Integrand size = 14, antiderivative size = 253 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx=-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}+\frac {i \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {i \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} i \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} i \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac {1}{6} i \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \]
-(1-I*x)^(5/6)*(1+I*x)^(1/6)/x-2/3*I*arctanh((1+I*x)^(1/6)/(1-I*x)^(1/6))+ 1/6*I*ln(1-(1+I*x)^(1/6)/(1-I*x)^(1/6)+(1+I*x)^(1/3)/(1-I*x)^(1/3))-1/6*I* ln(1+(1+I*x)^(1/6)/(1-I*x)^(1/6)+(1+I*x)^(1/3)/(1-I*x)^(1/3))+1/3*I*arctan (1/3*(1-2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))*3^(1/2)-1/3*I*arctan(1/3*( 1+2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))*3^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.25 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx=-\frac {i (1-i x)^{5/6} \left (-5 i+5 x+2 x \operatorname {Hypergeometric2F1}\left (\frac {5}{6},1,\frac {11}{6},\frac {i+x}{i-x}\right )\right )}{5 (1+i x)^{5/6} x} \]
((-1/5*I)*(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)
Time = 0.31 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {5585, 105, 104, 754, 27, 219, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx\) |
| \(\Big \downarrow \) 5585 |
| \(\displaystyle \int \frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x} x^2}dx\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle \frac {1}{3} i \int \frac {1}{\sqrt [6]{1-i x} (i x+1)^{5/6} x}dx-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle 2 i \int \frac {1}{\frac {i x+1}{1-i x}-1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 754 |
| \(\displaystyle 2 i \left (-\frac {1}{3} \int \frac {1}{1-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{2 \left (\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{2 \left (\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 i \left (-\frac {1}{3} \int \frac {1}{1-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle 2 i \left (-\frac {1}{6} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle 2 i \left (\frac {1}{6} \left (\frac {1}{2} \int -\frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 i \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle 2 i \left (\frac {1}{6} \left (3 \int \frac {1}{-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-3}d\left (\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-1\right )-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (3 \int \frac {1}{-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-3}d\left (\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle 2 i \left (\frac {1}{6} \left (-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\sqrt {3} \arctan \left (\frac {-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle 2 i \left (\frac {1}{6} \left (\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\sqrt {3} \arctan \left (\frac {-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\) |
-(((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/x) + (2*I)*(-1/3*ArcTanh[(1 + I*x)^(1/ 6)/(1 - I*x)^(1/6)] + (-(Sqrt[3]*ArcTan[(-1 + (2*(1 + I*x)^(1/6))/(1 - I*x )^(1/6))/Sqrt[3]]) + Log[1 - (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^( 1/3)/(1 - I*x)^(1/3)]/2)/6 + (-(Sqrt[3]*ArcTan[(1 + (2*(1 + I*x)^(1/6))/(1 - I*x)^(1/6))/Sqrt[3]]) - Log[1 + (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]/2)/6)
3.2.19.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[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] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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/(a*n)) Int[1/(r^2 - s^2*x^2), x] + 2*(r/(a*n)) Sum[u, {k, 1, (n - 2)/4}], x]] / ; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
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] && !Intege rQ[(I*n - 1)/2]
\[\int \frac {{\left (\frac {i x +1}{\sqrt {x^{2}+1}}\right )}^{\frac {1}{3}}}{x^{2}}d x\]
Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx=\frac {{\left (\sqrt {3} x - i \, x\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + {\left (\sqrt {3} x + i \, x\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) - {\left (\sqrt {3} x + i \, x\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) - {\left (\sqrt {3} x - i \, x\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) - 2 i \, x \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + 1\right ) + 2 i \, x \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - 1\right ) - 6 \, {\left (-i \, x + 1\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}}}{6 \, x} \]
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*log((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*sqrt(x^ 2 + 1)/(x + I))^(1/3))/x
\[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx=\int \frac {\sqrt [3]{\frac {i \left (x - i\right )}{\sqrt {x^{2} + 1}}}}{x^{2}}\, dx \]
\[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {1}{3}}}{x^{2}} \,d x } \]
\[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {1}{3}}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^2} \, dx=\int \frac {{\left (\frac {1+x\,1{}\mathrm {i}}{\sqrt {x^2+1}}\right )}^{1/3}}{x^2} \,d x \]