\(\int \frac {e^{\frac {1}{3} i \arctan (x)}}{x} \, dx\) [118]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [F]
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 14, antiderivative size = 430 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x} \, dx=\arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-2 \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{2} \sqrt {3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{2} \sqrt {3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{2} \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac {1}{2} \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {5170, 132, 65, 338, 301, 648, 632, 210, 642, 209, 95, 216, 212} \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x} \, dx=\arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-2 \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{2} \sqrt {3} \log \left (\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )+\frac {1}{2} \sqrt {3} \log \left (\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )+\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\frac {1}{2} \log \left (\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right ) \]

[In]

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

[Out]

ArcTan[Sqrt[3] - (2*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)] - ArcTan[Sqrt[3] + (2*(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]] - Sqrt[3]*ArcTan[(1 + (2*(1 + I*x)^(1/6))/
(1 - I*x)^(1/6))/Sqrt[3]] - 2*ArcTan[(1 - I*x)^(1/6)/(1 + I*x)^(1/6)] - 2*ArcTanh[(1 + I*x)^(1/6)/(1 - I*x)^(1
/6)] - (Sqrt[3]*Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3) - (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)])/2 + (Sqr
t[3]*Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3) + (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)])/2 + Log[1 - (1 + I*
x)^(1/6)/(1 - I*x)^(1/6) + (1 + I*x)^(1/3)/(1 - I*x)^(1/3)]/2 - Log[1 + (1 + I*x)^(1/6)/(1 - I*x)^(1/6) + (1 +
 I*x)^(1/3)/(1 - I*x)^(1/3)]/2

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[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] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[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)*ExpandTo
Sum[(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 209

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

Rule 210

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 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 216

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*C
os[(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] + Dis
t[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

Rule 301

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] +
Dist[2*(r^(m + 1)/(a*n*s^m)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] &&
IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 5170

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] &&  !IntegerQ[(I*n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x} x} \, dx \\ & = i \int \frac {1}{\sqrt [6]{1-i x} (1+i x)^{5/6}} \, dx+\int \frac {1}{\sqrt [6]{1-i x} (1+i x)^{5/6} x} \, dx \\ & = -\left (6 \text {Subst}\left (\int \frac {x^4}{\left (2-x^6\right )^{5/6}} \, dx,x,\sqrt [6]{1-i x}\right )\right )+6 \text {Subst}\left (\int \frac {1}{-1+x^6} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-2 \text {Subst}\left (\int \frac {1-\frac {x}{2}}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-2 \text {Subst}\left (\int \frac {1+\frac {x}{2}}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-6 \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \\ & = -2 \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {3}{2} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {3}{2} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \text {Subst}\left (\int \frac {-\frac {1}{2}+\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \text {Subst}\left (\int \frac {-\frac {1}{2}-\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \\ & = -2 \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac {1}{2} \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac {1}{2} \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{2} \sqrt {3} \text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{2} \sqrt {3} \text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \\ & = \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-2 \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{2} \sqrt {3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{2} \sqrt {3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{2} \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac {1}{2} \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )+\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \\ & = \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt {3}}\right )-2 \arctan \left (\frac {\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-\frac {1}{2} \sqrt {3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{2} \sqrt {3} \log \left (1+\frac {\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac {\sqrt {3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac {1}{2} \log \left (1-\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right )-\frac {1}{2} \log \left (1+\frac {\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+\frac {\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}\right ) \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.21 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x} \, dx=-\frac {3 (1-i x)^{5/6} \left (\sqrt [6]{2} (1+i x)^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {5}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2}-\frac {i x}{2}\right )+2 \operatorname {Hypergeometric2F1}\left (\frac {5}{6},1,\frac {11}{6},\frac {i+x}{i-x}\right )\right )}{5 (1+i x)^{5/6}} \]

[In]

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

[Out]

(-3*(1 - I*x)^(5/6)*(2^(1/6)*(1 + I*x)^(5/6)*Hypergeometric2F1[5/6, 5/6, 11/6, 1/2 - (I/2)*x] + 2*Hypergeometr
ic2F1[5/6, 1, 11/6, (I + x)/(I - x)]))/(5*(1 + I*x)^(5/6))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 339, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {1}{3} i \arctan (x)}}{x} \, dx=\frac {1}{2} \, {\left (\sqrt {3} + i\right )} \log \left (\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + \frac {1}{2} \, {\left (\sqrt {3} - i\right )} \log \left (\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} + 1\right )} \log \left (\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) + \frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2}\right ) + \frac {1}{2} \, {\left (i \, \sqrt {3} + 1\right )} \log \left (-\frac {1}{2} i \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2}\right ) - \frac {1}{2} \, {\left (\sqrt {3} - i\right )} \log \left (-\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) - \frac {1}{2} \, {\left (\sqrt {3} + i\right )} \log \left (-\frac {1}{2} \, \sqrt {3} + \left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) - \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + 1\right ) + i \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} + i\right ) - i \, \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - i\right ) + \log \left (\left (\frac {i \, \sqrt {x^{2} + 1}}{x + i}\right )^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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