\(\int x^3 \tan ^2(a+i \log (x)) \, dx\) [143]

Optimal result

Integrand size = 15, antiderivative size = 63 \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=2 e^{2 i a} x^2-\frac {x^4}{4}-\frac {2 e^{6 i a}}{e^{2 i a}+x^2}-4 e^{4 i a} \log \left (e^{2 i a}+x^2\right ) \]

[Out]

2*exp(2*I*a)*x^2-1/4*x^4-2*exp(6*I*a)/(exp(2*I*a)+x^2)-4*exp(4*I*a)*ln(exp(2*I*a)+x^2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4591, 456, 457, 78} \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=2 e^{2 i a} x^2-\frac {2 e^{6 i a}}{x^2+e^{2 i a}}-4 e^{4 i a} \log \left (x^2+e^{2 i a}\right )-\frac {x^4}{4} \]

[In]

Int[x^3*Tan[a + I*Log[x]]^2,x]

[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]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 456

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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 x^3}{\left (1+\frac {e^{2 i a}}{x^2}\right )^2} \, dx \\ & = \int \frac {x^3 \left (-i e^{2 i a}+i x^2\right )^2}{\left (e^{2 i a}+x^2\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {\left (-i e^{2 i a}+i x\right )^2 x}{\left (e^{2 i a}+x\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (4 e^{2 i a}-x+\frac {4 e^{6 i a}}{\left (e^{2 i a}+x\right )^2}-\frac {8 e^{4 i a}}{e^{2 i a}+x}\right ) \, dx,x,x^2\right ) \\ & = 2 e^{2 i a} x^2-\frac {x^4}{4}-\frac {2 e^{6 i a}}{e^{2 i a}+x^2}-4 e^{4 i a} \log \left (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. \(155\) vs. \(2(63)=126\).

Time = 0.19 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.46 \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=-\frac {x^4}{4}+2 x^2 \cos (2 a)-4 i \arctan \left (\frac {\left (1+x^2\right ) \cot (a)}{-1+x^2}\right ) \cos (4 a)-2 \cos (4 a) \log \left (1+x^4+2 x^2 \cos (2 a)\right )+2 i x^2 \sin (2 a)+4 \arctan \left (\frac {\left (1+x^2\right ) \cot (a)}{-1+x^2}\right ) \sin (4 a)-2 i \log \left (1+x^4+2 x^2 \cos (2 a)\right ) \sin (4 a)-\frac {2 (\cos (5 a)+i \sin (5 a))}{\left (1+x^2\right ) \cos (a)-i \left (-1+x^2\right ) \sin (a)} \]

[In]

Integrate[x^3*Tan[a + I*Log[x]]^2,x]

[Out]

-1/4*x^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])

Maple [A] (verified)

Time = 5.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83

method	result	size
risch	\(-\frac {9 x^{4}}{4}+\frac {2 x^{4}}{1+\frac {{\mathrm e}^{2 i a}}{x^{2}}}+4 \,{\mathrm e}^{2 i a} x^{2}-4 \,{\mathrm e}^{4 i a} \ln \left ({\mathrm e}^{2 i a}+x^{2}\right )\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=-\frac {x^{6} - 7 \, x^{4} e^{\left (2 i \, a\right )} - 8 \, x^{2} e^{\left (4 i \, a\right )} + 16 \, {\left (x^{2} e^{\left (4 i \, a\right )} + e^{\left (6 i \, a\right )}\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) + 8 \, e^{\left (6 i \, a\right )}}{4 \, {\left (x^{2} + e^{\left (2 i \, a\right )}\right )}} \]

[In]

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

[Out]

-1/4*(x^6 - 7*x^4*e^(2*I*a) - 8*x^2*e^(4*I*a) + 16*(x^2*e^(4*I*a) + e^(6*I*a))*log(x^2 + e^(2*I*a)) + 8*e^(6*I
*a))/(x^2 + e^(2*I*a))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=- \frac {x^{4}}{4} + 2 x^{2} e^{2 i a} - 4 e^{4 i a} \log {\left (x^{2} + e^{2 i a} \right )} - \frac {2 e^{6 i a}}{x^{2} + e^{2 i a}} \]

[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))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (46) = 92\).

