Integrand size = 14, antiderivative size = 59 \[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=-\frac {5 x^2}{16}+\frac {x^4}{16}+\frac {3}{8} x \sqrt {1-x^2} \arcsin (x)+\frac {1}{4} x \left (1-x^2\right )^{3/2} \arcsin (x)+\frac {3 \arcsin (x)^2}{16} \] Output:
-5/16*x^2+1/16*x^4+1/4*x*(-x^2+1)^(3/2)*arcsin(x)+3/16*arcsin(x)^2+3/8*x*a rcsin(x)*(-x^2+1)^(1/2)
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=\frac {1}{16} \left (-5 x^2+x^4-2 x \sqrt {1-x^2} \left (-5+2 x^2\right ) \arcsin (x)+3 \arcsin (x)^2\right ) \] Input:
Integrate[(1 - x^2)^(3/2)*ArcSin[x],x]
Output:
(-5*x^2 + x^4 - 2*x*Sqrt[1 - x^2]*(-5 + 2*x^2)*ArcSin[x] + 3*ArcSin[x]^2)/ 16
Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5158, 244, 2009, 5156, 15, 5152}
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.
\(\displaystyle \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx\) |
\(\Big \downarrow \) 5158 |
\(\displaystyle \frac {3}{4} \int \sqrt {1-x^2} \arcsin (x)dx-\frac {1}{4} \int x \left (1-x^2\right )dx+\frac {1}{4} x \left (1-x^2\right )^{3/2} \arcsin (x)\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {3}{4} \int \sqrt {1-x^2} \arcsin (x)dx-\frac {1}{4} \int \left (x-x^3\right )dx+\frac {1}{4} x \left (1-x^2\right )^{3/2} \arcsin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{4} \int \sqrt {1-x^2} \arcsin (x)dx+\frac {1}{4} x \left (1-x^2\right )^{3/2} \arcsin (x)+\frac {1}{4} \left (\frac {x^4}{4}-\frac {x^2}{2}\right )\) |
\(\Big \downarrow \) 5156 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {\arcsin (x)}{\sqrt {1-x^2}}dx-\frac {\int xdx}{2}+\frac {1}{2} x \sqrt {1-x^2} \arcsin (x)\right )+\frac {1}{4} x \left (1-x^2\right )^{3/2} \arcsin (x)+\frac {1}{4} \left (\frac {x^4}{4}-\frac {x^2}{2}\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {\arcsin (x)}{\sqrt {1-x^2}}dx+\frac {1}{2} \sqrt {1-x^2} x \arcsin (x)-\frac {x^2}{4}\right )+\frac {1}{4} x \left (1-x^2\right )^{3/2} \arcsin (x)+\frac {1}{4} \left (\frac {x^4}{4}-\frac {x^2}{2}\right )\) |
\(\Big \downarrow \) 5152 |
\(\displaystyle \frac {1}{4} x \left (1-x^2\right )^{3/2} \arcsin (x)+\frac {3}{4} \left (\frac {1}{2} \sqrt {1-x^2} x \arcsin (x)+\frac {\arcsin (x)^2}{4}-\frac {x^2}{4}\right )+\frac {1}{4} \left (\frac {x^4}{4}-\frac {x^2}{2}\right )\) |
Input:
Int[(1 - x^2)^(3/2)*ArcSin[x],x]
Output:
(-1/2*x^2 + x^4/4)/4 + (x*(1 - x^2)^(3/2)*ArcSin[x])/4 + (3*(-1/4*x^2 + (x *Sqrt[1 - x^2]*ArcSin[x])/2 + ArcSin[x]^2/4))/4
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[(1/(b*c*(n + 1)))*Simp[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]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSin[c*x])^n/2), x] + (Simp[(1/2 )*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int[(a + b*ArcSin[c*x])^n/Sqrt[ 1 - c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^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]
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] + (S imp[2*d*(p/(2*p + 1)) Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Simp[b*c*(n/(2*p + 1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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] && GtQ[n, 0] && GtQ[p, 0]
