3.646 \(\int x^2 \sin ^{-1}(x)^2 \, dx\)

Optimal. Leaf size=61 \[ -\frac{2 x^3}{27}+\frac{1}{3} x^3 \sin ^{-1}(x)^2+\frac{2}{9} \sqrt{1-x^2} x^2 \sin ^{-1}(x)+\frac{4}{9} \sqrt{1-x^2} \sin ^{-1}(x)-\frac{4 x}{9} \]

[Out]

(-4*x)/9 - (2*x^3)/27 + (4*Sqrt[1 - x^2]*ArcSin[x])/9 + (2*x^2*Sqrt[1 - x^2]*ArcSin[x])/9 + (x^3*ArcSin[x]^2)/
3

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Rubi [A]  time = 0.0938961, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4627, 4707, 4677, 8, 30} \[ -\frac{2 x^3}{27}+\frac{1}{3} x^3 \sin ^{-1}(x)^2+\frac{2}{9} \sqrt{1-x^2} x^2 \sin ^{-1}(x)+\frac{4}{9} \sqrt{1-x^2} \sin ^{-1}(x)-\frac{4 x}{9} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSin[x]^2,x]

[Out]

(-4*x)/9 - (2*x^3)/27 + (4*Sqrt[1 - x^2]*ArcSin[x])/9 + (2*x^2*Sqrt[1 - x^2]*ArcSin[x])/9 + (x^3*ArcSin[x]^2)/
3

Rule 4627

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

Rule 4707

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

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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rubi steps

\begin{align*} \int x^2 \sin ^{-1}(x)^2 \, dx &=\frac{1}{3} x^3 \sin ^{-1}(x)^2-\frac{2}{3} \int \frac{x^3 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx\\ &=\frac{2}{9} x^2 \sqrt{1-x^2} \sin ^{-1}(x)+\frac{1}{3} x^3 \sin ^{-1}(x)^2-\frac{2 \int x^2 \, dx}{9}-\frac{4}{9} \int \frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx\\ &=-\frac{2 x^3}{27}+\frac{4}{9} \sqrt{1-x^2} \sin ^{-1}(x)+\frac{2}{9} x^2 \sqrt{1-x^2} \sin ^{-1}(x)+\frac{1}{3} x^3 \sin ^{-1}(x)^2-\frac{4 \int 1 \, dx}{9}\\ &=-\frac{4 x}{9}-\frac{2 x^3}{27}+\frac{4}{9} \sqrt{1-x^2} \sin ^{-1}(x)+\frac{2}{9} x^2 \sqrt{1-x^2} \sin ^{-1}(x)+\frac{1}{3} x^3 \sin ^{-1}(x)^2\\ \end{align*}

Mathematica [A]  time = 0.0222281, size = 42, normalized size = 0.69 \[ \frac{1}{27} \left (-2 \left (x^2+6\right ) x+9 x^3 \sin ^{-1}(x)^2+6 \sqrt{1-x^2} \left (x^2+2\right ) \sin ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSin[x]^2,x]

[Out]

(-2*x*(6 + x^2) + 6*Sqrt[1 - x^2]*(2 + x^2)*ArcSin[x] + 9*x^3*ArcSin[x]^2)/27

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Maple [A]  time = 0.023, size = 37, normalized size = 0.6 \begin{align*}{\frac{{x}^{3} \left ( \arcsin \left ( x \right ) \right ) ^{2}}{3}}+{\frac{2\,\arcsin \left ( x \right ) \left ({x}^{2}+2 \right ) }{9}\sqrt{-{x}^{2}+1}}-{\frac{2\,{x}^{3}}{27}}-{\frac{4\,x}{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsin(x)^2,x)

[Out]

1/3*x^3*arcsin(x)^2+2/9*arcsin(x)*(x^2+2)*(-x^2+1)^(1/2)-2/27*x^3-4/9*x

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Maxima [A]  time = 1.44083, size = 63, normalized size = 1.03 \begin{align*} \frac{1}{3} \, x^{3} \arcsin \left (x\right )^{2} - \frac{2}{27} \, x^{3} + \frac{2}{9} \,{\left (\sqrt{-x^{2} + 1} x^{2} + 2 \, \sqrt{-x^{2} + 1}\right )} \arcsin \left (x\right ) - \frac{4}{9} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsin(x)^2,x, algorithm="maxima")

[Out]

1/3*x^3*arcsin(x)^2 - 2/27*x^3 + 2/9*(sqrt(-x^2 + 1)*x^2 + 2*sqrt(-x^2 + 1))*arcsin(x) - 4/9*x

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Fricas [A]  time = 2.44157, size = 109, normalized size = 1.79 \begin{align*} \frac{1}{3} \, x^{3} \arcsin \left (x\right )^{2} - \frac{2}{27} \, x^{3} + \frac{2}{9} \,{\left (x^{2} + 2\right )} \sqrt{-x^{2} + 1} \arcsin \left (x\right ) - \frac{4}{9} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsin(x)^2,x, algorithm="fricas")

[Out]

1/3*x^3*arcsin(x)^2 - 2/27*x^3 + 2/9*(x^2 + 2)*sqrt(-x^2 + 1)*arcsin(x) - 4/9*x

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Sympy [A]  time = 0.597459, size = 54, normalized size = 0.89 \begin{align*} \frac{x^{3} \operatorname{asin}^{2}{\left (x \right )}}{3} - \frac{2 x^{3}}{27} + \frac{2 x^{2} \sqrt{1 - x^{2}} \operatorname{asin}{\left (x \right )}}{9} - \frac{4 x}{9} + \frac{4 \sqrt{1 - x^{2}} \operatorname{asin}{\left (x \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asin(x)**2,x)

[Out]

x**3*asin(x)**2/3 - 2*x**3/27 + 2*x**2*sqrt(1 - x**2)*asin(x)/9 - 4*x/9 + 4*sqrt(1 - x**2)*asin(x)/9

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Giac [A]  time = 1.07623, size = 77, normalized size = 1.26 \begin{align*} \frac{1}{3} \,{\left (x^{2} - 1\right )} x \arcsin \left (x\right )^{2} + \frac{1}{3} \, x \arcsin \left (x\right )^{2} - \frac{2}{9} \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}} \arcsin \left (x\right ) - \frac{2}{27} \,{\left (x^{2} - 1\right )} x + \frac{2}{3} \, \sqrt{-x^{2} + 1} \arcsin \left (x\right ) - \frac{14}{27} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsin(x)^2,x, algorithm="giac")

[Out]

1/3*(x^2 - 1)*x*arcsin(x)^2 + 1/3*x*arcsin(x)^2 - 2/9*(-x^2 + 1)^(3/2)*arcsin(x) - 2/27*(x^2 - 1)*x + 2/3*sqrt
(-x^2 + 1)*arcsin(x) - 14/27*x