Integrand size = 15, antiderivative size = 136 \[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {1-x^3}}{2 x^2}-\frac {\sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}} \] Output:
-1/2*(-x^3+1)^(1/2)/x^2-1/6*(1/2*6^(1/2)+1/2*2^(1/2))*(1-x)*((x^2+x+1)/(1+ 3^(1/2)-x)^2)^(1/2)*EllipticF((1-3^(1/2)-x)/(1+3^(1/2)-x),I*3^(1/2)+2*I)*3 ^(3/4)/((1-x)/(1+3^(1/2)-x)^2)^(1/2)/(-x^3+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.15 \[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},x^3\right )}{2 x^2} \] Input:
Integrate[1/(x^3*Sqrt[1 - x^3]),x]
Output:
-1/2*Hypergeometric2F1[-2/3, 1/2, 1/3, x^3]/x^2
Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {847, 759}
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.
\(\displaystyle \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {1}{4} \int \frac {1}{\sqrt {1-x^3}}dx-\frac {\sqrt {1-x^3}}{2 x^2}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle -\frac {\sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}}-\frac {\sqrt {1-x^3}}{2 x^2}\) |
Input:
Int[1/(x^3*Sqrt[1 - x^3]),x]
Output:
-1/2*Sqrt[1 - x^3]/x^2 - (Sqrt[2 + Sqrt[3]]*(1 - x)*Sqrt[(1 + x + x^2)/(1 + Sqrt[3] - x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] - x)/(1 + Sqrt[3] - x)], - 7 - 4*Sqrt[3]])/(2*3^(1/4)*Sqrt[(1 - x)/(1 + Sqrt[3] - x)^2]*Sqrt[1 - x^3] )
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.61 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.11
method | result | size |
meijerg | \(-\frac {\operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {1}{2}\right ], \left [\frac {1}{3}\right ], x^{3}\right )}{2 x^{2}}\) | \(15\) |
default | \(-\frac {\sqrt {-x^{3}+1}}{2 x^{2}}-\frac {i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{6 \sqrt {-x^{3}+1}}\) | \(122\) |
elliptic | \(-\frac {\sqrt {-x^{3}+1}}{2 x^{2}}-\frac {i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{6 \sqrt {-x^{3}+1}}\) | \(122\) |
risch | \(\frac {x^{3}-1}{2 x^{2} \sqrt {-x^{3}+1}}-\frac {i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{6 \sqrt {-x^{3}+1}}\) | \(127\) |
Input:
int(1/x^3/(-x^3+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/x^2*hypergeom([-2/3,1/2],[1/3],x^3)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.19 \[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=\frac {-i \, x^{2} {\rm weierstrassPInverse}\left (0, 4, x\right ) - \sqrt {-x^{3} + 1}}{2 \, x^{2}} \] Input:
integrate(1/x^3/(-x^3+1)^(1/2),x, algorithm="fricas")
Output:
1/2*(-I*x^2*weierstrassPInverse(0, 4, x) - sqrt(-x^3 + 1))/x^2
Time = 0.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=\frac {\Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{2} \\ \frac {1}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} \] Input:
integrate(1/x**3/(-x**3+1)**(1/2),x)
Output:
gamma(-2/3)*hyper((-2/3, 1/2), (1/3,), x**3*exp_polar(2*I*pi))/(3*x**2*gam ma(1/3))
\[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=\int { \frac {1}{\sqrt {-x^{3} + 1} x^{3}} \,d x } \] Input:
integrate(1/x^3/(-x^3+1)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(-x^3 + 1)*x^3), x)
\[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=\int { \frac {1}{\sqrt {-x^{3} + 1} x^{3}} \,d x } \] Input:
integrate(1/x^3/(-x^3+1)^(1/2),x, algorithm="giac")
Output:
integrate(1/(sqrt(-x^3 + 1)*x^3), x)
Time = 0.06 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.38 \[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {1-x^3}}{2\,x^2}-\frac {\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3-1}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{2\,\sqrt {1-x^3}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:
int(1/(x^3*(1 - x^3)^(1/2)),x)
Output:
- (1 - x^3)^(1/2)/(2*x^2) - (((3^(1/2)*1i)/2 + 3/2)*(x^3 - 1)^(1/2)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*(-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/ 2)*ellipticF(asin((-(x - 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i) /2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(2*(1 - x^3)^(1/2)*(((3^(1/2)*1i)/2 - 1 /2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1 /2) + 1) + x^3)^(1/2))
\[ \int \frac {1}{x^3 \sqrt {1-x^3}} \, dx=-\left (\int \frac {\sqrt {-x^{3}+1}}{x^{6}-x^{3}}d x \right ) \] Input:
int(1/x^3/(-x^3+1)^(1/2),x)
Output:
- int(sqrt( - x**3 + 1)/(x**6 - x**3),x)