3.381 \(\int \frac{e^{\tanh ^{-1}(x)} x}{\sqrt{1-x}} \, dx\)

Optimal. Leaf size=42 \[ -\frac{2}{3} (x+1)^{3/2}-2 \sqrt{x+1}+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{2}}\right ) \]

[Out]

-2*Sqrt[1 + x] - (2*(1 + x)^(3/2))/3 + 2*Sqrt[2]*ArcTanh[Sqrt[1 + x]/Sqrt[2]]

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Rubi [A]  time = 0.0463722, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6129, 80, 50, 63, 206} \[ -\frac{2}{3} (x+1)^{3/2}-2 \sqrt{x+1}+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[x]*x)/Sqrt[1 - x],x]

[Out]

-2*Sqrt[1 + x] - (2*(1 + x)^(3/2))/3 + 2*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 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 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)} x}{\sqrt{1-x}} \, dx &=\int \frac{x \sqrt{1+x}}{1-x} \, dx\\ &=-\frac{2}{3} (1+x)^{3/2}+\int \frac{\sqrt{1+x}}{1-x} \, dx\\ &=-2 \sqrt{1+x}-\frac{2}{3} (1+x)^{3/2}+2 \int \frac{1}{(1-x) \sqrt{1+x}} \, dx\\ &=-2 \sqrt{1+x}-\frac{2}{3} (1+x)^{3/2}+4 \operatorname{Subst}\left (\int \frac{1}{2-x^2} \, dx,x,\sqrt{1+x}\right )\\ &=-2 \sqrt{1+x}-\frac{2}{3} (1+x)^{3/2}+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1+x}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0201375, size = 36, normalized size = 0.86 \[ 2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{2}}\right )-\frac{2}{3} \sqrt{x+1} (x+4) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[x]*x)/Sqrt[1 - x],x]

[Out]

(-2*Sqrt[1 + x]*(4 + x))/3 + 2*Sqrt[2]*ArcTanh[Sqrt[1 + x]/Sqrt[2]]

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Maple [A]  time = 0.042, size = 61, normalized size = 1.5 \begin{align*} -{\frac{2}{3\,x-3}\sqrt{-{x}^{2}+1}\sqrt{1-x} \left ( 3\,{\it Artanh} \left ( 1/2\,\sqrt{1+x}\sqrt{2} \right ) \sqrt{2}-\sqrt{1+x}x-4\,\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)*x/(1-x)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.81759, size = 204, normalized size = 4.86 \begin{align*} \frac{3 \, \sqrt{2}{\left (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 ) + 2 \, \sqrt{-x^{2} + 1}{\left (x + 4\right )} \sqrt{-x + 1}}{3 \,{\left (x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23088, size = 59, normalized size = 1.4 \begin{align*} -\frac{2}{3} \,{\left (x + 1\right )}^{\frac{3}{2}} - \sqrt{2} \log \left (\frac{\sqrt{2} - \sqrt{x + 1}}{\sqrt{2} + \sqrt{x + 1}}\right ) - 2 \, \sqrt{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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