\(\int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx\) [127]

Optimal result

Integrand size = 14, antiderivative size = 142 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx=-\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}-\frac {i (1-i x)^{2/3} \sqrt [3]{1+i x}}{3 x}-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}}-\frac {1}{3} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {\log (x)}{9} \]

[Out]

-1/2*(1-I*x)^(2/3)*(1+I*x)^(4/3)/x^2-1/3*I*(1-I*x)^(2/3)*(1+I*x)^(1/3)/x-1/3*ln((1-I*x)^(1/3)-(1+I*x)^(1/3))+1
/9*ln(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 = 142, 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, 98, 96, 93} \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx=-\frac {2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}}-\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}-\frac {i (1-i x)^{2/3} \sqrt [3]{1+i x}}{3 x}-\frac {1}{3} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {\log (x)}{9} \]

[In]

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

[Out]

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

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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[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^3} \, dx \\ & = -\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}+\frac {1}{3} i \int \frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x} x^2} \, dx \\ & = -\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}-\frac {i (1-i x)^{2/3} \sqrt [3]{1+i x}}{3 x}-\frac {2}{9} \int \frac {1}{\sqrt [3]{1-i x} (1+i x)^{2/3} x} \, dx \\ & = -\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}-\frac {i (1-i x)^{2/3} \sqrt [3]{1+i x}}{3 x}-\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 (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {\log (x)}{9} \\ \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 = 69, normalized size of antiderivative = 0.49 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx=\frac {(1-i x)^{2/3} \left (-3-8 i x+5 x^2+2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {i+x}{i-x}\right )\right )}{6 (1+i x)^{2/3} x^2} \]

[In]

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

[Out]

((1 - I*x)^(2/3)*(-3 - (8*I)*x + 5*x^2 + 2*x^2*Hypergeometric2F1[2/3, 1, 5/3, (I + x)/(I - x)]))/(6*(1 + I*x)^
(2/3)*x^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx=-\frac {4 \, x^{2} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - 1\right ) + 2 \, {\left (-i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 2 \, {\left (i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 3 \, {\left (5 \, x^{2} + 2 i \, x + 3\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}}}{18 \, x^{2}} \]

[In]

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

[Out]

-1/18*(4*x^2*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) - 1) + 2*(-I*sqrt(3)*x^2 - x^2)*log((I*sqrt(x^2 + 1)/(x + I))
^(2/3) + 1/2*I*sqrt(3) + 1/2) + 2*(I*sqrt(3)*x^2 - x^2)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*sqrt(3) +
1/2) + 3*(5*x^2 + 2*I*x + 3)*(I*sqrt(x^2 + 1)/(x + I))^(2/3))/x^2

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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