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

Optimal. Leaf size=36 \[ -\frac{2}{5} (1-x)^{5/2}+2 (1-x)^{3/2}-4 \sqrt{1-x} \]

[Out]

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

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Rubi [A]  time = 0.0415186, antiderivative size = 36, 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} \[ -\frac{2}{5} (1-x)^{5/2}+2 (1-x)^{3/2}-4 \sqrt{1-x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*x*Sqrt[1 + x],x]

[Out]

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

Mathematica [A]  time = 0.0083818, size = 21, normalized size = 0.58 \[ -\frac{2}{5} \sqrt{1-x} \left (x^2+3 x+6\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[x]*x*Sqrt[1 + x],x]

[Out]

(-2*Sqrt[1 - x]*(6 + 3*x + x^2))/5

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Maple [A]  time = 0.036, size = 28, normalized size = 0.8 \begin{align*}{\frac{ \left ( -2+2\,x \right ) \left ({x}^{2}+3\,x+6 \right ) }{5}\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)^(3/2)/(-x^2+1)^(1/2)*x,x)

[Out]

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

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Maxima [A]  time = 0.94335, size = 30, normalized size = 0.83 \begin{align*} \frac{2 \,{\left (x^{3} + 2 \, x^{2} + 3 \, x - 6\right )}}{5 \, \sqrt{-x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*x,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.7774, size = 66, normalized size = 1.83 \begin{align*} -\frac{2 \,{\left (x^{2} + 3 \, x + 6\right )} \sqrt{-x^{2} + 1}}{5 \, \sqrt{x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*x,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.28932, size = 51, normalized size = 1.42 \begin{align*} -\frac{2}{5} \,{\left (x - 1\right )}^{2} \sqrt{-x + 1} + 2 \,{\left (-x + 1\right )}^{\frac{3}{2}} + \frac{8}{5} \, \sqrt{2} - 4 \, \sqrt{-x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(3/2)/(-x^2+1)^(1/2)*x,x, algorithm="giac")

[Out]

-2/5*(x - 1)^2*sqrt(-x + 1) + 2*(-x + 1)^(3/2) + 8/5*sqrt(2) - 4*sqrt(-x + 1)