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

Optimal. Leaf size=25 \[ \frac{2}{3} (1-x)^{3/2}-4 \sqrt{1-x} \]

[Out]

-4*Sqrt[1 - x] + (2*(1 - x)^(3/2))/3

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Rubi [A]  time = 0.0234036, antiderivative size = 25, 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} \[ \frac{2}{3} (1-x)^{3/2}-4 \sqrt{1-x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]*Sqrt[1 + x],x]

[Out]

-4*Sqrt[1 - x] + (2*(1 - x)^(3/2))/3

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

Mathematica [A]  time = 0.005727, size = 16, normalized size = 0.64 \[ -\frac{2}{3} \sqrt{1-x} (x+5) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[x]*Sqrt[1 + x],x]

[Out]

(-2*Sqrt[1 - x]*(5 + x))/3

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Maple [A]  time = 0.029, size = 23, normalized size = 0.9 \begin{align*}{\frac{ \left ( -2+2\,x \right ) \left ( x+5 \right ) }{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)^(3/2)/(-x^2+1)^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.947353, size = 23, normalized size = 0.92 \begin{align*} \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)^(3/2)/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

2/3*(x^2 + 4*x - 5)/sqrt(-x + 1)

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Fricas [A]  time = 1.70803, size = 55, normalized size = 2.2 \begin{align*} -\frac{2 \, \sqrt{-x^{2} + 1}{\left (x + 5\right )}}{3 \, \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, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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)

[Out]

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

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Giac [A]  time = 1.23354, size = 32, normalized size = 1.28 \begin{align*} \frac{2}{3} \,{\left (-x + 1\right )}^{\frac{3}{2}} + \frac{8}{3} \, \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, algorithm="giac")

[Out]

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