Integrand size = 17, antiderivative size = 86 \[ \int \sqrt {1-\left (1-x^2\right )^3} \, dx=-\frac {\left (3-2 x^2\right ) \sqrt {3 x^2-3 x^4+x^6}}{8 x}-\frac {3 \sqrt {3 x^2-3 x^4+x^6} \text {arcsinh}\left (\frac {3-2 x^2}{\sqrt {3}}\right )}{16 x \sqrt {3-3 x^2+x^4}} \]
-1/8*(-2*x^2+3)*(x^6-3*x^4+3*x^2)^(1/2)/x-3/16*arcsinh(1/3*(-2*x^2+3)*3^(1 /2))*(x^6-3*x^4+3*x^2)^(1/2)/x/(x^4-3*x^2+3)^(1/2)
Time = 0.00 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \sqrt {1-\left (1-x^2\right )^3} \, dx=\frac {x \left (-18+30 x^2-18 x^4+4 x^6-3 \sqrt {3-3 x^2+x^4} \log \left (3-2 x^2+2 \sqrt {3-3 x^2+x^4}\right )\right )}{16 \sqrt {x^2 \left (3-3 x^2+x^4\right )}} \]
(x*(-18 + 30*x^2 - 18*x^4 + 4*x^6 - 3*Sqrt[3 - 3*x^2 + x^4]*Log[3 - 2*x^2 + 2*Sqrt[3 - 3*x^2 + x^4]]))/(16*Sqrt[x^2*(3 - 3*x^2 + x^4)])
Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2093, 1950, 1432, 1087, 1090, 222}
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.
\(\displaystyle \int \sqrt {1-\left (1-x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2093 |
\(\displaystyle \int \sqrt {x^6-3 x^4+3 x^2}dx\) |
\(\Big \downarrow \) 1950 |
\(\displaystyle \frac {\sqrt {x^6-3 x^4+3 x^2} \int x \sqrt {x^4-3 x^2+3}dx}{x \sqrt {x^4-3 x^2+3}}\) |
\(\Big \downarrow \) 1432 |
\(\displaystyle \frac {\sqrt {x^6-3 x^4+3 x^2} \int \sqrt {x^4-3 x^2+3}dx^2}{2 x \sqrt {x^4-3 x^2+3}}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {\sqrt {x^6-3 x^4+3 x^2} \left (\frac {3}{8} \int \frac {1}{\sqrt {x^4-3 x^2+3}}dx^2-\frac {1}{4} \left (3-2 x^2\right ) \sqrt {x^4-3 x^2+3}\right )}{2 x \sqrt {x^4-3 x^2+3}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {\sqrt {x^6-3 x^4+3 x^2} \left (\frac {1}{8} \sqrt {3} \int \frac {1}{\sqrt {\frac {x^4}{3}+1}}d\left (2 x^2-3\right )-\frac {1}{4} \left (3-2 x^2\right ) \sqrt {x^4-3 x^2+3}\right )}{2 x \sqrt {x^4-3 x^2+3}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\sqrt {x^6-3 x^4+3 x^2} \left (\frac {3}{8} \text {arcsinh}\left (\frac {2 x^2-3}{\sqrt {3}}\right )-\frac {1}{4} \left (3-2 x^2\right ) \sqrt {x^4-3 x^2+3}\right )}{2 x \sqrt {x^4-3 x^2+3}}\) |
(Sqrt[3*x^2 - 3*x^4 + x^6]*(-1/4*((3 - 2*x^2)*Sqrt[3 - 3*x^2 + x^4]) + (3* ArcSinh[(-3 + 2*x^2)/Sqrt[3]])/8))/(2*x*Sqrt[3 - 3*x^2 + x^4])
3.2.28.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
Int[Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Simp[Sqrt[a*x^q + b*x^n + c*x^(2*n - q)]/(x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]) Int[x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q]
Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && G eneralizedTrinomialQ[u, x] && !GeneralizedTrinomialMatchQ[u, x]
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {4 \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}\, x^{2}+3 \,\operatorname {arcsinh}\left (\frac {\sqrt {3}\, \left (2 x^{2}-3\right )}{3}\right ) x -6 \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}{16 x}\) | \(62\) |
trager | \(\frac {\left (2 x^{2}-3\right ) \sqrt {x^{6}-3 x^{4}+3 x^{2}}}{8 x}-\frac {3 \ln \left (\frac {-2 x^{3}+2 \sqrt {x^{6}-3 x^{4}+3 x^{2}}+3 x}{x}\right )}{16}\) | \(64\) |
risch | \(\frac {\left (2 x^{2}-3\right ) \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}{8 x}+\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x^{2}-\frac {3}{2}\right )}{3}\right ) \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}{16 \sqrt {x^{4}-3 x^{2}+3}\, x}\) | \(74\) |
default | \(\frac {\sqrt {x^{6}-3 x^{4}+3 x^{2}}\, \left (4 \sqrt {x^{4}-3 x^{2}+3}\, x^{2}+3 \,\operatorname {arcsinh}\left (\frac {\sqrt {3}\, \left (2 x^{2}-3\right )}{3}\right )-6 \sqrt {x^{4}-3 x^{2}+3}\right )}{16 x \sqrt {x^{4}-3 x^{2}+3}}\) | \(81\) |
1/16/x*(4*(x^2*(x^4-3*x^2+3))^(1/2)*x^2+3*arcsinh(1/3*3^(1/2)*(2*x^2-3))*x -6*(x^2*(x^4-3*x^2+3))^(1/2))
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \sqrt {1-\left (1-x^2\right )^3} \, dx=-\frac {12 \, x \log \left (-\frac {2 \, x^{3} - 3 \, x - 2 \, \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}}}{x}\right ) - 8 \, \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}} {\left (2 \, x^{2} - 3\right )} - 9 \, x}{64 \, x} \]
-1/64*(12*x*log(-(2*x^3 - 3*x - 2*sqrt(x^6 - 3*x^4 + 3*x^2))/x) - 8*sqrt(x ^6 - 3*x^4 + 3*x^2)*(2*x^2 - 3) - 9*x)/x
\[ \int \sqrt {1-\left (1-x^2\right )^3} \, dx=\int \sqrt {1 - \left (1 - x^{2}\right )^{3}}\, dx \]
\[ \int \sqrt {1-\left (1-x^2\right )^3} \, dx=\int { \sqrt {{\left (x^{2} - 1\right )}^{3} + 1} \,d x } \]
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \sqrt {1-\left (1-x^2\right )^3} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {x^{4} - 3 \, x^{2} + 3} {\left (2 \, x^{2} - 3\right )} - 3 \, \log \left (-2 \, x^{2} + 2 \, \sqrt {x^{4} - 3 \, x^{2} + 3} + 3\right )\right )} \mathrm {sgn}\left (x\right ) + \frac {3}{16} \, {\left (2 \, \sqrt {3} + \log \left (2 \, \sqrt {3} + 3\right )\right )} \mathrm {sgn}\left (x\right ) \]
1/16*(2*sqrt(x^4 - 3*x^2 + 3)*(2*x^2 - 3) - 3*log(-2*x^2 + 2*sqrt(x^4 - 3* x^2 + 3) + 3))*sgn(x) + 3/16*(2*sqrt(3) + log(2*sqrt(3) + 3))*sgn(x)
Timed out. \[ \int \sqrt {1-\left (1-x^2\right )^3} \, dx=\int \sqrt {{\left (x^2-1\right )}^3+1} \,d x \]