Time = 0.33 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {\arcsin \left (x \right ) \left (-2 \sqrt {-x^{2}+1}\, x^{3}+5 x \sqrt {-x^{2}+1}+3 \arcsin \left (x \right )\right )}{8}-\frac {3 \arcsin \left (x \right )^{2}}{16}+\frac {\left (2 x^{2}-5\right )^{2}}{64}\) | \(54\) |
Input:
int((-x^2+1)^(3/2)*arcsin(x),x,method=_RETURNVERBOSE)
Output:
1/8*arcsin(x)*(-2*(-x^2+1)^(1/2)*x^3+5*x*(-x^2+1)^(1/2)+3*arcsin(x))-3/16* arcsin(x)^2+1/64*(2*x^2-5)^2
Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66 \[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=\frac {1}{16} \, x^{4} - \frac {1}{8} \, {\left (2 \, x^{3} - 5 \, x\right )} \sqrt {-x^{2} + 1} \arcsin \left (x\right ) - \frac {5}{16} \, x^{2} + \frac {3}{16} \, \arcsin \left (x\right )^{2} \] Input:
integrate((-x^2+1)^(3/2)*arcsin(x),x, algorithm="fricas")
Output:
1/16*x^4 - 1/8*(2*x^3 - 5*x)*sqrt(-x^2 + 1)*arcsin(x) - 5/16*x^2 + 3/16*ar csin(x)^2
Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=\frac {x^{4}}{16} - \frac {x^{3} \sqrt {1 - x^{2}} \operatorname {asin}{\left (x \right )}}{4} - \frac {5 x^{2}}{16} + \frac {5 x \sqrt {1 - x^{2}} \operatorname {asin}{\left (x \right )}}{8} + \frac {3 \operatorname {asin}^{2}{\left (x \right )}}{16} \] Input:
integrate((-x**2+1)**(3/2)*asin(x),x)
Output:
x**4/16 - x**3*sqrt(1 - x**2)*asin(x)/4 - 5*x**2/16 + 5*x*sqrt(1 - x**2)*a sin(x)/8 + 3*asin(x)**2/16
Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=\frac {1}{16} \, x^{4} - \frac {5}{16} \, x^{2} + \frac {1}{8} \, {\left (2 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x + 3 \, \sqrt {-x^{2} + 1} x + 3 \, \arcsin \left (x\right )\right )} \arcsin \left (x\right ) - \frac {3}{16} \, \arcsin \left (x\right )^{2} \] Input:
integrate((-x^2+1)^(3/2)*arcsin(x),x, algorithm="maxima")
Output:
1/16*x^4 - 5/16*x^2 + 1/8*(2*(-x^2 + 1)^(3/2)*x + 3*sqrt(-x^2 + 1)*x + 3*a rcsin(x))*arcsin(x) - 3/16*arcsin(x)^2
Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=\frac {1}{4} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x \arcsin \left (x\right ) + \frac {3}{8} \, \sqrt {-x^{2} + 1} x \arcsin \left (x\right ) + \frac {1}{16} \, {\left (x^{2} - 1\right )}^{2} - \frac {3}{16} \, x^{2} + \frac {3}{16} \, \arcsin \left (x\right )^{2} + \frac {9}{128} \] Input:
integrate((-x^2+1)^(3/2)*arcsin(x),x, algorithm="giac")
Output:
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
Timed out. \[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=\int \mathrm {asin}\left (x\right )\,{\left (1-x^2\right )}^{3/2} \,d x \] Input:
int(asin(x)*(1 - x^2)^(3/2),x)
Output:
int(asin(x)*(1 - x^2)^(3/2), x)
\[ \int \left (1-x^2\right )^{3/2} \arcsin (x) \, dx=-\left (\int \sqrt {-x^{2}+1}\, \mathit {asin} \left (x \right ) x^{2}d x \right )+\int \sqrt {-x^{2}+1}\, \mathit {asin} \left (x \right )d x \] Input:
int((-x^2+1)^(3/2)*asin(x),x)
Output:
- int(sqrt( - x**2 + 1)*asin(x)*x**2,x) + int(sqrt( - x**2 + 1)*asin(x),x )