Integrand size = 19, antiderivative size = 65 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=\frac {7}{8} x \sqrt {1-x^2}+\frac {7}{60} (8-3 x) \left (1-x^2\right )^{3/2}+\frac {1}{5} (1-x)^2 \left (1-x^2\right )^{3/2}+\frac {7 \arcsin (x)}{8} \] Output:
7/8*x*(-x^2+1)^(1/2)+7/60*(8-3*x)*(-x^2+1)^(3/2)+1/5*(1-x)^2*(-x^2+1)^(3/2 )+7/8*arcsin(x)
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=-\frac {1}{120} \sqrt {1-x^2} \left (-136-15 x+112 x^2-90 x^3+24 x^4\right )-\frac {7}{4} \arctan \left (\frac {\sqrt {1-x^2}}{1+x}\right ) \] Input:
Integrate[(1 - x)^3*Sqrt[1 - x^2],x]
Output:
-1/120*(Sqrt[1 - x^2]*(-136 - 15*x + 112*x^2 - 90*x^3 + 24*x^4)) - (7*ArcT an[Sqrt[1 - x^2]/(1 + x)])/4
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {469, 469, 455, 211, 223}
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.
\(\displaystyle \int (1-x)^3 \sqrt {1-x^2} \, dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {7}{5} \int (1-x)^2 \sqrt {1-x^2}dx+\frac {1}{5} \left (1-x^2\right )^{3/2} (1-x)^2\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \int (1-x) \sqrt {1-x^2}dx+\frac {1}{4} (1-x) \left (1-x^2\right )^{3/2}\right )+\frac {1}{5} \left (1-x^2\right )^{3/2} (1-x)^2\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \left (\int \sqrt {1-x^2}dx+\frac {1}{3} \left (1-x^2\right )^{3/2}\right )+\frac {1}{4} (1-x) \left (1-x^2\right )^{3/2}\right )+\frac {1}{5} \left (1-x^2\right )^{3/2} (1-x)^2\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}}dx+\frac {1}{3} \left (1-x^2\right )^{3/2}+\frac {1}{2} x \sqrt {1-x^2}\right )+\frac {1}{4} (1-x) \left (1-x^2\right )^{3/2}\right )+\frac {1}{5} \left (1-x^2\right )^{3/2} (1-x)^2\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7}{5} \left (\frac {5}{4} \left (\frac {\arcsin (x)}{2}+\frac {1}{3} \left (1-x^2\right )^{3/2}+\frac {1}{2} x \sqrt {1-x^2}\right )+\frac {1}{4} (1-x) \left (1-x^2\right )^{3/2}\right )+\frac {1}{5} \left (1-x^2\right )^{3/2} (1-x)^2\) |
Input:
Int[(1 - x)^3*Sqrt[1 - x^2],x]
Output:
((1 - x)^2*(1 - x^2)^(3/2))/5 + (7*(((1 - x)*(1 - x^2)^(3/2))/4 + (5*((x*S qrt[1 - x^2])/2 + (1 - x^2)^(3/2)/3 + ArcSin[x]/2))/4))/5
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* ((n + p)/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* p + 1, 0] && IntegerQ[2*p]
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65
method | result | size |
risch | \(\frac {\left (24 x^{4}-90 x^{3}+112 x^{2}-15 x -136\right ) \left (x^{2}-1\right )}{120 \sqrt {-x^{2}+1}}+\frac {7 \arcsin \left (x \right )}{8}\) | \(42\) |
default | \(\frac {x^{2} \left (-x^{2}+1\right )^{\frac {3}{2}}}{5}+\frac {17 \left (-x^{2}+1\right )^{\frac {3}{2}}}{15}+\frac {7 x \sqrt {-x^{2}+1}}{8}+\frac {7 \arcsin \left (x \right )}{8}-\frac {3 x \left (-x^{2}+1\right )^{\frac {3}{2}}}{4}\) | \(55\) |
trager | \(\left (-\frac {1}{5} x^{4}+\frac {3}{4} x^{3}-\frac {14}{15} x^{2}+\frac {1}{8} x +\frac {17}{15}\right ) \sqrt {-x^{2}+1}+\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )}{8}\) | \(59\) |
