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

Optimal. Leaf size=38 \[ -\frac{2}{5} (1-x)^{5/2}+\frac{8}{3} (1-x)^{3/2}-8 \sqrt{1-x} \]

[Out]

-8*Sqrt[1 - x] + (8*(1 - x)^(3/2))/3 - (2*(1 - x)^(5/2))/5

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Rubi [A]  time = 0.0288797, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6129, 43} \[ -\frac{2}{5} (1-x)^{5/2}+\frac{8}{3} (1-x)^{3/2}-8 \sqrt{1-x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*(1 + x)^(3/2),x]

[Out]

-8*Sqrt[1 - x] + (8*(1 - x)^(3/2))/3 - (2*(1 - x)^(5/2))/5

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 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 \frac{(1+x)^2}{\sqrt{1-x}} \, dx\\ &=\int \left (\frac{4}{\sqrt{1-x}}-4 \sqrt{1-x}+(1-x)^{3/2}\right ) \, dx\\ &=-8 \sqrt{1-x}+\frac{8}{3} (1-x)^{3/2}-\frac{2}{5} (1-x)^{5/2}\\ \end{align*}

Mathematica [A]  time = 0.0089038, size = 23, normalized size = 0.61 \[ -\frac{2}{15} \sqrt{1-x} \left (3 x^2+14 x+43\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - x]*(43 + 14*x + 3*x^2))/15

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Maple [A]  time = 0.03, size = 30, normalized size = 0.8 \begin{align*}{\frac{ \left ( -2+2\,x \right ) \left ( 3\,{x}^{2}+14\,x+43 \right ) }{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)^(5/2)/(-x^2+1)^(1/2),x)

[Out]

2/15*(-1+x)*(3*x^2+14*x+43)*(1+x)^(1/2)/(-x^2+1)^(1/2)

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Maxima [A]  time = 0.953294, size = 32, normalized size = 0.84 \begin{align*} \frac{2 \,{\left (3 \, x^{3} + 11 \, x^{2} + 29 \, x - 43\right )}}{15 \, \sqrt{-x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(5/2)/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*x^3 + 11*x^2 + 29*x - 43)/sqrt(-x + 1)

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Fricas [A]  time = 1.67183, size = 73, normalized size = 1.92 \begin{align*} -\frac{2 \,{\left (3 \, x^{2} + 14 \, x + 43\right )} \sqrt{-x^{2} + 1}}{15 \, \sqrt{x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(5/2)/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(3*x^2 + 14*x + 43)*sqrt(-x^2 + 1)/sqrt(x + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**(5/2)/(-x**2+1)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.1949, size = 51, normalized size = 1.34 \begin{align*} -\frac{2}{5} \,{\left (x - 1\right )}^{2} \sqrt{-x + 1} + \frac{8}{3} \,{\left (-x + 1\right )}^{\frac{3}{2}} + \frac{64}{15} \, \sqrt{2} - 8 \, \sqrt{-x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(5/2)/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

-2/5*(x - 1)^2*sqrt(-x + 1) + 8/3*(-x + 1)^(3/2) + 64/15*sqrt(2) - 8*sqrt(-x + 1)