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

Optimal. Leaf size=59 \[ \frac {x^4}{16}-\frac {5 x^2}{16}+\frac {1}{4} \left (1-x^2\right )^{3/2} x \sin ^{-1}(x)+\frac {3}{8} \sqrt {1-x^2} x \sin ^{-1}(x)+\frac {3}{16} \sin ^{-1}(x)^2 \]

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Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4649, 4647, 4641, 30, 14} \[ \frac {x^4}{16}-\frac {5 x^2}{16}+\frac {1}{4} \left (1-x^2\right )^{3/2} x \sin ^{-1}(x)+\frac {3}{8} \sqrt {1-x^2} x \sin ^{-1}(x)+\frac {3}{16} \sin ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[(1 - x^2)^(3/2)*ArcSin[x],x]

[Out]

(-5*x^2)/16 + x^4/16 + (3*x*Sqrt[1 - x^2]*ArcSin[x])/8 + (x*(1 - x^2)^(3/2)*ArcSin[x])/4 + (3*ArcSin[x]^2)/16

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4641

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

Rule 4647

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

Rule 4649

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 42, normalized size = 0.71 \[ \frac {1}{16} \left (x^4-5 x^2-2 \sqrt {1-x^2} \left (2 x^2-5\right ) x \sin ^{-1}(x)+3 \sin ^{-1}(x)^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x^2 + x^4 - 2*x*Sqrt[1 - x^2]*(-5 + 2*x^2)*ArcSin[x] + 3*ArcSin[x]^2)/16

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

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.21, size = 39, normalized size = 0.66 \[ \frac {1}{16} \, x^{4} - \frac {1}{8} \, {\left (2 \, x^{3} - 5 \, x\right )} \sqrt {-x^{2} + 1} \arcsin \relax (x) - \frac {5}{16} \, x^{2} + \frac {3}{16} \, \arcsin \relax (x)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^4 - 1/8*(2*x^3 - 5*x)*sqrt(-x^2 + 1)*arcsin(x) - 5/16*x^2 + 3/16*arcsin(x)^2

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giac [A]  time = 1.05, size = 50, normalized size = 0.85 \[ \frac {1}{4} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x \arcsin \relax (x) + \frac {3}{8} \, \sqrt {-x^{2} + 1} x \arcsin \relax (x) + \frac {1}{16} \, {\left (x^{2} - 1\right )}^{2} - \frac {3}{16} \, x^{2} + \frac {3}{16} \, \arcsin \relax (x)^{2} + \frac {9}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-x^2 + 1)^(3/2)*x*arcsin(x) + 3/8*sqrt(-x^2 + 1)*x*arcsin(x) + 1/16*(x^2 - 1)^2 - 3/16*x^2 + 3/16*arcsin(
x)^2 + 9/128

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maple [A]  time = 0.37, size = 58, normalized size = 0.98

method	result	size
default	\(\frac {\arcsin \relax (x ) \left (-2 \sqrt {-x^{2}+1}\, x^{3}+5 \sqrt {-x^{2}+1}\, x +3 \arcsin \relax (x )\right )}{8}-\frac {3 \arcsin \relax (x )^{2}}{16}+\frac {\left (x^{2}-1\right )^{2}}{16}-\frac {3 x^{2}}{16}+\frac {3}{16}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arcsin(x)*(-2*(-x^2+1)^(1/2)*x^3+5*(-x^2+1)^(1/2)*x+3*arcsin(x))-3/16*arcsin(x)^2+1/16*(x^2-1)^2-3/16*x^2+
3/16

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maxima [A]  time = 1.02, size = 50, normalized size = 0.85 \[ \frac {1}{16} \, x^{4} - \frac {5}{16} \, x^{2} + \frac {1}{8} \, {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x + 3 \, \sqrt {-x^{2} + 1} x + 3 \, \arcsin \relax (x)\right )} \arcsin \relax (x) - \frac {3}{16} \, \arcsin \relax (x)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^4 - 5/16*x^2 + 1/8*(2*(-x^2 + 1)^(3/2)*x + 3*sqrt(-x^2 + 1)*x + 3*arcsin(x))*arcsin(x) - 3/16*arcsin(x)
^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {asin}\relax (x)\,{\left (1-x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x)*(1 - x^2)^(3/2),x)

[Out]

int(asin(x)*(1 - x^2)^(3/2), x)

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sympy [A]  time = 2.08, size = 53, normalized size = 0.90 \[ \frac {x^{4}}{16} - \frac {x^{3} \sqrt {1 - x^{2}} \operatorname {asin}{\relax (x )}}{4} - \frac {5 x^{2}}{16} + \frac {5 x \sqrt {1 - x^{2}} \operatorname {asin}{\relax (x )}}{8} + \frac {3 \operatorname {asin}^{2}{\relax (x )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4/16 - x**3*sqrt(1 - x**2)*asin(x)/4 - 5*x**2/16 + 5*x*sqrt(1 - x**2)*asin(x)/8 + 3*asin(x)**2/16

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