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

Optimal. Leaf size=23 \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 - x]/Sqrt[2]])

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Rubi [A]  time = 0.028107, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6129, 63, 206} \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]/(1 + x)^(3/2),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 - x]/Sqrt[2]])

Rule 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

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

Mathematica [A]  time = 0.0043876, size = 23, normalized size = 1. \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[x]/(1 + x)^(3/2),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 - x]/Sqrt[2]])

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Maple [B]  time = 0.062, size = 40, normalized size = 1.7 \begin{align*} -{\sqrt{2}\sqrt{-{x}^{2}+1}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1-x}} \right ){\frac{1}{\sqrt{1-x}}}{\frac{1}{\sqrt{1+x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(1+x)^(1/2)*(-x^2+1)^(1/2)/(1-x)^(1/2)*2^(1/2)*arctanh(1/2*(1-x)^(1/2)*2^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{2} + 1} \sqrt{x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.78585, size = 122, normalized size = 5.3 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-\frac{x^{2} + 2 \, \sqrt{2} \sqrt{-x^{2} + 1} \sqrt{x + 1} - 2 \, x - 3}{x^{2} + 2 \, x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.20302, size = 50, normalized size = 2.17 \begin{align*} -\frac{1}{2} \, \sqrt{2} \log \left (\sqrt{2} + \sqrt{-x + 1}\right ) + \frac{1}{2} \, \sqrt{2} \log \left (\sqrt{2} - \sqrt{-x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*log(sqrt(2) + sqrt(-x + 1)) + 1/2*sqrt(2)*log(sqrt(2) - sqrt(-x + 1))