\(\int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx\) [149]

Optimal result

Integrand size = 15, antiderivative size = 55 \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=-\frac {2 e^{-2 i a}}{1+\frac {e^{2 i a}}{x^2}}+\frac {1}{2 x^2}-2 e^{-2 i a} \log \left (1+\frac {e^{2 i a}}{x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4591, 455, 45} \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=-\frac {2 e^{-2 i a}}{1+\frac {e^{2 i a}}{x^2}}-2 e^{-2 i a} \log \left (1+\frac {e^{2 i a}}{x^2}\right )+\frac {1}{2 x^2} \]

[In]

Int[Tan[a + I*Log[x]]^2/x^3,x]

[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)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 4591

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(e*x)^m*((I - I*E^(2*I*a*d)*
x^(2*I*b*d))/(1 + E^(2*I*a*d)*x^(2*I*b*d)))^p, x] /; FreeQ[{a, b, d, e, m, p}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (i-\frac {i e^{2 i a}}{x^2}\right )^2}{\left (1+\frac {e^{2 i a}}{x^2}\right )^2 x^3} \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {\left (i-i e^{2 i a} x\right )^2}{\left (1+e^{2 i a} x\right )^2} \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \left (-1-\frac {4}{\left (1+e^{2 i a} x\right )^2}+\frac {4}{1+e^{2 i a} x}\right ) \, dx,x,\frac {1}{x^2}\right )\right ) \\ & = -\frac {2 e^{-2 i a}}{1+\frac {e^{2 i a}}{x^2}}+\frac {1}{2 x^2}-2 e^{-2 i a} \log \left (1+\frac {e^{2 i a}}{x^2}\right ) \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(150\) vs. \(2(55)=110\).

Time = 0.13 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.73 \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=\frac {1}{2 x^2}-2 i \arctan \left (\frac {\left (1+x^2\right ) \cot (a)}{-1+x^2}\right ) \cos (2 a)+4 \cos (2 a) \log (x)-\cos (2 a) \log \left (1+x^4+2 x^2 \cos (2 a)\right )+\frac {2 \cos (a)-2 i \sin (a)}{\left (1+x^2\right ) \cos (a)-i \left (-1+x^2\right ) \sin (a)}-2 \arctan \left (\frac {\left (1+x^2\right ) \cot (a)}{-1+x^2}\right ) \sin (2 a)-4 i \log (x) \sin (2 a)+i \log \left (1+x^4+2 x^2 \cos (2 a)\right ) \sin (2 a) \]

[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]

Maple [A] (verified)

Time = 1.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {1}{2 x^{2}}+\frac {2}{x^{2} \left (1+\frac {{\mathrm e}^{2 i a}}{x^{2}}\right )}-2 \,{\mathrm e}^{-2 i a} \ln \left ({\mathrm e}^{2 i a}+x^{2}\right )+4 \,{\mathrm e}^{-2 i a} \ln \left (x \right )\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.35 \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=\frac {5 \, x^{2} e^{\left (2 i \, a\right )} - 4 \, {\left (x^{4} + x^{2} e^{\left (2 i \, a\right )}\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) + 8 \, {\left (x^{4} + x^{2} e^{\left (2 i \, a\right )}\right )} \log \left (x\right ) + e^{\left (4 i \, a\right )}}{2 \, {\left (x^{4} e^{\left (2 i \, a\right )} + x^{2} e^{\left (4 i \, a\right )}\right )}} \]

[In]

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

[Out]

1/2*(5*x^2*e^(2*I*a) - 4*(x^4 + x^2*e^(2*I*a))*log(x^2 + e^(2*I*a)) + 8*(x^4 + x^2*e^(2*I*a))*log(x) + e^(4*I*
a))/(x^4*e^(2*I*a) + x^2*e^(4*I*a))

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.11 \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=- \frac {- 5 x^{2} - e^{2 i a}}{2 x^{4} + 2 x^{2} e^{2 i a}} + 4 e^{- 2 i a} \log {\left (x \right )} - 2 e^{- 2 i a} \log {\left (x^{2} + e^{2 i a} \right )} \]

[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))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (41) = 82\).

Time = 0.46 (sec) , antiderivative size = 178, normalized size of antiderivative = 3.24 \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=-\frac {2 \, \log \left (-x^{2} - e^{\left (2 i \, a\right )}\right )}{\frac {e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}} + \frac {4 \, \log \left (x\right )}{\frac {e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}} - \frac {2}{\frac {e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}} - \frac {2 \, e^{\left (2 i \, a\right )} \log \left (-x^{2} - e^{\left (2 i \, a\right )}\right )}{x^{2} {\left (\frac {e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} + \frac {4 \, e^{\left (2 i \, a\right )} \log \left (x\right )}{x^{2} {\left (\frac {e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} + \frac {e^{\left (2 i \, a\right )}}{2 \, x^{2} {\left (\frac {e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} + \frac {e^{\left (4 i \, a\right )}}{2 \, x^{4} {\left (\frac {e^{\left (4 i \, a\right )}}{x^{2}} + e^{\left (2 i \, a\right )}\right )}} \]

[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)))

Mupad [B] (verification not implemented)

Time = 26.60 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^2(a+i \log (x))}{x^3} \, dx=-2\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\ln \left (x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )+4\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\ln \left (x\right )+\frac {\frac {5\,x^2}{2}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}}{2}}{x^4+{\mathrm {e}}^{a\,2{}\mathrm {i}}\,x^2} \]

[In]

int(tan(a + log(x)*1i)^2/x^3,x)

[Out]

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