meijerg | \(\frac {i \left (-2 i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-2 i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{4 \sqrt {\pi }}-\frac {3 \left (\frac {4 \sqrt {\pi }}{3}-\frac {2 \sqrt {\pi }\, \left (-2 x^{2}+2\right ) \sqrt {-x^{2}+1}}{3}\right )}{4 \sqrt {\pi }}-\frac {3 i \left (-\frac {i \sqrt {\pi }\, x \left (-6 x^{2}+3\right ) \sqrt {-x^{2}+1}}{6}+\frac {i \sqrt {\pi }\, \arcsin \left (x \right )}{2}\right )}{4 \sqrt {\pi }}+\frac {-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (-x^{2}+1\right )^{\frac {3}{2}} \left (3 x^{2}+2\right )}{15}}{4 \sqrt {\pi }}\) | \(135\) |
Input:
int((1-x)^3*(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/120*(24*x^4-90*x^3+112*x^2-15*x-136)*(x^2-1)/(-x^2+1)^(1/2)+7/8*arcsin(x )
Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=-\frac {1}{120} \, {\left (24 \, x^{4} - 90 \, x^{3} + 112 \, x^{2} - 15 \, x - 136\right )} \sqrt {-x^{2} + 1} - \frac {7}{4} \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \] Input:
integrate((1-x)^3*(-x^2+1)^(1/2),x, algorithm="fricas")
Output:
-1/120*(24*x^4 - 90*x^3 + 112*x^2 - 15*x - 136)*sqrt(-x^2 + 1) - 7/4*arcta n((sqrt(-x^2 + 1) - 1)/x)
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=- \frac {x^{4} \sqrt {1 - x^{2}}}{5} + \frac {3 x^{3} \sqrt {1 - x^{2}}}{4} - \frac {14 x^{2} \sqrt {1 - x^{2}}}{15} + \frac {x \sqrt {1 - x^{2}}}{8} + \frac {17 \sqrt {1 - x^{2}}}{15} + \frac {7 \operatorname {asin}{\left (x \right )}}{8} \] Input:
integrate((1-x)**3*(-x**2+1)**(1/2),x)
Output:
-x**4*sqrt(1 - x**2)/5 + 3*x**3*sqrt(1 - x**2)/4 - 14*x**2*sqrt(1 - x**2)/ 15 + x*sqrt(1 - x**2)/8 + 17*sqrt(1 - x**2)/15 + 7*asin(x)/8
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=\frac {1}{5} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x^{2} - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x + \frac {17}{15} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} + \frac {7}{8} \, \sqrt {-x^{2} + 1} x + \frac {7}{8} \, \arcsin \left (x\right ) \] Input:
integrate((1-x)^3*(-x^2+1)^(1/2),x, algorithm="maxima")
Output:
1/5*(-x^2 + 1)^(3/2)*x^2 - 3/4*(-x^2 + 1)^(3/2)*x + 17/15*(-x^2 + 1)^(3/2) + 7/8*sqrt(-x^2 + 1)*x + 7/8*arcsin(x)
Time = 0.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.54 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=-\frac {1}{120} \, {\left ({\left (2 \, {\left (3 \, {\left (4 \, x - 15\right )} x + 56\right )} x - 15\right )} x - 136\right )} \sqrt {-x^{2} + 1} + \frac {7}{8} \, \arcsin \left (x\right ) \] Input:
integrate((1-x)^3*(-x^2+1)^(1/2),x, algorithm="giac")
Output:
-1/120*((2*(3*(4*x - 15)*x + 56)*x - 15)*x - 136)*sqrt(-x^2 + 1) + 7/8*arc sin(x)
Time = 6.61 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.54 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=\frac {7\,\mathrm {asin}\left (x\right )}{8}+\sqrt {1-x^2}\,\left (-\frac {x^4}{5}+\frac {3\,x^3}{4}-\frac {14\,x^2}{15}+\frac {x}{8}+\frac {17}{15}\right ) \] Input:
int(-(1 - x^2)^(1/2)*(x - 1)^3,x)
Output:
(7*asin(x))/8 + (1 - x^2)^(1/2)*(x/8 - (14*x^2)/15 + (3*x^3)/4 - x^4/5 + 1 7/15)
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int (1-x)^3 \sqrt {1-x^2} \, dx=\frac {7 \mathit {asin} \left (x \right )}{8}-\frac {\sqrt {-x^{2}+1}\, x^{4}}{5}+\frac {3 \sqrt {-x^{2}+1}\, x^{3}}{4}-\frac {14 \sqrt {-x^{2}+1}\, x^{2}}{15}+\frac {\sqrt {-x^{2}+1}\, x}{8}+\frac {17 \sqrt {-x^{2}+1}}{15}-\frac {17}{15} \] Input:
int((1-x)^3*(-x^2+1)^(1/2),x)
Output:
(105*asin(x) - 24*sqrt( - x**2 + 1)*x**4 + 90*sqrt( - x**2 + 1)*x**3 - 112 *sqrt( - x**2 + 1)*x**2 + 15*sqrt( - x**2 + 1)*x + 136*sqrt( - x**2 + 1) - 136)/120