\(\int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx\) [126]

Optimal result

Integrand size = 14, antiderivative size = 111 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=-\frac {(1-i x)^{2/3} \sqrt [3]{1+i x}}{x}+\frac {2 i \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{\sqrt {3}}+i \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )-\frac {1}{3} i \log (x) \]

[Out]

-(1-I*x)^(2/3)*(1+I*x)^(1/3)/x+I*ln((1-I*x)^(1/3)-(1+I*x)^(1/3))-1/3*I*ln(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 = 0.02 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5170, 96, 93} \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\frac {2 i \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{\sqrt {3}}-\frac {(1-i x)^{2/3} \sqrt [3]{1+i x}}{x}+i \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )-\frac {1}{3} i \log (x) \]

[In]

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

[Out]

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

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 96

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

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=-\frac {i (1-i x)^{2/3} \left (-i+x+x \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {i+x}{i-x}\right )\right )}{(1+i x)^{2/3} x} \]

[In]

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

[Out]

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

Maple [F]

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

[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 = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\frac {{\left (\sqrt {3} x - i \, x\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) - {\left (\sqrt {3} x + i \, x\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 2 i \, x \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - 1\right ) - 3 \, {\left (-i \, x + 1\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}}}{3 \, x} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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