Integrand size = 8, antiderivative size = 61 \[ \int x^2 \arcsin (x)^2 \, dx=-\frac {4 x}{9}-\frac {2 x^3}{27}+\frac {4}{9} \sqrt {1-x^2} \arcsin (x)+\frac {2}{9} x^2 \sqrt {1-x^2} \arcsin (x)+\frac {1}{3} x^3 \arcsin (x)^2 \] Output:
-4/9*x-2/27*x^3+1/3*x^3*arcsin(x)^2+4/9*arcsin(x)*(-x^2+1)^(1/2)+2/9*x^2*a rcsin(x)*(-x^2+1)^(1/2)
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69 \[ \int x^2 \arcsin (x)^2 \, dx=\frac {1}{27} \left (-2 x \left (6+x^2\right )+6 \sqrt {1-x^2} \left (2+x^2\right ) \arcsin (x)+9 x^3 \arcsin (x)^2\right ) \] Input:
Integrate[x^2*ArcSin[x]^2,x]
Output:
(-2*x*(6 + x^2) + 6*Sqrt[1 - x^2]*(2 + x^2)*ArcSin[x] + 9*x^3*ArcSin[x]^2) /27
Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5138, 5210, 15, 5182, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \arcsin (x)^2 \, dx\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {1}{3} x^3 \arcsin (x)^2-\frac {2}{3} \int \frac {x^3 \arcsin (x)}{\sqrt {1-x^2}}dx\) |
\(\Big \downarrow \) 5210 |
\(\displaystyle \frac {1}{3} x^3 \arcsin (x)^2-\frac {2}{3} \left (\frac {2}{3} \int \frac {x \arcsin (x)}{\sqrt {1-x^2}}dx+\frac {\int x^2dx}{3}-\frac {1}{3} \sqrt {1-x^2} x^2 \arcsin (x)\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{3} x^3 \arcsin (x)^2-\frac {2}{3} \left (\frac {2}{3} \int \frac {x \arcsin (x)}{\sqrt {1-x^2}}dx-\frac {1}{3} \sqrt {1-x^2} x^2 \arcsin (x)+\frac {x^3}{9}\right )\) |
\(\Big \downarrow \) 5182 |
\(\displaystyle \frac {1}{3} x^3 \arcsin (x)^2-\frac {2}{3} \left (\frac {2}{3} \left (\int 1dx-\sqrt {1-x^2} \arcsin (x)\right )-\frac {1}{3} \sqrt {1-x^2} x^2 \arcsin (x)+\frac {x^3}{9}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{3} x^3 \arcsin (x)^2-\frac {2}{3} \left (-\frac {1}{3} \sqrt {1-x^2} x^2 \arcsin (x)+\frac {2}{3} \left (x-\sqrt {1-x^2} \arcsin (x)\right )+\frac {x^3}{9}\right )\) |
Input:
Int[x^2*ArcSin[x]^2,x]
Output:
(x^3*ArcSin[x]^2)/3 - (2*(x^3/9 - (x^2*Sqrt[1 - x^2]*ArcSin[x])/3 + (2*(x - Sqrt[1 - x^2]*ArcSin[x]))/3))/3
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[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]
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] + Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] I nt[(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]
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] + (Simp[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] + S imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(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]
Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.61
method | result | size |
default | \(\frac {x^{3} \arcsin \left (x \right )^{2}}{3}+\frac {2 \arcsin \left (x \right ) \left (x^{2}+2\right ) \sqrt {-x^{2}+1}}{9}-\frac {2 x^{3}}{27}-\frac {4 x}{9}\) | \(37\) |
orering | \(\frac {\left (19 x^{4}+24 x^{2}-48\right ) \arcsin \left (x \right )^{2}}{27 x}-\frac {\left (6 x^{4}+17 x^{2}-30\right ) \left (2 x \arcsin \left (x \right )^{2}+\frac {2 x^{2} \arcsin \left (x \right )}{\sqrt {-x^{2}+1}}\right )}{27 x^{2}}+\frac {\left (x^{2}+6\right ) \left (-1+x \right ) \left (1+x \right ) \left (2 \arcsin \left (x \right )^{2}+\frac {8 x \arcsin \left (x \right )}{\sqrt {-x^{2}+1}}+\frac {2 x^{2}}{-x^{2}+1}+\frac {2 x^{3} \arcsin \left (x \right )}{\left (-x^{2}+1\right )^{\frac {3}{2}}}\right )}{27 x}\) | \(131\) |
Input:
int(x^2*arcsin(x)^2,x,method=_RETURNVERBOSE)
Output:
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
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.59 \[ \int x^2 \arcsin (x)^2 \, dx=\frac {1}{3} \, x^{3} \arcsin \left (x\right )^{2} - \frac {2}{27} \, x^{3} + \frac {2}{9} \, {\left (x^{2} + 2\right )} \sqrt {-x^{2} + 1} \arcsin \left (x\right ) - \frac {4}{9} \, x \] Input:
integrate(x^2*arcsin(x)^2,x, algorithm="fricas")
Output:
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
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int x^2 \arcsin (x)^2 \, dx=\frac {x^{3} \operatorname {asin}^{2}{\left (x \right )}}{3} - \frac {2 x^{3}}{27} + \frac {2 x^{2} \sqrt {1 - x^{2}} \operatorname {asin}{\left (x \right )}}{9} - \frac {4 x}{9} + \frac {4 \sqrt {1 - x^{2}} \operatorname {asin}{\left (x \right )}}{9} \] Input:
integrate(x**2*asin(x)**2,x)
Output:
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
Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int x^2 \arcsin (x)^2 \, dx=\frac {1}{3} \, x^{3} \arcsin \left (x\right )^{2} - \frac {2}{27} \, x^{3} + \frac {2}{9} \, {\left (\sqrt {-x^{2} + 1} x^{2} + 2 \, \sqrt {-x^{2} + 1}\right )} \arcsin \left (x\right ) - \frac {4}{9} \, x \] Input:
integrate(x^2*arcsin(x)^2,x, algorithm="maxima")
Output:
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
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int x^2 \arcsin (x)^2 \, dx=\frac {1}{3} \, {\left (x^{2} - 1\right )} x \arcsin \left (x\right )^{2} + \frac {1}{3} \, x \arcsin \left (x\right )^{2} - \frac {2}{9} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} \arcsin \left (x\right ) - \frac {2}{27} \, {\left (x^{2} - 1\right )} x + \frac {2}{3} \, \sqrt {-x^{2} + 1} \arcsin \left (x\right ) - \frac {14}{27} \, x \] Input:
integrate(x^2*arcsin(x)^2,x, algorithm="giac")
Output:
1/3*(x^2 - 1)*x*arcsin(x)^2 + 1/3*x*arcsin(x)^2 - 2/9*(-x^2 + 1)^(3/2)*arc sin(x) - 2/27*(x^2 - 1)*x + 2/3*sqrt(-x^2 + 1)*arcsin(x) - 14/27*x
Timed out. \[ \int x^2 \arcsin (x)^2 \, dx=\int x^2\,{\mathrm {asin}\left (x\right )}^2 \,d x \] Input:
int(x^2*asin(x)^2,x)
Output:
int(x^2*asin(x)^2, x)
\[ \int x^2 \arcsin (x)^2 \, dx=\int \mathit {asin} \left (x \right )^{2} x^{2}d x \] Input:
int(x^2*asin(x)^2,x)
Output:
int(asin(x)**2*x**2,x)