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]

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

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Rubi [A]  time = 0.04, antiderivative size = 36, normalized size of antiderivative = 1.00, 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 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]))))

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])

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.47, size = 24, normalized size = 0.67 \[ -\frac {2 \, {\left (x^{2} + 3 \, x + 6\right )} \sqrt {-x^{2} + 1}}{5 \, \sqrt {x + 1}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.03, size = 28, normalized size = 0.78 \[ \frac {2 \left (-1+x \right ) \left (x^{2}+3 x +6\right ) \sqrt {1+x}}{5 \sqrt {-x^{2}+1}} \]

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 = 0.31, size = 22, normalized size = 0.61 \[ \frac {2 \, {\left (x^{3} + 2 \, x^{2} + 3 \, x - 6\right )}}{5 \, \sqrt {-x + 1}} \]

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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mupad [B]  time = 0.87, size = 45, normalized size = 1.25 \[ -\frac {2\,x^2\,\sqrt {1-x^2}+6\,x\,\sqrt {1-x^2}+12\,\sqrt {1-x^2}}{5\,\sqrt {x+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(x + 1)^(3/2))/(1 - x^2)^(1/2),x)

[Out]

-(2*x^2*(1 - x^2)^(1/2) + 6*x*(1 - x^2)^(1/2) + 12*(1 - x^2)^(1/2))/(5*(x + 1)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (x + 1\right )^{\frac {3}{2}}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \]

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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