\(\int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx\) [140]

Optimal result

Integrand size = 14, antiderivative size = 163 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx=\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1-i x}}{\sqrt {3} \sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )+\frac {3}{2} \log \left (\sqrt [3]{1-i x}-\sqrt [3]{1+i x}\right )+\frac {1}{2} \log (1+i x)-\frac {\log (x)}{2} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx=-\frac {3 (1-i x)^{2/3} \left (\sqrt [3]{2} (1+i x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {2}{3},\frac {5}{3},\frac {1}{2}-\frac {i x}{2}\right )+2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {i+x}{i-x}\right )\right )}{4 (1+i x)^{2/3}} \] Input:

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

Output:

(-3*(1 - I*x)^(2/3)*(2^(1/3)*(1 + I*x)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5 
/3, 1/2 - (I/2)*x] + 2*Hypergeometric2F1[2/3, 1, 5/3, (I + x)/(I - x)]))/( 
4*(1 + I*x)^(2/3))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5585, 140, 72, 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.

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

\(\Big \downarrow \) 5585

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

\(\Big \downarrow \) 140

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

\(\Big \downarrow \) 72

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

\(\Big \downarrow \) 102

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 72

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> 
 With[{q = Rt[-d/b, 3]}, Simp[Sqrt[3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b* 
x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a + b* 
x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; F 
reeQ[{a, b, c, d}, x] && NegQ[d/b]
 

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 140

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*d^(m + n)*f^p   Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] 
, x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x 
)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 
0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n, -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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {2}{3} i \arctan (x)}}{x} \, dx=\frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \log \left (\frac {\sqrt {3} {\left (i \, x - 1\right )} + x + 2 i \, \sqrt {x^{2} + 1} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i}{2 \, {\left (x + i\right )}}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \log \left (\frac {\sqrt {3} {\left (-i \, x + 1\right )} + x + 2 i \, \sqrt {x^{2} + 1} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i}{2 \, {\left (x + i\right )}}\right ) + \log \left (-\frac {x - i \, \sqrt {x^{2} + 1} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i}{x + i}\right ) \] Input:

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

Output:

1/2*(I*sqrt(3) - 1)*log(1/2*(sqrt(3)*(I*x - 1) + x + 2*I*sqrt(x^2 + 1)*(I* 
sqrt(x^2 + 1)/(x + I))^(1/3) + I)/(x + I)) + 1/2*(-I*sqrt(3) - 1)*log(1/2* 
(sqrt(3)*(-I*x + 1) + x + 2*I*sqrt(x^2 + 1)*(I*sqrt(x^2 + 1)/(x + I))^(1/3 
) + I)/(x + I)) + log(-(x - I*sqrt(x^2 + 1)*(I*sqrt(x^2 + 1)/(x + I))^(1/3 
) + I)/(x + I))
 

Sympy [F]

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

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

Output:

Integral((I*(x - I)/sqrt(x**2 + 1))**(2/3)/x, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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