\(\int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx\) [122]

Optimal result

Integrand size = 14, antiderivative size = 177 \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=-\frac {11}{27} i (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}+\frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x+\frac {22 i \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{27 \sqrt {3}}+\frac {11}{27} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {11}{81} i \log (1+i x) \]

[Out]

-11/27*I*(1-I*x)^(2/3)*(1+I*x)^(1/3)-1/9*I*(1-I*x)^(2/3)*(1+I*x)^(4/3)+1/3*(1-I*x)^(2/3)*(1+I*x)^(4/3)*x+11/27
*I*ln(1+(1-I*x)^(1/3)/(1+I*x)^(1/3))+11/81*I*ln(1+I*x)+22/81*I*arctan(1/3*3^(1/2)-2/3*(1-I*x)^(1/3)/(1+I*x)^(1
/3)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5170, 92, 81, 52, 62} \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\frac {22 i \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{27 \sqrt {3}}+\frac {1}{3} (1-i x)^{2/3} x (1+i x)^{4/3}-\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}-\frac {11}{27} i (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {11}{27} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {11}{81} i \log (1+i x) \]

[In]

Int[E^(((2*I)/3)*ArcTan[x])*x^2,x]

[Out]

((-11*I)/27)*(1 - I*x)^(2/3)*(1 + I*x)^(1/3) - (I/9)*(1 - I*x)^(2/3)*(1 + I*x)^(4/3) + ((1 - I*x)^(2/3)*(1 + I
*x)^(4/3)*x)/3 + (((22*I)/27)*ArcTan[1/Sqrt[3] - (2*(1 - I*x)^(1/3))/(Sqrt[3]*(1 + I*x)^(1/3))])/Sqrt[3] + ((1
1*I)/27)*Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3)] + ((11*I)/81)*Log[1 + I*x]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 62

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-d/b, 3]}, Simp[Sqrt[
3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a
+ b*x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && NegQ[d/b]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 5170

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt [3]{1+i x} x^2}{\sqrt [3]{1-i x}} \, dx \\ & = \frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x+\frac {1}{3} \int \frac {\left (-1-\frac {2 i x}{3}\right ) \sqrt [3]{1+i x}}{\sqrt [3]{1-i x}} \, dx \\ & = -\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}+\frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x-\frac {11}{27} \int \frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}} \, dx \\ & = -\frac {11}{27} i (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}+\frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x-\frac {22}{81} \int \frac {1}{\sqrt [3]{1-i x} (1+i x)^{2/3}} \, dx \\ & = -\frac {11}{27} i (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac {1}{9} i (1-i x)^{2/3} (1+i x)^{4/3}+\frac {1}{3} (1-i x)^{2/3} (1+i x)^{4/3} x+\frac {22 i \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{27 \sqrt {3}}+\frac {11}{27} i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {11}{81} i \log (1+i x) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.41 \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\frac {1}{18} (1-i x)^{2/3} \left (2 \sqrt [3]{1+i x} \left (-i+4 x+3 i x^2\right )-11 i \sqrt [3]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {1}{2}-\frac {i x}{2}\right )\right ) \]

[In]

Integrate[E^(((2*I)/3)*ArcTan[x])*x^2,x]

[Out]

((1 - I*x)^(2/3)*(2*(1 + I*x)^(1/3)*(-I + 4*x + (3*I)*x^2) - (11*I)*2^(1/3)*Hypergeometric2F1[-1/3, 2/3, 5/3,
1/2 - (I/2)*x]))/18

Maple [F]

\[\int {\left (\frac {i x +1}{\sqrt {x^{2}+1}}\right )}^{\frac {2}{3}} x^{2}d x\]

[In]

int(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x^2,x)

[Out]

int(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.66 \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=-\frac {11}{81} \, {\left (\sqrt {3} + i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {11}{81} \, {\left (\sqrt {3} - i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {1}{27} \, {\left (9 \, x^{3} - 3 i \, x^{2} - 2 \, x - 14 i\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {22}{81} i \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + 1\right ) \]

[In]

integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x^2,x, algorithm="fricas")

[Out]

-11/81*(sqrt(3) + I)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1/2*I*sqrt(3) - 1/2) + 11/81*(sqrt(3) - I)*log((I*s
qrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*sqrt(3) - 1/2) + 1/27*(9*x^3 - 3*I*x^2 - 2*x - 14*I)*(I*sqrt(x^2 + 1)/(x +
I))^(2/3) + 22/81*I*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1)

Sympy [F(-1)]

Timed out. \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\text {Timed out} \]

[In]

integrate(((1+I*x)/(x**2+1)**(1/2))**(2/3)*x**2,x)

[Out]

Timed out

Maxima [F]

\[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\int { x^{2} \left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}} \,d x } \]

[In]

integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x^2,x, algorithm="maxima")

[Out]

integrate(x^2*((I*x + 1)/sqrt(x^2 + 1))^(2/3), x)

Giac [F]

\[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\int { x^{2} \left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}} \,d x } \]

[In]

integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)*x^2,x, algorithm="giac")

[Out]

integrate(x^2*((I*x + 1)/sqrt(x^2 + 1))^(2/3), x)

Mupad [F(-1)]

Timed out. \[ \int e^{\frac {2}{3} i \arctan (x)} x^2 \, dx=\int x^2\,{\left (\frac {1+x\,1{}\mathrm {i}}{\sqrt {x^2+1}}\right )}^{2/3} \,d x \]

[In]

int(x^2*((x*1i + 1)/(x^2 + 1)^(1/2))^(2/3),x)

[Out]

int(x^2*((x*1i + 1)/(x^2 + 1)^(1/2))^(2/3), x)