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

Optimal. Leaf size=278 \[ \frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}+\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{\log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )}{12 \sqrt{3}}-\frac{\log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )}{12 \sqrt{3}}-\frac{1}{18} \tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{18} \tan ^{-1}\left (\sqrt{3}+\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{9} \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \]

[Out]

((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/6 + ((1 - I*x)^(5/6)*(1 + I*x)^(7/6))/2 - ArcTan[Sqrt[3] - (2*(1 - I*x)^(1/6
))/(1 + I*x)^(1/6)]/18 + ArcTan[Sqrt[3] + (2*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)]/18 + ArcTan[(1 - I*x)^(1/6)/(1
+ I*x)^(1/6)]/9 + Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3) - (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)]/(12*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)]/(12*Sqrt[3])

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Rubi [A]  time = 0.338368, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.917, Rules used = {5062, 80, 50, 63, 331, 295, 634, 618, 204, 628, 203} \[ \frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}+\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{\log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}-\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )}{12 \sqrt{3}}-\frac{\log \left (\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}+\frac{\sqrt{3} \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}+1\right )}{12 \sqrt{3}}-\frac{1}{18} \tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{18} \tan ^{-1}\left (\sqrt{3}+\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{9} \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - I*x)^(5/6)*(1 + I*x)^(1/6))/6 + ((1 - I*x)^(5/6)*(1 + I*x)^(7/6))/2 - ArcTan[Sqrt[3] - (2*(1 - I*x)^(1/6
))/(1 + I*x)^(1/6)]/18 + ArcTan[Sqrt[3] + (2*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)]/18 + ArcTan[(1 - I*x)^(1/6)/(1
+ I*x)^(1/6)]/9 + Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3) - (Sqrt[3]*(1 - I*x)^(1/6))/(1 + I*x)^(1/6)]/(12*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)]/(12*Sqrt[3])

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

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] + Dist[(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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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])

Rubi steps

\begin{align*} \int e^{\frac{1}{3} i \tan ^{-1}(x)} x \, dx &=\int \frac{\sqrt [6]{1+i x} x}{\sqrt [6]{1-i x}} \, dx\\ &=\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}-\frac{1}{6} i \int \frac{\sqrt [6]{1+i x}}{\sqrt [6]{1-i x}} \, dx\\ &=\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}-\frac{1}{18} i \int \frac{1}{\sqrt [6]{1-i x} (1+i x)^{5/6}} \, dx\\ &=\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^4}{\left (2-x^6\right )^{5/6}} \, dx,x,\sqrt [6]{1-i x}\right )\\ &=\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^4}{1+x^6} \, dx,x,\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )\\ &=\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}+\frac{1}{9} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{9} \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 )+\frac{1}{9} \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 )\\ &=\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}+\frac{1}{9} \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{36} \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}{36} \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{\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 )}{12 \sqrt{3}}-\frac{\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 )}{12 \sqrt{3}}\\ &=\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}+\frac{1}{9} \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{\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 )}{12 \sqrt{3}}-\frac{\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 )}{12 \sqrt{3}}-\frac{1}{18} \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 )-\frac{1}{18} \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 )\\ &=\frac{1}{6} (1-i x)^{5/6} \sqrt [6]{1+i x}+\frac{1}{2} (1-i x)^{5/6} (1+i x)^{7/6}-\frac{1}{18} \tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{18} \tan ^{-1}\left (\sqrt{3}+\frac{2 \sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{1}{9} \tan ^{-1}\left (\frac{\sqrt [6]{1-i x}}{\sqrt [6]{1+i x}}\right )+\frac{\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 )}{12 \sqrt{3}}-\frac{\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 )}{12 \sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.022786, size = 57, normalized size = 0.21 \[ \frac{1}{10} (1-i x)^{5/6} \left (2 \sqrt [6]{2} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{5}{6},\frac{11}{6},\frac{1}{2}-\frac{i x}{2}\right )+5 (1+i x)^{7/6}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - I*x)^(5/6)*(5*(1 + I*x)^(7/6) + 2*2^(1/6)*Hypergeometric2F1[-1/6, 5/6, 11/6, 1/2 - (I/2)*x]))/10

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{{(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}}x\, 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)

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

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Fricas [A]  time = 1.78683, size = 645, normalized size = 2.32 \begin{align*} -\frac{1}{36} \,{\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}{36} \,{\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}{36} \,{\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}{36} \,{\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}{6} \,{\left (3 \, x^{2} - i \, x + 4\right )} \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} - \frac{1}{18} i \, \log \left (\left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} + i\right ) + \frac{1}{18} i \, \log \left (\left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{1}{3}} - i\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]

-1/36*(sqrt(3) + I)*log(1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) + 1/2*I) - 1/36*(sqrt(3) - I)*log(1/2*sq
rt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2*I) + 1/36*(sqrt(3) - I)*log(-1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x +
I))^(1/3) + 1/2*I) + 1/36*(sqrt(3) + I)*log(-1/2*sqrt(3) + (I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/2*I) + 1/6*(3*
x^2 - I*x + 4)*(I*sqrt(x^2 + 1)/(x + I))^(1/3) - 1/18*I*log((I*sqrt(x^2 + 1)/(x + I))^(1/3) + I) + 1/18*I*log(
(I*sqrt(x^2 + 1)/(x + I))^(1/3) - I)

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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

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