3.118 \(\int \frac{e^{\frac{1}{3} i \tan ^{-1}(x)}}{x} \, dx\)

Optimal. Leaf size=430 \[ -\frac{1}{2} \sqrt{3} \log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )+\frac{1}{2} \sqrt{3} \log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )+\frac{1}{2} \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\frac{1}{2} \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\tan ^{-1}\left (\sqrt{3}+\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-\sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-2 \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.458226, antiderivative size = 430, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 13, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.929, Rules used = {5062, 105, 63, 331, 295, 634, 618, 204, 628, 203, 93, 210, 206} \[ -\frac{1}{2} \sqrt{3} \log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )+\frac{1}{2} \sqrt{3} \log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )+\frac{1}{2} \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}-\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )-\frac{1}{2} \log \left (\frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}}+\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}+1\right )+\tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\tan ^{-1}\left (\sqrt{3}+\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-\sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-2 \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right ) \]

Antiderivative was successfully verified.

[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 5062

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]

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 63

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 331

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 295

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)*Int[1/(r^2 + s^2*x^2), x])/(a*n*s^m) +
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 634

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 618

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 204

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

Rule 628

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 203

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

Rule 93

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 210

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*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
Dist[(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 206

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

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{3} i \tan ^{-1}(x)}}{x} \, dx &=\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 \operatorname{Subst}\left (\int \frac{x^4}{\left (2-x^6\right )^{5/6}} \, dx,x,\sqrt [6]{1-i x}\right )\right )+6 \operatorname{Subst}\left (\int \frac{1}{-1+x^6} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )\right )-2 \operatorname{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 \operatorname{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 \operatorname{Subst}\left (\int \frac{x^4}{1+x^6} \, dx,x,\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )\\ &=-2 \tanh ^{-1}\left (\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )+\frac{1}{2} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}\right )-2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \operatorname{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 \operatorname{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 \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \tanh ^{-1}\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} \operatorname{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} \operatorname{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 \operatorname{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 \operatorname{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} \operatorname{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} \operatorname{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} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-\sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-2 \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \tanh ^{-1}\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 )+\operatorname{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 )+\operatorname{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 )\\ &=\tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-\tan ^{-1}\left (\sqrt{3}+\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-\sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [6]{1+i x}}{\sqrt [6]{1-i x}}}{\sqrt{3}}\right )-2 \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )-2 \tanh ^{-1}\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]  time = 0.0305776, size = 90, normalized size = 0.21 \[ -\frac{3 (1-i x)^{5/6} \left (\sqrt [6]{2} (1+i x)^{5/6} \text{Hypergeometric2F1}\left (\frac{5}{6},\frac{5}{6},\frac{11}{6},\frac{1}{2}-\frac{i x}{2}\right )+2 \text{Hypergeometric2F1}\left (\frac{5}{6},1,\frac{11}{6},\frac{x+i}{-x+i}\right )\right )}{5 (1+i x)^{5/6}} \]

Warning: Unable to verify antiderivative.

[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]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\sqrt [3]{{(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, x + 1}{\sqrt{x^{2} + 1}}\right )^{\frac{1}{3}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 1.7904, size = 1108, normalized size = 2.58 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, x + 1}{\sqrt{x^{2} + 1}}\right )^{\frac{1}{3}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)