3.130 \(\int e^{\frac{2}{3} \tanh ^{-1}(x)} \, dx\)

Optimal. Leaf size=84 \[ -(1-x)^{2/3} \sqrt [3]{x+1}+\frac{1}{3} \log (x+1)+\log \left (\frac{\sqrt [3]{1-x}}{\sqrt [3]{x+1}}+1\right )+\frac{2 \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{1-x}}{\sqrt{3} \sqrt [3]{x+1}}\right )}{\sqrt{3}} \]

[Out]

-((1 - x)^(2/3)*(1 + x)^(1/3)) + (2*ArcTan[1/Sqrt[3] - (2*(1 - x)^(1/3))/(Sqrt[3]*(1 + x)^(1/3))])/Sqrt[3] + L
og[1 + x]/3 + Log[1 + (1 - x)^(1/3)/(1 + x)^(1/3)]

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Rubi [A]  time = 0.0171154, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6125, 50, 60} \[ -(1-x)^{2/3} \sqrt [3]{x+1}+\frac{1}{3} \log (x+1)+\log \left (\frac{\sqrt [3]{1-x}}{\sqrt [3]{x+1}}+1\right )+\frac{2 \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{1-x}}{\sqrt{3} \sqrt [3]{x+1}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - x)^(2/3)*(1 + x)^(1/3)) + (2*ArcTan[1/Sqrt[3] - (2*(1 - x)^(1/3))/(Sqrt[3]*(1 + x)^(1/3))])/Sqrt[3] + L
og[1 + x]/3 + Log[1 + (1 - x)^(1/3)/(1 + x)^(1/3)]

Rule 6125

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

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

Mathematica [A]  time = 0.135186, size = 87, normalized size = 1.04 \[ -\frac{2 e^{\frac{2}{3} \tanh ^{-1}(x)}}{e^{2 \tanh ^{-1}(x)}+1}+\frac{2}{3} \log \left (e^{\frac{2}{3} \tanh ^{-1}(x)}+1\right )-\frac{1}{3} \log \left (-e^{\frac{2}{3} \tanh ^{-1}(x)}+e^{\frac{4}{3} \tanh ^{-1}(x)}+1\right )+\frac{2 \tan ^{-1}\left (\frac{2 e^{\frac{2}{3} \tanh ^{-1}(x)}-1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int \left ({(1+x){\frac{1}{\sqrt{-{x}^{2}+1}}}} \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+x)/(-x^2+1)^(1/2))^(2/3),x)

[Out]

int(((1+x)/(-x^2+1)^(1/2))^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\frac{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+x)/(-x^2+1)^(1/2))^(2/3),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.67409, size = 387, normalized size = 4.61 \begin{align*}{\left (x - 1\right )} \left (-\frac{\sqrt{-x^{2} + 1}}{x - 1}\right )^{\frac{2}{3}} + \frac{2}{3} \, \sqrt{3} \arctan \left (\frac{2}{3} \, \sqrt{3} \left (-\frac{\sqrt{-x^{2} + 1}}{x - 1}\right )^{\frac{2}{3}} - \frac{1}{3} \, \sqrt{3}\right ) + \frac{2}{3} \, \log \left (\left (-\frac{\sqrt{-x^{2} + 1}}{x - 1}\right )^{\frac{2}{3}} + 1\right ) - \frac{1}{3} \, \log \left (-\frac{{\left (x - 1\right )} \left (-\frac{\sqrt{-x^{2} + 1}}{x - 1}\right )^{\frac{2}{3}} - x + \sqrt{-x^{2} + 1} \left (-\frac{\sqrt{-x^{2} + 1}}{x - 1}\right )^{\frac{1}{3}} + 1}{x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - 1)*(-sqrt(-x^2 + 1)/(x - 1))^(2/3) + 2/3*sqrt(3)*arctan(2/3*sqrt(3)*(-sqrt(-x^2 + 1)/(x - 1))^(2/3) - 1/3
*sqrt(3)) + 2/3*log((-sqrt(-x^2 + 1)/(x - 1))^(2/3) + 1) - 1/3*log(-((x - 1)*(-sqrt(-x^2 + 1)/(x - 1))^(2/3) -
x + sqrt(-x^2 + 1)*(-sqrt(-x^2 + 1)/(x - 1))^(1/3) + 1)/(x - 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+x)/(-x**2+1)**(1/2))**(2/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\frac{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+x)/(-x^2+1)^(1/2))^(2/3),x, algorithm="giac")

[Out]

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