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

Optimal. Leaf size=253 \[ -\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}+\frac{1}{6} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\frac{1}{6} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\frac{i \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{i \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{2}{3} i \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \]

[Out]

-(((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x) + (I*ArcTan[(1 - (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]])/Sqrt[3]
- (I*ArcTan[(1 + (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]])/Sqrt[3] - ((2*I)/3)*ArcTanh[(1 + I*x)^(1/6)/(
1 - I*x)^(1/6)] + (I/6)*Log[1 - (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)] - (I/6)*Log
[1 + (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]

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Rubi [A]  time = 0.155645, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {5062, 94, 93, 210, 634, 618, 204, 628, 206} \[ -\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}+\frac{1}{6} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\frac{1}{6} i \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\frac{i \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{i \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{2}{3} i \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x) + (I*ArcTan[(1 - (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]])/Sqrt[3]
- (I*ArcTan[(1 + (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]])/Sqrt[3] - ((2*I)/3)*ArcTanh[(1 + I*x)^(1/6)/(
1 - I*x)^(1/6)] + (I/6)*Log[1 - (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)] - (I/6)*Log
[1 + (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]

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]

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 210

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt[-(a/b
), n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r +
s*Cos[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{3} i \tan ^{-1}(x)}}{x^2} \, dx &=\int \frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x} x^2} \, dx\\ &=-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}+\frac{1}{3} i \int \frac{1}{\sqrt [6]{1-i x} (1+i x)^{5/6} x} \, dx\\ &=-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}+2 i \operatorname{Subst}\left (\int \frac{1}{-1+x^6} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\\ &=-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-\frac{2}{3} i \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac{2}{3} i \operatorname{Subst}\left (\int \frac{1-\frac{x}{2}}{1-x+x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac{2}{3} i \operatorname{Subst}\left (\int \frac{1+\frac{x}{2}}{1+x+x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\\ &=-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-\frac{2}{3} i \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac{1}{6} i \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac{1}{6} i \operatorname{Subst}\left (\int \frac{1+2 x}{1+x+x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\\ &=-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}-\frac{2}{3} i \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac{1}{6} i \log \left (1-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac{1}{6} i \log \left (1+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )+i \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+i \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\\ &=-\frac{(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}+\frac{i \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{i \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{2}{3} i \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac{1}{6} i \log \left (1-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac{1}{6} i \log \left (1+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.013165, size = 64, normalized size = 0.25 \[ -\frac{i (1-i x)^{5/6} \left (2 x \text{Hypergeometric2F1}\left (\frac{5}{6},1,\frac{11}{6},\frac{x+i}{-x+i}\right )+5 x-5 i\right )}{5 (1+i x)^{5/6} x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I/5)*(1 - I*x)^(5/6)*(-5*I + 5*x + 2*x*Hypergeometric2F1[5/6, 1, 11/6, (I + x)/(I - x)]))/((1 + I*x)^(5/6)*
x)

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Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}\sqrt [3]{{(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, x + 1}{\sqrt{x^{2} + 1}}\right )^{\frac{1}{3}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.73545, size = 635, normalized size = 2.51 \begin{align*} \frac{{\left (\sqrt{3} x - i \, x\right )} \log \left (\frac{1}{2} i \, \sqrt{3} + \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} + \frac{1}{2}\right ) +{\left (\sqrt{3} x + i \, x\right )} \log \left (\frac{1}{2} i \, \sqrt{3} + \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} - \frac{1}{2}\right ) -{\left (\sqrt{3} x + i \, x\right )} \log \left (-\frac{1}{2} i \, \sqrt{3} + \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} + \frac{1}{2}\right ) -{\left (\sqrt{3} x - i \, x\right )} \log \left (-\frac{1}{2} i \, \sqrt{3} + \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} - \frac{1}{2}\right ) - 2 i \, x \log \left (\left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} + 1\right ) + 2 i \, x \log \left (\left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} - 1\right ) - 6 \,{\left (-i \, x + 1\right )} \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}}}{6 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((sqrt(3)*x - I*x)*log(1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) + (sqrt(3)*x + I*x)*log(1/2*
I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) - (sqrt(3)*x + I*x)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(
x + I))^(1/3) + 1/2) - (sqrt(3)*x - I*x)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) - 2*I*x*l
og((I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1) + 2*I*x*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1) - 6*(-I*x + 1)*(I*sqr
t(x^2 + 1)/(x + I))^(1/3))/x

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, x + 1}{\sqrt{x^{2} + 1}}\right )^{\frac{1}{3}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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