i \left (x^2-1\right ) \sin (a)}-2 i \sin (4 a) \log \left (2 x^2 \cos (2 a)+x^4+1\right )-4 i \cos (4 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cot (a)}{x^2-1}\right )+4 \sin (4 a) \tan ^{-1}\left (\frac{\left (x^2+1\right ) \cot (a)}{x^2-1}\right )-\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Tan[a + I*Log[x]]^2,x]

[Out]

-x^4/4 + 2*x^2*Cos[2*a] - (4*I)*ArcTan[((1 + x^2)*Cot[a])/(-1 + x^2)]*Cos[4*a] - 2*Cos[4*a]*Log[1 + x^4 + 2*x^
2*Cos[2*a]] + (2*I)*x^2*Sin[2*a] + 4*ArcTan[((1 + x^2)*Cot[a])/(-1 + x^2)]*Sin[4*a] - (2*I)*Log[1 + x^4 + 2*x^
2*Cos[2*a]]*Sin[4*a] - (2*(Cos[5*a] + I*Sin[5*a]))/((1 + x^2)*Cos[a] - I*(-1 + x^2)*Sin[a])

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Maple [A]  time = 0.078, size = 62, normalized size = 1. \begin{align*} -{\frac{9\,{x}^{4}}{4}}+2\,{\frac{{x}^{4}}{ \left ({{\rm e}^{i \left ( a+i\ln \left ( x \right ) \right ) }} \right ) ^{2}+1}}+4\,{x}^{2} \left ({{\rm e}^{ia}} \right ) ^{2}-4\, \left ({{\rm e}^{ia}} \right ) ^{4}\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^3*tan(a+I*ln(x))^2,x)

[Out]

-9/4*x^4+2*x^4/(exp(I*(a+I*ln(x)))^2+1)+4*x^2*exp(I*a)^2-4*exp(I*a)^4*ln(exp(I*a)^2+x^2)

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Maxima [B]  time = 1.15875, size = 312, normalized size = 4.95 \begin{align*} -\frac{x^{6} - x^{4}{\left (7 \, \cos \left (2 \, a\right ) + 7 i \, \sin \left (2 \, a\right )\right )} -{\left (16 \,{\left (-i \, \cos \left (4 \, a\right ) + \sin \left (4 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + 8 \, \cos \left (4 \, a\right ) + 8 i \, \sin \left (4 \, a\right )\right )} x^{2} -{\left (16 \,{\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \cos \left (4 \, a\right ) +{\left (16 \, \cos \left (2 \, a\right ) + 16 i \, \sin \left (2 \, a\right )\right )} \sin \left (4 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) +{\left (x^{2}{\left (8 \, \cos \left (4 \, a\right ) + 8 i \, \sin \left (4 \, a\right )\right )} +{\left (8 \, \cos \left (2 \, a\right ) + 8 i \, \sin \left (2 \, a\right )\right )} \cos \left (4 \, a\right ) - 8 \,{\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \sin \left (4 \, 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 ) + 8 \, \cos \left (6 \, a\right ) + 8 i \, \sin \left (6 \, a\right )}{4 \, x^{2} + 4 \, \cos \left (2 \, a\right ) + 4 i \, \sin \left (2 \, a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^6 - x^4*(7*cos(2*a) + 7*I*sin(2*a)) - (16*(-I*cos(4*a) + sin(4*a))*arctan2(sin(2*a), x^2 + cos(2*a)) + 8*c
os(4*a) + 8*I*sin(4*a))*x^2 - (16*(-I*cos(2*a) + sin(2*a))*cos(4*a) + (16*cos(2*a) + 16*I*sin(2*a))*sin(4*a))*
arctan2(sin(2*a), x^2 + cos(2*a)) + (x^2*(8*cos(4*a) + 8*I*sin(4*a)) + (8*cos(2*a) + 8*I*sin(2*a))*cos(4*a) -
8*(-I*cos(2*a) + sin(2*a))*sin(4*a))*log(x^4 + 2*x^2*cos(2*a) + cos(2*a)^2 + sin(2*a)^2) + 8*cos(6*a) + 8*I*si
n(6*a))/(4*x^2 + 4*cos(2*a) + 4*I*sin(2*a))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, x^{4} +{\left (e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 1\right )}{\rm integral}\left (-\frac{x^{3} e^{\left (2 i \, a - 2 \, \log \left (x\right )\right )} + 9 \, x^{3}}{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^3*tan(a+I*log(x))^2,x, algorithm="fricas")

[Out]

(2*x^4 + (e^(2*I*a - 2*log(x)) + 1)*integral(-(x^3*e^(2*I*a - 2*log(x)) + 9*x^3)/(e^(2*I*a - 2*log(x)) + 1), x
))/(e^(2*I*a - 2*log(x)) + 1)

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Sympy [A]  time = 0.672236, size = 54, normalized size = 0.86 \begin{align*} - \frac{x^{4}}{4} + 2 x^{2} e^{2 i a} - 4 e^{4 i a} \log{\left (x^{2} + e^{2 i a} \right )} - \frac{2 e^{6 i a}}{x^{2} + e^{2 i a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*tan(a+I*ln(x))**2,x)

[Out]

-x**4/4 + 2*x**2*exp(2*I*a) - 4*exp(4*I*a)*log(x**2 + exp(2*I*a)) - 2*exp(6*I*a)/(x**2 + exp(2*I*a))

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Giac [B]  time = 1.2509, size = 352, normalized size = 5.59 \begin{align*} -\frac{x^{6}}{4 \,{\left (x^{2} + \frac{e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} + \frac{3 \, x^{4} 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 \, x^{2} 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{17 \, x^{2} e^{\left (4 i \, a\right )}}{4 \,{\left (x^{2} + \frac{e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} - \frac{8 \, e^{\left (6 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{e^{\left (6 i \, a\right )}}{x^{2} + \frac{e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}} - \frac{4 \, e^{\left (8 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{3 \, e^{\left (8 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^3*tan(a+I*log(x))^2,x, algorithm="giac")

[Out]

-1/4*x^6/(x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a)) + 3/2*x^4*e^(2*I*a)/(x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a)) - 4*x^2*e
^(4*I*a)*log(-x^2 - e^(2*I*a))/(x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a)) + 17/4*x^2*e^(4*I*a)/(x^2 + e^(4*I*a)/x^2 +
2*e^(2*I*a)) - 8*e^(6*I*a)*log(-x^2 - e^(2*I*a))/(x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a)) + e^(6*I*a)/(x^2 + e^(4*
I*a)/x^2 + 2*e^(2*I*a)) - 4*e^(8*I*a)*log(-x^2 - e^(2*I*a))/((x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a))*x^2) - 3/2*e^
(8*I*a)/((x^2 + e^(4*I*a)/x^2 + 2*e^(2*I*a))*x^2)