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

Optimal. Leaf size=49 \[ \frac{2}{7} (1-x)^{7/2}-2 (1-x)^{5/2}+\frac{16}{3} (1-x)^{3/2}-8 \sqrt{1-x} \]

[Out]

-8*Sqrt[1 - x] + (16*(1 - x)^(3/2))/3 - 2*(1 - x)^(5/2) + (2*(1 - x)^(7/2))/7

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Rubi [A]  time = 0.0479878, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {6129, 77} \[ \frac{2}{7} (1-x)^{7/2}-2 (1-x)^{5/2}+\frac{16}{3} (1-x)^{3/2}-8 \sqrt{1-x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*x*(1 + x)^(3/2),x]

[Out]

-8*Sqrt[1 - x] + (16*(1 - x)^(3/2))/3 - 2*(1 - x)^(5/2) + (2*(1 - x)^(7/2))/7

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(x)} x (1+x)^{3/2} \, dx &=\int \frac{x (1+x)^2}{\sqrt{1-x}} \, dx\\ &=\int \left (\frac{4}{\sqrt{1-x}}-8 \sqrt{1-x}+5 (1-x)^{3/2}-(1-x)^{5/2}\right ) \, dx\\ &=-8 \sqrt{1-x}+\frac{16}{3} (1-x)^{3/2}-2 (1-x)^{5/2}+\frac{2}{7} (1-x)^{7/2}\\ \end{align*}

Mathematica [A]  time = 0.0111833, size = 28, normalized size = 0.57 \[ -\frac{2}{21} \sqrt{1-x} \left (3 x^3+12 x^2+23 x+46\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - x]*(46 + 23*x + 12*x^2 + 3*x^3))/21

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Maple [A]  time = 0.03, size = 35, normalized size = 0.7 \begin{align*}{\frac{ \left ( -2+2\,x \right ) \left ( 3\,{x}^{3}+12\,{x}^{2}+23\,x+46 \right ) }{21}\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,x)

[Out]

2/21*(-1+x)*(3*x^3+12*x^2+23*x+46)*(1+x)^(1/2)/(-x^2+1)^(1/2)

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Maxima [A]  time = 0.94956, size = 39, normalized size = 0.8 \begin{align*} \frac{2 \,{\left (3 \, x^{4} + 9 \, x^{3} + 11 \, x^{2} + 23 \, x - 46\right )}}{21 \, \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,x, algorithm="maxima")

[Out]

2/21*(3*x^4 + 9*x^3 + 11*x^2 + 23*x - 46)/sqrt(-x + 1)

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Fricas [A]  time = 1.80191, size = 85, normalized size = 1.73 \begin{align*} -\frac{2 \,{\left (3 \, x^{3} + 12 \, x^{2} + 23 \, x + 46\right )} \sqrt{-x^{2} + 1}}{21 \, \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,x, algorithm="fricas")

[Out]

-2/21*(3*x^3 + 12*x^2 + 23*x + 46)*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,x)

[Out]

Timed out

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Giac [A]  time = 1.28964, size = 70, normalized size = 1.43 \begin{align*} -\frac{2}{7} \,{\left (x - 1\right )}^{3} \sqrt{-x + 1} - 2 \,{\left (x - 1\right )}^{2} \sqrt{-x + 1} + \frac{16}{3} \,{\left (-x + 1\right )}^{\frac{3}{2}} + \frac{64}{21} \, \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,x, algorithm="giac")

[Out]

-2/7*(x - 1)^3*sqrt(-x + 1) - 2*(x - 1)^2*sqrt(-x + 1) + 16/3*(-x + 1)^(3/2) + 64/21*sqrt(2) - 8*sqrt(-x + 1)