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

Optimal. Leaf size=11 \[ \frac{2}{3} (x+1)^{3/2} \]

[Out]

(2*(1 + x)^(3/2))/3

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Rubi [A]  time = 0.0200117, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6127, 26, 32} \[ \frac{2}{3} (x+1)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*Sqrt[1 - x],x]

[Out]

(2*(1 + x)^(3/2))/3

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 26

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(j_))^(p_.), x_Symbol] :> Dist[(-(b^2/d))^m, Int[
u/(a - b*x^n)^m, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[j, 2*n] && EqQ[p, -m] && EqQ[b^2*c + a^2*d,
0] && GtQ[a, 0] && LtQ[d, 0]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0040061, size = 11, normalized size = 1. \[ \frac{2}{3} (x+1)^{3/2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[x]*Sqrt[1 - x],x]

[Out]

(2*(1 + x)^(3/2))/3

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Maple [B]  time = 0.029, size = 24, normalized size = 2.2 \begin{align*}{\frac{2\, \left ( 1+x \right ) ^{2}}{3}\sqrt{1-x}{\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.980432, size = 31, normalized size = 2.82 \begin{align*} \frac{2 \,{\left (x^{2} - x - 2\right )}}{3 \, \sqrt{x + 1}} + 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)*(1-x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.87658, size = 68, normalized size = 6.18 \begin{align*} -\frac{2 \, \sqrt{-x^{2} + 1}{\left (x + 1\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)*(1-x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19436, size = 18, normalized size = 1.64 \begin{align*} \frac{2}{3} \,{\left (x + 1\right )}^{\frac{3}{2}} - \frac{4}{3} \, \sqrt{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x + 1)^(3/2) - 4/3*sqrt(2)