\(\int x^3 \tan (a+i \log (x)) \, dx\) [135]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int x^3 \tan (a+i \log (x)) \, dx=-i e^{2 i a} x^2+\frac {i x^4}{4}+i e^{4 i a} \log \left (e^{2 i a}+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4591, 456, 457, 78} \[ \int x^3 \tan (a+i \log (x)) \, dx=-i e^{2 i a} x^2+i e^{4 i a} \log \left (x^2+e^{2 i a}\right )+\frac {i x^4}{4} \]

[In]

Int[x^3*Tan[a + I*Log[x]],x]

[Out]

(-I)*E^((2*I)*a)*x^2 + (I/4)*x^4 + I*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 ) x^3}{1+\frac {e^{2 i a}}{x^2}} \, dx \\ & = \int \frac {x^3 \left (-i e^{2 i a}+i x^2\right )}{e^{2 i a}+x^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {\left (-i e^{2 i a}+i x\right ) x}{e^{2 i a}+x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-2 i e^{2 i a}+i x+\frac {2 i e^{4 i a}}{e^{2 i a}+x}\right ) \, dx,x,x^2\right ) \\ & = -i e^{2 i a} x^2+\frac {i x^4}{4}+i 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. \(132\) vs. \(2(47)=94\).

Time = 0.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.81 \[ \int x^3 \tan (a+i \log (x)) \, dx=\frac {i x^4}{4}-i x^2 \cos (2 a)+\arctan \left (\frac {\left (1+x^2\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right ) \cos (4 a)+\frac {1}{2} i \cos (4 a) \log \left (1+x^4+2 x^2 \cos (2 a)\right )+x^2 \sin (2 a)+i \arctan \left (\frac {\left (1+x^2\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right ) \sin (4 a)-\frac {1}{2} \log \left (1+x^4+2 x^2 \cos (2 a)\right ) \sin (4 a) \]

[In]

Integrate[x^3*Tan[a + I*Log[x]],x]

[Out]

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

Maple [A] (verified)

Time = 4.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79

method	result	size
risch	\(-i {\mathrm e}^{2 i a} x^{2}+\frac {i x^{4}}{4}+i {\mathrm e}^{4 i a} \ln \left ({\mathrm e}^{2 i a}+x^{2}\right )\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.64 \[ \int x^3 \tan (a+i \log (x)) \, dx=\frac {1}{4} i \, x^{4} - i \, x^{2} e^{\left (2 i \, a\right )} + i \, e^{\left (4 i \, a\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int x^3 \tan (a+i \log (x)) \, dx=\frac {i x^{4}}{4} - i x^{2} e^{2 i a} + i e^{4 i a} \log {\left (x^{2} + e^{2 i a} \right )} \]

[In]

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

[Out]

I*x**4/4 - I*x**2*exp(2*I*a) + I*exp(4*I*a)*log(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. 88 vs. \(2 (30) = 60\).

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.87 \[ \int x^3 \tan (a+i \log (x)) \, dx=\frac {1}{4} i \, x^{4} + x^{2} {\left (-i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} - {\left (\cos \left (4 \, a\right ) + i \, \sin \left (4 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + \frac {1}{2} \, {\left (i \, \cos \left (4 \, a\right ) - \sin \left (4 \, a\right )\right )} \log \left (x^{4} + 2 \, x^{2} \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right ) \]

[In]

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

[Out]

1/4*I*x^4 + x^2*(-I*cos(2*a) + sin(2*a)) - (cos(4*a) + I*sin(4*a))*arctan2(sin(2*a), x^2 + cos(2*a)) + 1/2*(I*
cos(4*a) - sin(4*a))*log(x^4 + 2*x^2*cos(2*a) + cos(2*a)^2 + sin(2*a)^2)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int x^3 \tan (a+i \log (x)) \, dx=\frac {1}{4} i \, x^{4} - i \, x^{2} e^{\left (2 i \, a\right )} - \frac {1}{2} \, \pi e^{\left (4 i \, a\right )} + i \, e^{\left (4 i \, a\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 25.98 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int x^3 \tan (a+i \log (x)) \, dx={\mathrm {e}}^{a\,4{}\mathrm {i}}\,\ln \left (x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}-x^2\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,1{}\mathrm {i}+\frac {x^4\,1{}\mathrm {i}}{4} \]

[In]

int(x^3*tan(a + log(x)*1i),x)

[Out]

exp(a*4i)*log(exp(a*2i) + x^2)*1i - x^2*exp(a*2i)*1i + (x^4*1i)/4