\(\int e^{\frac {2}{3} i \arctan (x)} \, dx\) [124]

Optimal result

Integrand size = 10, antiderivative size = 116 \[ \int e^{\frac {2}{3} i \arctan (x)} \, dx=i (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac {2 i \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{\sqrt {3}}-i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{3} i \log (1+i x) \]

[Out]

I*(1-I*x)^(2/3)*(1+I*x)^(1/3)-I*ln(1+(1-I*x)^(1/3)/(1+I*x)^(1/3))-1/3*I*ln(1+I*x)-2/3*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 = 0.01 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5169, 52, 62} \[ \int e^{\frac {2}{3} i \arctan (x)} \, dx=-\frac {2 i \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{\sqrt {3}}+i (1-i x)^{2/3} \sqrt [3]{1+i x}-i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{3} i \log (1+i x) \]

[In]

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

[Out]

I*(1 - I*x)^(2/3)*(1 + I*x)^(1/3) - ((2*I)*ArcTan[1/Sqrt[3] - (2*(1 - I*x)^(1/3))/(Sqrt[3]*(1 + I*x)^(1/3))])/
Sqrt[3] - I*Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3)] - (I/3)*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 5169

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}} \, dx \\ & = i (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac {2}{3} \int \frac {1}{\sqrt [3]{1-i x} (1+i x)^{2/3}} \, dx \\ & = i (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac {2 i \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{\sqrt {3}}-i \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac {1}{3} 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 = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29 \[ \int e^{\frac {2}{3} i \arctan (x)} \, dx=-\frac {3}{2} i e^{\frac {8}{3} i \arctan (x)} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},2,\frac {7}{3},-e^{2 i \arctan (x)}\right ) \]

[In]

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

[Out]

((-3*I)/2)*E^(((8*I)/3)*ArcTan[x])*Hypergeometric2F1[4/3, 2, 7/3, -E^((2*I)*ArcTan[x])]

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int e^{\frac {2}{3} i \arctan (x)} \, dx=\frac {1}{3} \, {\left (\sqrt {3} + i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) - \frac {1}{3} \, {\left (\sqrt {3} - i\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} - \frac {1}{2}\right ) + {\left (x + i\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {2}{3} i \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + 1\right ) \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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