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

Optimal. Leaf size=37 \[ \frac{\sqrt{x+1}}{1-x}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{2}}\right )}{\sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0370145, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6127, 627, 47, 63, 206} \[ \frac{\sqrt{x+1}}{1-x}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]/(1 - x)^(3/2),x]

[Out]

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

Rule 6127

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 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{\sqrt{1-x^2}}{(1-x)^{5/2}} \, dx\\ &=\int \frac{\sqrt{1+x}}{(1-x)^2} \, dx\\ &=\frac{\sqrt{1+x}}{1-x}-\frac{1}{2} \int \frac{1}{(1-x) \sqrt{1+x}} \, dx\\ &=\frac{\sqrt{1+x}}{1-x}-\operatorname{Subst}\left (\int \frac{1}{2-x^2} \, dx,x,\sqrt{1+x}\right )\\ &=\frac{\sqrt{1+x}}{1-x}-\frac{\tanh ^{-1}\left (\frac{\sqrt{1+x}}{\sqrt{2}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0233544, size = 36, normalized size = 0.97 \[ -\frac{\sqrt{x+1}}{x-1}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.043, size = 69, normalized size = 1.9 \begin{align*}{\frac{1}{2\, \left ( -1+x \right ) ^{2}}\sqrt{-{x}^{2}+1}\sqrt{1-x} \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+x}} \right ) x-{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+x}} \right ) \sqrt{2}+2\,\sqrt{1+x} \right ){\frac{1}{\sqrt{1+x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.9207, size = 212, normalized size = 5.73 \begin{align*} \frac{\sqrt{2}{\left (x^{2} - 2 \, x + 1\right )} \log \left (-\frac{x^{2} + 2 \, \sqrt{2} \sqrt{-x^{2} + 1} \sqrt{-x + 1} + 2 \, x - 3}{x^{2} - 2 \, x + 1}\right ) + 4 \, \sqrt{-x^{2} + 1} \sqrt{-x + 1}}{4 \,{\left (x^{2} - 2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*(x^2 - 2*x + 1)*log(-(x^2 + 2*sqrt(2)*sqrt(-x^2 + 1)*sqrt(-x + 1) + 2*x - 3)/(x^2 - 2*x + 1)) + 4
*sqrt(-x^2 + 1)*sqrt(-x + 1))/(x^2 - 2*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18524, size = 57, normalized size = 1.54 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (\frac{\sqrt{2} - \sqrt{x + 1}}{\sqrt{2} + \sqrt{x + 1}}\right ) - \frac{\sqrt{x + 1}}{x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log((sqrt(2) - sqrt(x + 1))/(sqrt(2) + sqrt(x + 1))) - sqrt(x + 1)/(x - 1)