\(\int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx\) [668]

Optimal result

Integrand size = 19, antiderivative size = 73 \[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=-\frac {3 x^2}{8}+\frac {3}{4} x \sqrt {1-x^2} \arcsin (x)-\frac {3 \arcsin (x)^2}{8}+\frac {3}{4} x^2 \arcsin (x)^2-\frac {1}{2} x \sqrt {1-x^2} \arcsin (x)^3+\frac {\arcsin (x)^4}{8} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {4795, 4737, 4723, 30} \[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=-\frac {1}{2} x \sqrt {1-x^2} \arcsin (x)^3+\frac {3}{4} x^2 \arcsin (x)^2+\frac {3}{4} x \sqrt {1-x^2} \arcsin (x)+\frac {\arcsin (x)^4}{8}-\frac {3 \arcsin (x)^2}{8}-\frac {3 x^2}{8} \]

[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 30

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

Rule 4723

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 4737

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

Rule 4795

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=\frac {1}{8} \left (-3 x^2+6 x \sqrt {1-x^2} \arcsin (x)+\left (-3+6 x^2\right ) \arcsin (x)^2-4 x \sqrt {1-x^2} \arcsin (x)^3+\arcsin (x)^4\right ) \]

[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] (verified)

Time = 0.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {\arcsin \left (x \right )^{3} \left (-x \sqrt {-x^{2}+1}+\arcsin \left (x \right )\right )}{2}+\frac {3 \arcsin \left (x \right )^{2} \left (x^{2}-1\right )}{4}+\frac {3 \arcsin \left (x \right ) \left (x \sqrt {-x^{2}+1}+\arcsin \left (x \right )\right )}{4}-\frac {3 \arcsin \left (x \right )^{2}}{8}-\frac {3 x^{2}}{8}-\frac {3 \arcsin \left (x \right )^{4}}{8}\)	\(69\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=\frac {1}{8} \, \arcsin \left (x\right )^{4} + \frac {3}{8} \, {\left (2 \, x^{2} - 1\right )} \arcsin \left (x\right )^{2} - \frac {3}{8} \, x^{2} - \frac {1}{4} \, {\left (2 \, x \arcsin \left (x\right )^{3} - 3 \, x \arcsin \left (x\right )\right )} \sqrt {-x^{2} + 1} \]

[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] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=\frac {3 x^{2} \operatorname {asin}^{2}{\left (x \right )}}{4} - \frac {3 x^{2}}{8} - \frac {x \sqrt {1 - x^{2}} \operatorname {asin}^{3}{\left (x \right )}}{2} + \frac {3 x \sqrt {1 - x^{2}} \operatorname {asin}{\left (x \right )}}{4} + \frac {\operatorname {asin}^{4}{\left (x \right )}}{8} - \frac {3 \operatorname {asin}^{2}{\left (x \right )}}{8} \]

[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

Maxima [F]

\[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=\int { \frac {x^{2} \arcsin \left (x\right )^{3}}{\sqrt {-x^{2} + 1}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=-\frac {1}{2} \, \sqrt {-x^{2} + 1} x \arcsin \left (x\right )^{3} + \frac {1}{8} \, \arcsin \left (x\right )^{4} + \frac {3}{4} \, {\left (x^{2} - 1\right )} \arcsin \left (x\right )^{2} + \frac {3}{4} \, \sqrt {-x^{2} + 1} x \arcsin \left (x\right ) - \frac {3}{8} \, x^{2} + \frac {3}{8} \, \arcsin \left (x\right )^{2} + \frac {3}{16} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \arcsin (x)^3}{\sqrt {1-x^2}} \, dx=\int \frac {x^2\,{\mathrm {asin}\left (x\right )}^3}{\sqrt {1-x^2}} \,d x \]

[In]

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

[Out]

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