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

Optimal. Leaf size=23 \[ \frac{4}{3} (x+1)^{3/2}-\frac{2}{5} (x+1)^{5/2} \]

[Out]

(4*(1 + x)^(3/2))/3 - (2*(1 + x)^(5/2))/5

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Rubi [A]  time = 0.033876, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6127, 627, 43} \[ \frac{4}{3} (x+1)^{3/2}-\frac{2}{5} (x+1)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*(1 - x)^(3/2),x]

[Out]

(4*(1 + x)^(3/2))/3 - (2*(1 + x)^(5/2))/5

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(x)} (1-x)^{3/2} \, dx &=\int \sqrt{1-x} \sqrt{1-x^2} \, dx\\ &=\int (1-x) \sqrt{1+x} \, dx\\ &=\int \left (2 \sqrt{1+x}-(1+x)^{3/2}\right ) \, dx\\ &=\frac{4}{3} (1+x)^{3/2}-\frac{2}{5} (1+x)^{5/2}\\ \end{align*}

Mathematica [A]  time = 0.0069197, size = 16, normalized size = 0.7 \[ -\frac{2}{15} (x+1)^{3/2} (3 x-7) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + x)^(3/2)*(-7 + 3*x))/15

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Maple [A]  time = 0.03, size = 29, normalized size = 1.3 \begin{align*} -{\frac{2\, \left ( 3\,x-7 \right ) \left ( 1+x \right ) ^{2}}{15}\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)^(3/2),x)

[Out]

-2/15*(1+x)^2*(3*x-7)*(1-x)^(1/2)/(-x^2+1)^(1/2)

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Maxima [B]  time = 0.973383, size = 49, normalized size = 2.13 \begin{align*} -\frac{2 \,{\left (x^{3} - 2 \, x^{2} + 3 \, x + 6\right )}}{5 \, \sqrt{x + 1}} - \frac{2 \,{\left (x^{2} - 4 \, x - 5\right )}}{3 \, \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)^(3/2),x, algorithm="maxima")

[Out]

-2/5*(x^3 - 2*x^2 + 3*x + 6)/sqrt(x + 1) - 2/3*(x^2 - 4*x - 5)/sqrt(x + 1)

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Fricas [B]  time = 1.80654, size = 81, normalized size = 3.52 \begin{align*} \frac{2 \,{\left (3 \, x^{2} - 4 \, x - 7\right )} \sqrt{-x^{2} + 1} \sqrt{-x + 1}}{15 \,{\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)^(3/2),x, algorithm="fricas")

[Out]

2/15*(3*x^2 - 4*x - 7)*sqrt(-x^2 + 1)*sqrt(-x + 1)/(x - 1)

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

[Out]

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

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Giac [A]  time = 1.20362, size = 27, normalized size = 1.17 \begin{align*} -\frac{2}{5} \,{\left (x + 1\right )}^{\frac{5}{2}} + \frac{4}{3} \,{\left (x + 1\right )}^{\frac{3}{2}} - \frac{16}{15} \, \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)^(3/2),x, algorithm="giac")

[Out]

-2/5*(x + 1)^(5/2) + 4/3*(x + 1)^(3/2) - 16/15*sqrt(2)