Integrand size = 14, antiderivative size = 111 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=-\frac {(1-i x)^{2/3} \sqrt [3]{1+i x}}{x}+\frac {2 i \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )}{\sqrt {3}}+i \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )-\frac {1}{3} i \log (x) \] Output:
-(1-I*x)^(2/3)*(1+I*x)^(1/3)/x+2/3*I*arctan(1/3*3^(1/2)+2/3*(1-I*x)^(1/3)* 3^(1/2)/(1+I*x)^(1/3))*3^(1/2)+I*ln((1-I*x)^(1/3)-(1+I*x)^(1/3))-1/3*I*ln( x)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=-\frac {i (1-i x)^{2/3} \left (-i+x+x \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {i+x}{i-x}\right )\right )}{(1+i x)^{2/3} x} \] Input:
Integrate[E^(((2*I)/3)*ArcTan[x])/x^2,x]
Output:
((-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)
Time = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5585, 105, 102}
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.
\(\displaystyle \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 5585 |
\(\displaystyle \int \frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x} x^2}dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2}{3} i \int \frac {1}{\sqrt [3]{1-i x} (i x+1)^{2/3} x}dx-\frac {(1-i x)^{2/3} \sqrt [3]{1+i x}}{x}\) |
\(\Big \downarrow \) 102 |
\(\displaystyle \frac {2}{3} i \left (\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )-\frac {\log (x)}{2}\right )-\frac {(1-i x)^{2/3} \sqrt [3]{1+i x}}{x}\) |
Input:
Int[E^(((2*I)/3)*ArcTan[x])/x^2,x]
Output:
-(((1 - I*x)^(2/3)*(1 + I*x)^(1/3))/x) + ((2*I)/3)*(Sqrt[3]*ArcTan[1/Sqrt[ 3] + (2*(1 - I*x)^(1/3))/(Sqrt[3]*(1 + I*x)^(1/3))] + (3*Log[(1 - I*x)^(1/ 3) - (1 + I*x)^(1/3)])/2 - Log[x]/2)
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.) *(x_))), x_] :> 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]
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[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]
\[\int \frac {{\left (\frac {i x +1}{\sqrt {x^{2}+1}}\right )}^{\frac {2}{3}}}{x^{2}}d x\]
Input:
int(((1+I*x)/(x^2+1)^(1/2))^(2/3)/x^2,x)
Output:
int(((1+I*x)/(x^2+1)^(1/2))^(2/3)/x^2,x)
Time = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\frac {{\left (\sqrt {3} x - i \, x\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) - {\left (\sqrt {3} x + i \, x\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 2 i \, x \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - 1\right ) - 3 \, {\left (-i \, x + 1\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}}}{3 \, x} \] Input:
integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)/x^2,x, algorithm="fricas")
Output:
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*sqrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*sq rt(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
\[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\int \frac {\left (\frac {i \left (x - i\right )}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}}}{x^{2}}\, dx \] Input:
integrate(((1+I*x)/(x**2+1)**(1/2))**(2/3)/x**2,x)
Output:
Integral((I*(x - I)/sqrt(x**2 + 1))**(2/3)/x**2, x)
\[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}}}{x^{2}} \,d x } \] Input:
integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)/x^2,x, algorithm="maxima")
Output:
integrate(((I*x + 1)/sqrt(x^2 + 1))^(2/3)/x^2, x)
\[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, x + 1}{\sqrt {x^{2} + 1}}\right )^{\frac {2}{3}}}{x^{2}} \,d x } \] Input:
integrate(((1+I*x)/(x^2+1)^(1/2))^(2/3)/x^2,x, algorithm="giac")
Output:
integrate(((I*x + 1)/sqrt(x^2 + 1))^(2/3)/x^2, x)
Timed out. \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\int \frac {{\left (\frac {1+x\,1{}\mathrm {i}}{\sqrt {x^2+1}}\right )}^{2/3}}{x^2} \,d x \] Input:
int(((x*1i + 1)/(x^2 + 1)^(1/2))^(2/3)/x^2,x)
Output:
int(((x*1i + 1)/(x^2 + 1)^(1/2))^(2/3)/x^2, x)
\[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^2} \, dx=\int \frac {\left (i x +1\right )^{\frac {2}{3}}}{\left (x^{2}+1\right )^{\frac {1}{3}} x^{2}}d x \] Input:
int(((1+I*x)/(x^2+1)^(1/2))^(2/3)/x^2,x)
Output:
int((i*x + 1)**(2/3)/((x**2 + 1)**(1/3)*x**2),x)