\(\int e^{\frac {1}{3} i \arctan (x)} \, dx\) [132]

Optimal result

Integrand size = 10, antiderivative size = 196 \[ \int e^{\frac {1}{3} i \arctan (x)} \, dx=i (1-i x)^{5/6} \sqrt [6]{1+i x}-\frac {1}{3} i \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{3} i \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {2}{3} i \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\frac {i \text {arctanh}\left (\frac {\sqrt {3} \sqrt [6]{1-i x}}{\left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right ) \sqrt [6]{1+i x}}\right )}{\sqrt {3}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.17 \[ \int e^{\frac {1}{3} i \arctan (x)} \, dx=-\frac {12}{7} i e^{\frac {7}{3} i \arctan (x)} \operatorname {Hypergeometric2F1}\left (\frac {7}{6},2,\frac {13}{6},-e^{2 i \arctan (x)}\right ) \] Input:

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

Output:

((-12*I)/7)*E^(((7*I)/3)*ArcTan[x])*Hypergeometric2F1[7/6, 2, 13/6, -E^((2 
*I)*ArcTan[x])]
 

Rubi [A] (warning: unable to verify)

Time = 0.59 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.21, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {5584, 60, 73, 854, 824, 27, 216, 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.

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

\(\Big \downarrow \) 5584

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 824

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

I*(1 - I*x)^(5/6)*(1 + I*x)^(1/6) + (2*I)*(ArcTan[(1 - I*x)^(1/6)/(1 + I*x 
)^(1/6)]/3 + (-ArcTan[Sqrt[3] - (2*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)] + (Sq 
rt[3]*Log[1 + (1 - I*x)^(1/3) - (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)] 
)/2)/6 + (ArcTan[Sqrt[3] + (2*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)] - (Sqrt[3] 
*Log[1 + (1 - I*x)^(1/3) + (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)])/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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 824

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator 
[Rt[a/b, n]], s = Denominator[Rt[a/b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k 
- 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 
 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k 
- 1)*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] 
; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))   Int[1/(r^2 + s^2*x^2), x] + 2*(r^(m 
+ 1)/(a*n*s^m))   Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGt 
Q[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]
 

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

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 5584

Int[E^(ArcTan[(a_.)*(x_)]*(n_.)), x_Symbol] :> Int[(1 - I*a*x)^(I*(n/2))/(1 
 + I*a*x)^(I*(n/2)), x] /; FreeQ[{a, n}, x] &&  !IntegerQ[(I*n - 1)/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.99 \[ \int e^{\frac {1}{3} i \arctan (x)} \, dx=\frac {1}{6} \, {\left (-i \, \sqrt {3} + 1\right )} \log \left (\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + \frac {1}{6} \, {\left (-i \, \sqrt {3} - 1\right )} \log \left (\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + \frac {1}{6} \, {\left (i \, \sqrt {3} + 1\right )} \log \left (-\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + \frac {1}{6} \, {\left (i \, \sqrt {3} - 1\right )} \log \left (-\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + {\left (x + i\right )} \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{3} \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i\right ) - \frac {1}{3} \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - i\right ) \] Input:

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

Output:

1/6*(-I*sqrt(3) + 1)*log(1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1 
/2*I) + 1/6*(-I*sqrt(3) - 1)*log(1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^( 
1/3) - 1/2*I) + 1/6*(I*sqrt(3) + 1)*log(-1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x 
 + I))^(1/3) + 1/2*I) + 1/6*(I*sqrt(3) - 1)*log(-1/2*sqrt(3) + (I*sqrt(x^2 
 + 1)/(x + I))^(1/3) - 1/2*I) + (x + I)*(I*sqrt(x^2 + 1)/(x + I))^(1/3) + 
1/3*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) + I) - 1/3*log((I*sqrt(x^2 + 1)/(x 
 + I))^(1/3) - I)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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