\(\int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx\) [142]

Optimal result

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

-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-2 
/9*3^(1/2)*arctan(1/3*3^(1/2)+2/3*(1-I*x)^(1/3)*3^(1/2)/(1+I*x)^(1/3))-1/3 
*ln((1-I*x)^(1/3)-(1+I*x)^(1/3))+1/9*ln(x)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.49 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx=\frac {(1-i x)^{2/3} \left (-3-8 i x+5 x^2+2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {i+x}{i-x}\right )\right )}{6 (1+i x)^{2/3} x^2} \] Input:

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

Output:

((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)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5585, 107, 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.

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

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x} x^3}dx\)

\(\Big \downarrow \) 107

\(\displaystyle \frac {1}{3} i \int \frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x} x^2}dx-\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {1}{3} i \left (\frac {2}{3} i \int \frac {1}{\sqrt [3]{1-i x} (i x+1)^{2/3} x}dx-\frac {(1-i x)^{2/3} \sqrt [3]{1+i x}}{x}\right )-\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}\)

\(\Big \downarrow \) 102

\(\displaystyle \frac {1}{3} i \left (\frac {2}{3} i \left (\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )-\frac {\log (x)}{2}\right )-\frac {(1-i x)^{2/3} \sqrt [3]{1+i x}}{x}\right )-\frac {(1-i x)^{2/3} (1+i x)^{4/3}}{2 x^2}\)

Input:

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

Output:

-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*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))
 

Defintions of rubi rules used

rule 102

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]
 

rule 105

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]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> 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] + Simp[(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] || SumSimplerQ[m, 1])
 

rule 5585

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx=-\frac {4 \, x^{2} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - 1\right ) + 2 \, {\left (-i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} + \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 2 \, {\left (i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}} - \frac {1}{2} i \, \sqrt {3} + \frac {1}{2}\right ) + 3 \, {\left (5 \, x^{2} + 2 i \, x + 3\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {2}{3}}}{18 \, x^{2}} \] Input:

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

Output:

-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*s 
qrt(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. \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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