\(\int x^2 \tan (a+i \log (x)) \, dx\) [136]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 43 \[ \int x^2 \tan (a+i \log (x)) \, dx=-2 i e^{2 i a} x+\frac {i x^3}{3}+2 i e^{3 i a} \arctan \left (e^{-i a} x\right ) \]

[Out]

-2*I*exp(2*I*a)*x+1/3*I*x^3+2*I*exp(3*I*a)*arctan(x/exp(I*a))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4591, 456, 470, 327, 209} \[ \int x^2 \tan (a+i \log (x)) \, dx=2 i e^{3 i a} \arctan \left (e^{-i a} x\right )-2 i e^{2 i a} x+\frac {i x^3}{3} \]

[In]

Int[x^2*Tan[a + I*Log[x]],x]

[Out]

(-2*I)*E^((2*I)*a)*x + (I/3)*x^3 + (2*I)*E^((3*I)*a)*ArcTan[x/E^(I*a)]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, x]

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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
 n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
 && NeQ[m + n*(p + 1) + 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 ) x^2}{1+\frac {e^{2 i a}}{x^2}} \, dx \\ & = \int \frac {x^2 \left (-i e^{2 i a}+i x^2\right )}{e^{2 i a}+x^2} \, dx \\ & = \frac {i x^3}{3}-\left (2 i e^{2 i a}\right ) \int \frac {x^2}{e^{2 i a}+x^2} \, dx \\ & = -2 i e^{2 i a} x+\frac {i x^3}{3}+\left (2 i e^{4 i a}\right ) \int \frac {1}{e^{2 i a}+x^2} \, dx \\ & = -2 i e^{2 i a} x+\frac {i x^3}{3}+2 i e^{3 i a} \arctan \left (e^{-i a} x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.53 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {i x^3}{3}-2 i x \cos (2 a)+2 i \arctan (x \cos (a)-i x \sin (a)) \cos (3 a)+2 x \sin (2 a)-2 \arctan (x \cos (a)-i x \sin (a)) \sin (3 a) \]

[In]

Integrate[x^2*Tan[a + I*Log[x]],x]

[Out]

(I/3)*x^3 - (2*I)*x*Cos[2*a] + (2*I)*ArcTan[x*Cos[a] - I*x*Sin[a]]*Cos[3*a] + 2*x*Sin[2*a] - 2*ArcTan[x*Cos[a]
 - I*x*Sin[a]]*Sin[3*a]

Maple [A] (verified)

Time = 3.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77

method result size
risch \(\frac {i x^{3}}{3}-2 i {\mathrm e}^{2 i a} x +2 i \arctan \left (x \,{\mathrm e}^{-i a}\right ) {\mathrm e}^{3 i a}\) \(33\)

[In]

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

[Out]

1/3*I*x^3-2*I*exp(2*I*a)*x+2*I*arctan(x*exp(-I*a))*exp(3*I*a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {1}{3} i \, x^{3} - 2 i \, x e^{\left (2 i \, a\right )} - e^{\left (3 i \, a\right )} \log \left (x + i \, e^{\left (i \, a\right )}\right ) + e^{\left (3 i \, a\right )} \log \left (x - i \, e^{\left (i \, a\right )}\right ) \]

[In]

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

[Out]

1/3*I*x^3 - 2*I*x*e^(2*I*a) - e^(3*I*a)*log(x + I*e^(I*a)) + e^(3*I*a)*log(x - I*e^(I*a))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {i x^{3}}{3} - 2 i x e^{2 i a} + \left (\log {\left (x e^{2 i a} - i e^{3 i a} \right )} - \log {\left (x e^{2 i a} + i e^{3 i a} \right )}\right ) e^{3 i a} \]

[In]

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

[Out]

I*x**3/3 - 2*I*x*exp(2*I*a) + (log(x*exp(2*I*a) - I*exp(3*I*a)) - log(x*exp(2*I*a) + I*exp(3*I*a)))*exp(3*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. 149 vs. \(2 (26) = 52\).

Time = 0.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 3.47 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {1}{3} i \, x^{3} - 2 \, x {\left (i \, \cos \left (2 \, a\right ) - \sin \left (2 \, a\right )\right )} - {\left (i \, \cos \left (3 \, a\right ) - \sin \left (3 \, a\right )\right )} \arctan \left (\frac {2 \, x \cos \left (a\right )}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}, \frac {x^{2} - \cos \left (a\right )^{2} - \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + \frac {1}{2} \, {\left (\cos \left (3 \, a\right ) + i \, \sin \left (3 \, a\right )\right )} \log \left (\frac {x^{2} + \cos \left (a\right )^{2} + 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) \]

[In]

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

[Out]

1/3*I*x^3 - 2*x*(I*cos(2*a) - sin(2*a)) - (I*cos(3*a) - sin(3*a))*arctan2(2*x*cos(a)/(x^2 + cos(a)^2 - 2*x*sin
(a) + sin(a)^2), (x^2 - cos(a)^2 - sin(a)^2)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2)) + 1/2*(cos(3*a) + I*sin
(3*a))*log((x^2 + cos(a)^2 + 2*x*sin(a) + sin(a)^2)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2))

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.60 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {1}{3} i \, x^{3} + 2 i \, \arctan \left (x e^{\left (-i \, a\right )}\right ) e^{\left (3 i \, a\right )} - 2 i \, x e^{\left (2 i \, a\right )} \]

[In]

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

[Out]

1/3*I*x^3 + 2*I*arctan(x*e^(-I*a))*e^(3*I*a) - 2*I*x*e^(2*I*a)

Mupad [B] (verification not implemented)

Time = 26.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int x^2 \tan (a+i \log (x)) \, dx={\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\right )}^{3/2}\,\mathrm {atan}\left (\frac {x}{\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}}}}\right )\,2{}\mathrm {i}+\frac {x^3\,1{}\mathrm {i}}{3}-x\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,2{}\mathrm {i} \]

[In]

int(x^2*tan(a + log(x)*1i),x)

[Out]

exp(a*2i)^(3/2)*atan(x/exp(a*2i)^(1/2))*2i + (x^3*1i)/3 - x*exp(a*2i)*2i