\(\int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx\) [125]

Optimal result

Integrand size = 14, antiderivative size = 163 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx=\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {1}{2} \log (1+i x)-\frac {\log (x)}{2} \]

[Out]

3/2*ln(1+(1-I*x)^(1/3)/(1+I*x)^(1/3))+3/2*ln((1-I*x)^(1/3)-(1+I*x)^(1/3))+1/2*ln(1+I*x)-1/2*ln(x)+arctan(1/3*3
^(1/2)-2/3*(1-I*x)^(1/3)/(1+I*x)^(1/3)*3^(1/2))*3^(1/2)+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 = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5170, 132, 62, 93} \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx=\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {1}{2} \log (1+i x)-\frac {\log (x)}{2} \]

[In]

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

[Out]

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

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 93

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

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[b*d^(m
+ n)*f^p, Int[(a + b*x)^(m - 1)/(c + d*x)^m, x], x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandTo
Sum[(a + b*x)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n,
-1]))

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}}{\sqrt [3]{1-i x} x} \, dx \\ & = i \int \frac {1}{\sqrt [3]{1-i x} (1+i x)^{2/3}} \, dx+\int \frac {1}{\sqrt [3]{1-i x} (1+i x)^{2/3} x} \, dx \\ & = \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {1}{2} \log (1+i x)-\frac {\log (x)}{2} \\ \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 = 90, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx=-\frac {3 (1-i x)^{2/3} \left (\sqrt [3]{2} (1+i x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {2}{3},\frac {5}{3},\frac {1}{2}-\frac {i x}{2}\right )+2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {i+x}{i-x}\right )\right )}{4 (1+i x)^{2/3}} \]

[In]

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

[Out]

(-3*(1 - I*x)^(2/3)*(2^(1/3)*(1 + I*x)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, 1/2 - (I/2)*x] + 2*Hypergeometri
c2F1[2/3, 1, 5/3, (I + x)/(I - x)]))/(4*(1 + I*x)^(2/3))

Maple [F]

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

[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 = 145, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx=\frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \log \left (\frac {\sqrt {3} {\left (i \, x - 1\right )} + x + 2 i \, \sqrt {x^{2} + 1} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i}{2 \, {\left (x + i\right )}}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \log \left (\frac {\sqrt {3} {\left (-i \, x + 1\right )} + x + 2 i \, \sqrt {x^{2} + 1} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i}{2 \, {\left (x + i\right )}}\right ) + \log \left (-\frac {x - i \, \sqrt {x^{2} + 1} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i}{x + i}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((I*(x - I)/sqrt(x**2 + 1))**(2/3)/x, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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