Integrand size = 15, antiderivative size = 134 \[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=-\frac {2}{5} x \sqrt {1-x^3}-\frac {4 \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}} \] Output:
-2/5*x*(-x^3+1)^(1/2)-4/15*(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 = 10.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.24 \[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=\frac {2}{5} x \left (-\sqrt {1-x^3}+\operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},x^3\right )\right ) \] Input:
Integrate[x^3/Sqrt[1 - x^3],x]
Output:
(2*x*(-Sqrt[1 - x^3] + Hypergeometric2F1[1/3, 1/2, 4/3, x^3]))/5
Time = 0.21 (sec) , antiderivative size = 134, 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 = {843, 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.
\(\displaystyle \int \frac {x^3}{\sqrt {1-x^3}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {2}{5} \int \frac {1}{\sqrt {1-x^3}}dx-\frac {2}{5} x \sqrt {1-x^3}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle -\frac {4 \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}}-\frac {2}{5} \sqrt {1-x^3} x\) |
Input:
Int[x^3/Sqrt[1 - x^3],x]
Output:
(-2*x*Sqrt[1 - x^3])/5 - (4*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]])/(5*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^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && 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 | \(\frac {x^{4} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], x^{3}\right )}{4}\) | \(15\) |
default | \(-\frac {2 x \sqrt {-x^{3}+1}}{5}-\frac {4 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{15 \sqrt {-x^{3}+1}}\) | \(120\) |
elliptic | \(-\frac {2 x \sqrt {-x^{3}+1}}{5}-\frac {4 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{15 \sqrt {-x^{3}+1}}\) | \(120\) |
risch | \(\frac {2 x \left (x^{3}-1\right )}{5 \sqrt {-x^{3}+1}}-\frac {4 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{15 \sqrt {-x^{3}+1}}\) | \(125\) |
Input:
int(x^3/(-x^3+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*hypergeom([1/2,4/3],[7/3],x^3)
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.14 \[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=-\frac {2}{5} \, \sqrt {-x^{3} + 1} x - \frac {4}{5} i \, {\rm weierstrassPInverse}\left (0, 4, x\right ) \] Input:
integrate(x^3/(-x^3+1)^(1/2),x, algorithm="fricas")
Output:
-2/5*sqrt(-x^3 + 1)*x - 4/5*I*weierstrassPInverse(0, 4, x)
Time = 0.41 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.23 \[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=\frac {x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \] Input:
integrate(x**3/(-x**3+1)**(1/2),x)
Output:
x**4*gamma(4/3)*hyper((1/2, 4/3), (7/3,), x**3*exp_polar(2*I*pi))/(3*gamma (7/3))
\[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=\int { \frac {x^{3}}{\sqrt {-x^{3} + 1}} \,d x } \] Input:
integrate(x^3/(-x^3+1)^(1/2),x, algorithm="maxima")
Output:
integrate(x^3/sqrt(-x^3 + 1), x)
\[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=\int { \frac {x^{3}}{\sqrt {-x^{3} + 1}} \,d x } \] Input:
integrate(x^3/(-x^3+1)^(1/2),x, algorithm="giac")
Output:
integrate(x^3/sqrt(-x^3 + 1), x)
Time = 0.15 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.38 \[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=-\frac {2\,x\,\sqrt {1-x^3}}{5}-\frac {4\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3-1}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{5\,\sqrt {1-x^3}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:
int(x^3/(1 - x^3)^(1/2),x)
Output:
- (2*x*(1 - x^3)^(1/2))/5 - (4*((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)*1 i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(5*(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))
\[ \int \frac {x^3}{\sqrt {1-x^3}} \, dx=-\frac {2 \sqrt {-x^{3}+1}\, x}{5}-\frac {2 \left (\int \frac {\sqrt {-x^{3}+1}}{x^{3}-1}d x \right )}{5} \] Input:
int(x^3/(-x^3+1)^(1/2),x)
Output:
( - 2*(sqrt( - x**3 + 1)*x + int(sqrt( - x**3 + 1)/(x**3 - 1),x)))/5