3.127 \(\int \frac {e^{\frac {2}{3} i \tan ^{-1}(x)}}{x^3} \, dx\)

Optimal. Leaf size=142 \[ -\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}-\frac {i (1-i x)^{2/3} \sqrt [3]{1+i x}}{3 x}-\frac {1}{3} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {\log (x)}{9}-\frac {2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}} \]

[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]  time = 0.04, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5062, 96, 94, 91} \[ -\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}-\frac {i (1-i x)^{2/3} \sqrt [3]{1+i x}}{3 x}-\frac {1}{3} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {\log (x)}{9}-\frac {2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - I*x)^(2/3)*(1 + I*x)^(4/3))/(2*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 91

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 94

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[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

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 5062

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*} \int \frac {e^{\frac {2}{3} i \tan ^{-1}(x)}}{x^3} \, dx &=\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 \tan ^{-1}\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]  time = 0.02, size = 69, normalized size = 0.49 \[ \frac {(1-i x)^{2/3} \left (2 x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {x+i}{i-x}\right )+5 x^2-8 i x-3\right )}{6 (1+i x)^{2/3} x^2} \]

Warning: Unable to verify antiderivative.

[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)

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 138, normalized size = 0.97 \[ -\frac {4 \, x^{2} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - 1\right ) + 2 \, {\left (-i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 2 \, {\left (i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 3 \, {\left (5 \, x^{2} + 2 i \, x + 3\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}}}{18 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i x +1}{\sqrt {x^{2}+1}}\right )^{\frac {2}{3}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {1+x\,1{}\mathrm {i}}{\sqrt {x^2+1}}\right )}^{2/3}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________