3.123 \(\int e^{\frac{2}{3} i \tan ^{-1}(x)} x \, dx\)

Optimal. Leaf size=140 \[ \frac{1}{2} (1-i x)^{2/3} (1+i x)^{4/3}+\frac{1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac{1}{3} \log \left (1+\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac{1}{9} \log (1+i x)-\frac{2 \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{1-i x}}{\sqrt{3} \sqrt [3]{1+i x}}\right )}{3 \sqrt{3}} \]

[Out]

((1 - I*x)^(2/3)*(1 + I*x)^(1/3))/3 + ((1 - I*x)^(2/3)*(1 + I*x)^(4/3))/2 - (2*ArcTan[1/Sqrt[3] - (2*(1 - I*x)
^(1/3))/(Sqrt[3]*(1 + I*x)^(1/3))])/(3*Sqrt[3]) - Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3)]/3 - Log[1 + I*x]/9

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Rubi [A]  time = 0.0359681, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5062, 80, 50, 60} \[ \frac{1}{2} (1-i x)^{2/3} (1+i x)^{4/3}+\frac{1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}-\frac{1}{3} \log \left (1+\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac{1}{9} \log (1+i x)-\frac{2 \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{1-i x}}{\sqrt{3} \sqrt [3]{1+i x}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[E^(((2*I)/3)*ArcTan[x])*x,x]

[Out]

((1 - I*x)^(2/3)*(1 + I*x)^(1/3))/3 + ((1 - I*x)^(2/3)*(1 + I*x)^(4/3))/2 - (2*ArcTan[1/Sqrt[3] - (2*(1 - I*x)
^(1/3))/(Sqrt[3]*(1 + I*x)^(1/3))])/(3*Sqrt[3]) - Log[1 + (1 - I*x)^(1/3)/(1 + I*x)^(1/3)]/3 - Log[1 + I*x]/9

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 60

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(d/b), 3]}, Simp[(Sq
rt[3]*q*ArcTan[1/Sqrt[3] - (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3))])/d, x] + (Simp[(3*q*Log[(q*(a + b*
x)^(1/3))/(c + d*x)^(1/3) + 1])/(2*d), x] + Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && NegQ[d/b]

Rubi steps

\begin{align*} \int e^{\frac{2}{3} i \tan ^{-1}(x)} x \, dx &=\int \frac{\sqrt [3]{1+i x} x}{\sqrt [3]{1-i x}} \, dx\\ &=\frac{1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac{1}{3} i \int \frac{\sqrt [3]{1+i x}}{\sqrt [3]{1-i x}} \, dx\\ &=\frac{1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac{1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac{2}{9} i \int \frac{1}{\sqrt [3]{1-i x} (1+i x)^{2/3}} \, dx\\ &=\frac{1}{3} (1-i x)^{2/3} \sqrt [3]{1+i x}+\frac{1}{2} (1-i x)^{2/3} (1+i x)^{4/3}-\frac{2 \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{1-i x}}{\sqrt{3} \sqrt [3]{1+i x}}\right )}{3 \sqrt{3}}-\frac{1}{3} \log \left (1+\frac{\sqrt [3]{1-i x}}{\sqrt [3]{1+i x}}\right )-\frac{1}{9} \log (1+i x)\\ \end{align*}

Mathematica [C]  time = 0.0198766, size = 54, normalized size = 0.39 \[ \frac{1}{2} (1-i x)^{2/3} \left (\sqrt [3]{2} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{2}{3},\frac{5}{3},\frac{1}{2}-\frac{i x}{2}\right )+(1+i x)^{4/3}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(((2*I)/3)*ArcTan[x])*x,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]  time = 1.6965, size = 362, normalized size = 2.59 \begin{align*} -\frac{1}{9} \,{\left (i \, \sqrt{3} - 1\right )} \log \left (\left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{2}{3}} + \frac{1}{2} i \, \sqrt{3} - \frac{1}{2}\right ) - \frac{1}{9} \,{\left (-i \, \sqrt{3} - 1\right )} \log \left (\left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{2}{3}} - \frac{1}{2} i \, \sqrt{3} - \frac{1}{2}\right ) + \frac{1}{6} \,{\left (3 \, x^{2} - 2 i \, x + 5\right )} \left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{2}{3}} - \frac{2}{9} \, \log \left (\left (\frac{i \, \sqrt{x^{2} + 1}}{x + i}\right )^{\frac{2}{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))^(2/3)*x,x, algorithm="fricas")

[Out]

-1/9*(I*sqrt(3) - 1)*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1/2*I*sqrt(3) - 1/2) - 1/9*(-I*sqrt(3) - 1)*log((I*
sqrt(x^2 + 1)/(x + I))^(2/3) - 1/2*I*sqrt(3) - 1/2) + 1/6*(3*x^2 - 2*I*x + 5)*(I*sqrt(x^2 + 1)/(x + I))^(2/3)
- 2/9*log((I*sqrt(x^2 + 1)/(x + I))^(2/3) + 1)

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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))**(2/3)*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{2}{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))^(2/3)*x,x, algorithm="giac")

[Out]

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