3.668 \(\int \frac{x^2 \sin ^{-1}(x)^3}{\sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=73 \[ -\frac{3 x^2}{8}-\frac{1}{2} x \sqrt{1-x^2} \sin ^{-1}(x)^3+\frac{3}{4} x^2 \sin ^{-1}(x)^2+\frac{3}{4} x \sqrt{1-x^2} \sin ^{-1}(x)+\frac{1}{8} \sin ^{-1}(x)^4-\frac{3}{8} \sin ^{-1}(x)^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.154527, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {4707, 4641, 4627, 30} \[ -\frac{3 x^2}{8}-\frac{1}{2} x \sqrt{1-x^2} \sin ^{-1}(x)^3+\frac{3}{4} x^2 \sin ^{-1}(x)^2+\frac{3}{4} x \sqrt{1-x^2} \sin ^{-1}(x)+\frac{1}{8} \sin ^{-1}(x)^4-\frac{3}{8} \sin ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 30

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

Rubi steps

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

Mathematica [A]  time = 0.032461, size = 60, normalized size = 0.82 \[ \frac{1}{8} \left (-3 x^2-4 x \sqrt{1-x^2} \sin ^{-1}(x)^3+\left (6 x^2-3\right ) \sin ^{-1}(x)^2+6 x \sqrt{1-x^2} \sin ^{-1}(x)+\sin ^{-1}(x)^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x^2 + 6*x*Sqrt[1 - x^2]*ArcSin[x] + (-3 + 6*x^2)*ArcSin[x]^2 - 4*x*Sqrt[1 - x^2]*ArcSin[x]^3 + ArcSin[x]^4
)/8

________________________________________________________________________________________

Maple [A]  time = 0.053, size = 69, normalized size = 1. \begin{align*}{\frac{ \left ( \arcsin \left ( x \right ) \right ) ^{3}}{2} \left ( -x\sqrt{-{x}^{2}+1}+\arcsin \left ( x \right ) \right ) }+{\frac{3\, \left ( \arcsin \left ( x \right ) \right ) ^{2} \left ({x}^{2}-1 \right ) }{4}}+{\frac{3\,\arcsin \left ( x \right ) }{4} \left ( x\sqrt{-{x}^{2}+1}+\arcsin \left ( x \right ) \right ) }-{\frac{3\, \left ( \arcsin \left ( x \right ) \right ) ^{2}}{8}}-{\frac{3\,{x}^{2}}{8}}-{\frac{3\, \left ( \arcsin \left ( x \right ) \right ) ^{4}}{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \arcsin \left (x\right )^{3}}{\sqrt{-x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsin(x)^3/sqrt(-x^2 + 1), x)

________________________________________________________________________________________

Fricas [A]  time = 2.45903, size = 151, normalized size = 2.07 \begin{align*} \frac{1}{8} \, \arcsin \left (x\right )^{4} + \frac{3}{8} \,{\left (2 \, x^{2} - 1\right )} \arcsin \left (x\right )^{2} - \frac{3}{8} \, x^{2} - \frac{1}{4} \,{\left (2 \, x \arcsin \left (x\right )^{3} - 3 \, x \arcsin \left (x\right )\right )} \sqrt{-x^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.27156, size = 66, normalized size = 0.9 \begin{align*} \frac{3 x^{2} \operatorname{asin}^{2}{\left (x \right )}}{4} - \frac{3 x^{2}}{8} - \frac{x \sqrt{1 - x^{2}} \operatorname{asin}^{3}{\left (x \right )}}{2} + \frac{3 x \sqrt{1 - x^{2}} \operatorname{asin}{\left (x \right )}}{4} + \frac{\operatorname{asin}^{4}{\left (x \right )}}{8} - \frac{3 \operatorname{asin}^{2}{\left (x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.11686, size = 81, normalized size = 1.11 \begin{align*} -\frac{1}{2} \, \sqrt{-x^{2} + 1} x \arcsin \left (x\right )^{3} + \frac{1}{8} \, \arcsin \left (x\right )^{4} + \frac{3}{4} \,{\left (x^{2} - 1\right )} \arcsin \left (x\right )^{2} + \frac{3}{4} \, \sqrt{-x^{2} + 1} x \arcsin \left (x\right ) - \frac{3}{8} \, x^{2} + \frac{3}{8} \, \arcsin \left (x\right )^{2} + \frac{3}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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