\(\int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx\) [135]

Optimal result

Integrand size = 14, antiderivative size = 228 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}-\frac {i (1-i x)^{5/6} \sqrt [6]{1+i x}}{6 x}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {\arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )}{6 \sqrt {3}}+\frac {1}{9} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {1}{18} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\left (1+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \sqrt [6]{1-i x}}\right ) \] Output:

-1/2*(1-I*x)^(5/6)*(1+I*x)^(7/6)/x^2-1/6*I*(1-I*x)^(5/6)*(1+I*x)^(1/6)/x-1 
/18*3^(1/2)*arctan(1/3*(1-2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))+1/18*3^( 
1/2)*arctan(1/3*(1+2*(1+I*x)^(1/6)/(1-I*x)^(1/6))*3^(1/2))+1/9*arctanh((1+ 
I*x)^(1/6)/(1-I*x)^(1/6))+1/18*arctanh((1+I*x)^(1/6)/(1+(1+I*x)^(1/3)/(1-I 
*x)^(1/3))/(1-I*x)^(1/6))
 

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.32 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=\frac {(1-i x)^{5/6} \left (5 \left (-3-7 i x+4 x^2\right )+2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{6},1,\frac {11}{6},\frac {i+x}{i-x}\right )\right )}{30 (1+i x)^{5/6} x^2} \] Input:

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

Output:

((1 - I*x)^(5/6)*(5*(-3 - (7*I)*x + 4*x^2) + 2*x^2*Hypergeometric2F1[5/6, 
1, 11/6, (I + x)/(I - x)]))/(30*(1 + I*x)^(5/6)*x^2)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.29, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {5585, 107, 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.

\(\displaystyle \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx\)

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x} x^3}dx\)

\(\Big \downarrow \) 107

\(\displaystyle \frac {1}{6} i \int \frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x} x^2}dx-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {1}{6} i \left (\frac {1}{3} i \int \frac {1}{\sqrt [6]{1-i x} (i x+1)^{5/6} x}dx-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{6} i \left (2 i \int \frac {1}{\frac {i x+1}{1-i x}-1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 754

\(\displaystyle \frac {1}{6} i \left (2 i \left (-\frac {1}{3} \int \frac {1}{1-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{2 \left (\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{2 \left (\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} i \left (2 i \left (-\frac {1}{3} \int \frac {1}{1-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{6} i \left (2 i \left (-\frac {1}{6} \int \frac {2-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{6} \int \frac {\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+2}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{6} i \left (2 i \left (\frac {1}{6} \left (\frac {1}{2} \int -\frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{6} i \left (2 i \left (\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (-\frac {3}{2} \int \frac {1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{6} i \left (2 i \left (\frac {1}{6} \left (3 \int \frac {1}{-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-3}d\left (\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-1\right )-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )+\frac {1}{6} \left (3 \int \frac {1}{-\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-3}d\left (\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1\right )-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{6} i \left (2 i \left (\frac {1}{6} \left (-\frac {1}{2} \int \frac {1-\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\sqrt {3} \arctan \left (\frac {-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\frac {1}{2} \int \frac {\frac {2 \sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}{\frac {\sqrt [3]{i x+1}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}+1}d\frac {\sqrt [6]{i x+1}}{\sqrt [6]{1-i x}}-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{6} i \left (2 i \left (\frac {1}{6} \left (\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\sqrt {3} \arctan \left (\frac {-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )\right )+\frac {1}{6} \left (-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )\right )-\frac {1}{3} \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-\frac {(1-i x)^{5/6} \sqrt [6]{1+i x}}{x}\right )-\frac {(1-i x)^{5/6} (1+i x)^{7/6}}{2 x^2}\)

Input:

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

Output:

-1/2*((1 - I*x)^(5/6)*(1 + I*x)^(7/6))/x^2 + (I/6)*(-(((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]]) + L 
og[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))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 104

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]
 

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 217

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])
 

rule 219

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])
 

rule 754

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]
 

rule 1083

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]
 

rule 1103

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]
 

rule 1142

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]
 

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=\frac {2 \, x^{2} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + 1\right ) - 2 \, x^{2} \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - 1\right ) + {\left (i \, \sqrt {3} x^{2} + x^{2}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + {\left (i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) + {\left (-i \, \sqrt {3} x^{2} + x^{2}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + {\left (-i \, \sqrt {3} x^{2} - x^{2}\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) - 6 \, {\left (4 \, x^{2} + i \, x + 3\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}}}{36 \, x^{2}} \] Input:

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

Output:

1/36*(2*x^2*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1) - 2*x^2*log((I*sqrt(x 
^2 + 1)/(x + I))^(1/3) - 1) + (I*sqrt(3)*x^2 + x^2)*log(1/2*I*sqrt(3) + (I 
*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) + (I*sqrt(3)*x^2 - x^2)*log(1/2*I*sqr 
t(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) + (-I*sqrt(3)*x^2 + x^2)*log 
(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2) + (-I*sqrt(3)*x^2 
 - x^2)*log(-1/2*I*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2) - 6*(4 
*x^2 + I*x + 3)*(I*sqrt(x^2 + 1)/(x + I))^(1/3))/x^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x^3} \, dx=\int \frac {\left (i x +1\right )^{\frac {1}{3}}}{\left (x^{2}+1\right )^{\frac {1}{6}} x^{3}}d x \] Input:

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

Output:

int((i*x + 1)**(1/3)/((x**2 + 1)**(1/6)*x**3),x)