\(\int e^{\frac {2}{3} i \arctan (x)} x \, dx\) [123]

Optimal result

Integrand size = 12, antiderivative size = 140 \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\frac {1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}}-\frac {1}{3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{9} \log (1+i x) \]

[Out]

1/3*(1-I*x)^(2/3)*(1+I*x)^(1/3)+1/2*(1-I*x)^(2/3)*(1+I*x)^(4/3)-1/3*ln(1+(1-I*x)^(1/3)/(1+I*x)^(1/3))-1/9*ln(1
+I*x)-2/9*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 = 0.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5170, 81, 52, 62} \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}}+\frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}+\frac {1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac {1}{3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{9} \log (1+i x) \]

[In]

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

[Out]

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

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 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}{\sqrt [3]{1-i x}} \, dx \\ & = \frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {1}{3} i \int \frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}} \, dx \\ & = \frac {1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {2}{9} i \int \frac {1}{\sqrt [3]{1-i x} (1+i x)^{2/3}} \, dx \\ & = \frac {1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}}-\frac {1}{3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{9} \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 = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.39 \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=\frac {1}{2} (1-i x)^{2/3} \left ((1+i x)^{4/3}+\sqrt [3]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {1}{2}-\frac {i x}{2}\right )\right ) \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int e^{\frac {2}{3} i \arctan (x)} x \, dx=-\frac {1}{9} \, {\left (i \, \sqrt {3} - 1\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) - \frac {1}{9} \, {\left (-i \, \sqrt {3} - 1\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + \frac {1}{6} \, {\left (3 \, x^{2} - 2 i \, x + 5\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {2}{9} \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + 1\right ) \]

[In]

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

[Out]

-1/9*(I*sqrt(3) - 1)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1/2*I*sqrt(3) - 1/2) - 1/9*(-I*sqrt(3) - 1)*log((I*
sqrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*sqrt(3) - 1/2) + 1/6*(3*x^2 - 2*I*x + 5)*(I*sqrt(x^2 + 1)/(x + I))^(2/3)
- 2/9*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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