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

Optimal. Leaf size=20 \[ \frac{\sqrt{1-x}}{\sqrt{x+1}}+\sin ^{-1}(x) \]

[Out]

Sqrt[1 - x]/Sqrt[1 + x] + ArcSin[x]

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Rubi [A]  time = 0.0335342, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6129, 78, 41, 216} \[ \frac{\sqrt{1-x}}{\sqrt{x+1}}+\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[x]*x)/(1 + x)^2,x]

[Out]

Sqrt[1 - x]/Sqrt[1 + x] + ArcSin[x]

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
 n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 41

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(x)} x}{(1+x)^2} \, dx &=\int \frac{x}{\sqrt{1-x} (1+x)^{3/2}} \, dx\\ &=\frac{\sqrt{1-x}}{\sqrt{1+x}}+\int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx\\ &=\frac{\sqrt{1-x}}{\sqrt{1+x}}+\int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=\frac{\sqrt{1-x}}{\sqrt{1+x}}+\sin ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.034002, size = 20, normalized size = 1. \[ \frac{\sqrt{1-x}}{\sqrt{x+1}}+\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - x]/Sqrt[1 + x] + ArcSin[x]

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Maple [A]  time = 0.032, size = 24, normalized size = 1.2 \begin{align*} \arcsin \left ( x \right ) +{\frac{1}{1+x}\sqrt{- \left ( 1+x \right ) ^{2}+2\,x+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42316, size = 24, normalized size = 1.2 \begin{align*} \frac{\sqrt{-x^{2} + 1}}{x + 1} + \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-x^2 + 1)/(x + 1) + arcsin(x)

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Fricas [B]  time = 1.87501, size = 105, normalized size = 5.25 \begin{align*} -\frac{2 \,{\left (x + 1\right )} \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) - x - \sqrt{-x^{2} + 1} - 1}{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23876, size = 32, normalized size = 1.6 \begin{align*} -\frac{2}{\frac{\sqrt{-x^{2} + 1} - 1}{x} - 1} + \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/((sqrt(-x^2 + 1) - 1)/x - 1) + arcsin(x)