Time = 0.21 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.44 \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=-\frac {x^{6} - 7 \, x^{4} {\left (\cos \left (2 \, a\right ) + i \, \sin \left (2 \, a\right )\right )} - 8 \, {\left (2 \, {\left (-i \, \cos \left (4 \, a\right ) + \sin \left (4 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + \cos \left (4 \, a\right ) + i \, \sin \left (4 \, a\right )\right )} x^{2} - 16 \, {\left ({\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \cos \left (4 \, a\right ) + {\left (\cos \left (2 \, a\right ) + i \, \sin \left (2 \, a\right )\right )} \sin \left (4 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + 8 \, {\left (x^{2} {\left (\cos \left (4 \, a\right ) + i \, \sin \left (4 \, a\right )\right )} + {\left (\cos \left (2 \, a\right ) + i \, \sin \left (2 \, a\right )\right )} \cos \left (4 \, a\right ) - {\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \sin \left (4 \, a\right )\right )} \log \left (x^{4} + 2 \, x^{2} \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right ) + 8 \, \cos \left (6 \, a\right ) + 8 i \, \sin \left (6 \, a\right )}{4 \, {\left (x^{2} + \cos \left (2 \, a\right ) + i \, \sin \left (2 \, a\right )\right )}} \]

[In]

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

[Out]

-1/4*(x^6 - 7*x^4*(cos(2*a) + I*sin(2*a)) - 8*(2*(-I*cos(4*a) + sin(4*a))*arctan2(sin(2*a), x^2 + cos(2*a)) +
cos(4*a) + I*sin(4*a))*x^2 - 16*((-I*cos(2*a) + sin(2*a))*cos(4*a) + (cos(2*a) + I*sin(2*a))*sin(4*a))*arctan2
(sin(2*a), x^2 + cos(2*a)) + 8*(x^2*(cos(4*a) + I*sin(4*a)) + (cos(2*a) + I*sin(2*a))*cos(4*a) - (-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*sin(6*a))/(x^2 +
cos(2*a) + I*sin(2*a))

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. 261 vs. \(2 (46) = 92\).

Time = 0.46 (sec) , antiderivative size = 261, normalized size of antiderivative = 4.14 \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=-\frac {x^{6}}{4 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} + \frac {3 \, x^{4} e^{\left (2 i \, a\right )}}{2 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} - \frac {4 \, x^{2} e^{\left (4 i \, a\right )} \log \left (-x^{2} - e^{\left (2 i \, a\right )}\right )}{x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}} + \frac {17 \, x^{2} e^{\left (4 i \, a\right )}}{4 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )}} - \frac {8 \, e^{\left (6 i \, a\right )} \log \left (-x^{2} - e^{\left (2 i \, a\right )}\right )}{x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}} + \frac {e^{\left (6 i \, a\right )}}{x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}} - \frac {4 \, e^{\left (8 i \, a\right )} \log \left (-x^{2} - e^{\left (2 i \, a\right )}\right )}{{\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )} x^{2}} - \frac {3 \, e^{\left (8 i \, a\right )}}{2 \, {\left (x^{2} + \frac {e^{\left (4 i \, a\right )}}{x^{2}} + 2 \, e^{\left (2 i \, a\right )}\right )} x^{2}} \]

[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)

Mupad [B] (verification not implemented)

Time = 27.49 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int x^3 \tan ^2(a+i \log (x)) \, dx=-\frac {2\,{\mathrm {e}}^{a\,6{}\mathrm {i}}}{x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}}-4\,{\mathrm {e}}^{a\,4{}\mathrm {i}}\,\ln \left (x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )+2\,x^2\,{\mathrm {e}}^{a\,2{}\mathrm {i}}-\frac {x^4}{4} \]

[In]

int(x^3*tan(a + log(x)*1i)^2,x)

[Out]

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