3.655 \(\int x (1-x^2)^{3/2} \sin ^{-1}(x) \, dx\)

Optimal. Leaf size=37 \[ \frac {x^5}{25}-\frac {2 x^3}{15}-\frac {1}{5} \left (1-x^2\right )^{5/2} \sin ^{-1}(x)+\frac {x}{5} \]

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4677, 194} \[ \frac {x^5}{25}-\frac {2 x^3}{15}-\frac {1}{5} \left (1-x^2\right )^{5/2} \sin ^{-1}(x)+\frac {x}{5} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 - x^2)^(3/2)*ArcSin[x],x]

[Out]

x/5 - (2*x^3)/15 + x^5/25 - ((1 - x^2)^(5/2)*ArcSin[x])/5

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x \left (1-x^2\right )^{3/2} \sin ^{-1}(x) \, dx &=-\frac {1}{5} \left (1-x^2\right )^{5/2} \sin ^{-1}(x)+\frac {1}{5} \int \left (1-x^2\right )^2 \, dx\\ &=-\frac {1}{5} \left (1-x^2\right )^{5/2} \sin ^{-1}(x)+\frac {1}{5} \int \left (1-2 x^2+x^4\right ) \, dx\\ &=\frac {x}{5}-\frac {2 x^3}{15}+\frac {x^5}{25}-\frac {1}{5} \left (1-x^2\right )^{5/2} \sin ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.95 \[ \frac {1}{5} \left (\frac {x^5}{5}-\frac {2 x^3}{3}-\left (1-x^2\right )^{5/2} \sin ^{-1}(x)+x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(1 - x^2)^(3/2)*ArcSin[x],x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (1-x^2\right )^{3/2} \sin ^{-1}(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x*(1 - x^2)^(3/2)*ArcSin[x],x]

[Out]

Could not integrate

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fricas [A]  time = 0.73, size = 37, normalized size = 1.00 \[ \frac {1}{25} \, x^{5} - \frac {2}{15} \, x^{3} - \frac {1}{5} \, {\left (x^{4} - 2 \, x^{2} + 1\right )} \sqrt {-x^{2} + 1} \arcsin \relax (x) + \frac {1}{5} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*x^5 - 2/15*x^3 - 1/5*(x^4 - 2*x^2 + 1)*sqrt(-x^2 + 1)*arcsin(x) + 1/5*x

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giac [A]  time = 0.86, size = 34, normalized size = 0.92 \[ \frac {1}{25} \, x^{5} - \frac {1}{5} \, {\left (x^{2} - 1\right )}^{2} \sqrt {-x^{2} + 1} \arcsin \relax (x) - \frac {2}{15} \, x^{3} + \frac {1}{5} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*x^5 - 1/5*(x^2 - 1)^2*sqrt(-x^2 + 1)*arcsin(x) - 2/15*x^3 + 1/5*x

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maple [A]  time = 0.35, size = 37, normalized size = 1.00

method	result	size
default	\(-\frac {\left (x^{2}-1\right )^{2} \sqrt {-x^{2}+1}\, \arcsin \relax (x )}{5}+\frac {\left (3 x^{4}-10 x^{2}+15\right ) x}{75}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-x^2+1)^(3/2)*arcsin(x),x,method=_RETURNVERBOSE)

[Out]

-1/5*(x^2-1)^2*(-x^2+1)^(1/2)*arcsin(x)+1/75*(3*x^4-10*x^2+15)*x

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maxima [A]  time = 0.98, size = 27, normalized size = 0.73 \[ \frac {1}{25} \, x^{5} - \frac {1}{5} \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}} \arcsin \relax (x) - \frac {2}{15} \, x^{3} + \frac {1}{5} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*x^5 - 1/5*(-x^2 + 1)^(5/2)*arcsin(x) - 2/15*x^3 + 1/5*x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int x\,\mathrm {asin}\relax (x)\,{\left (1-x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asin(x)*(1 - x^2)^(3/2),x)

[Out]

int(x*asin(x)*(1 - x^2)^(3/2), x)

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sympy [B]  time = 3.45, size = 63, normalized size = 1.70 \[ \frac {x^{5}}{25} - \frac {x^{4} \sqrt {1 - x^{2}} \operatorname {asin}{\relax (x )}}{5} - \frac {2 x^{3}}{15} + \frac {2 x^{2} \sqrt {1 - x^{2}} \operatorname {asin}{\relax (x )}}{5} + \frac {x}{5} - \frac {\sqrt {1 - x^{2}} \operatorname {asin}{\relax (x )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-x**2+1)**(3/2)*asin(x),x)

[Out]

x**5/25 - x**4*sqrt(1 - x**2)*asin(x)/5 - 2*x**3/15 + 2*x**2*sqrt(1 - x**2)*asin(x)/5 + x/5 - sqrt(1 - x**2)*a
sin(x)/5

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