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

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Rubi [A]  time = 0.02, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [B]  time = 0.54, size = 146, normalized size = 1.74 \[ {\left (x - 1\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {2}{3} \, \log \left (\left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} + 1\right ) - \frac {1}{3} \, \log \left (-\frac {{\left (x - 1\right )} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {2}{3}} - x + \sqrt {-x^{2} + 1} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}} + 1}{x - 1}\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {2}{3}}\,{d x} \]

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)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \left (\frac {1+x}{\sqrt {-x^{2}+1}}\right )^{\frac {2}{3}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {2}{3}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {x+1}{\sqrt {1-x^2}}\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {x + 1}{\sqrt {1 - x^{2}}}\right )^{\frac {2}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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