\(\int \frac {1}{x^2 \sqrt {1-x^3}} \, dx\) [266]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 15, antiderivative size = 270 \[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {1-x^3}}{1+\sqrt {3}-x}-\frac {\sqrt {1-x^3}}{x}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}}-\frac {\sqrt {2} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}} \] Output:

-(-x^3+1)^(1/2)/(1+3^(1/2)-x)-(-x^3+1)^(1/2)/x+1/2*3^(1/4)*(1/2*6^(1/2)-1/ 
2*2^(1/2))*(1-x)*((x^2+x+1)/(1+3^(1/2)-x)^2)^(1/2)*EllipticE((1-3^(1/2)-x) 
/(1+3^(1/2)-x),I*3^(1/2)+2*I)/((1-x)/(1+3^(1/2)-x)^2)^(1/2)/(-x^3+1)^(1/2) 
-1/3*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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.07 \[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},x^3\right )}{x} \] Input:

Integrate[1/(x^2*Sqrt[1 - x^3]),x]
 

Output:

-(Hypergeometric2F1[-1/3, 1/2, 2/3, x^3]/x)
 

Rubi [A] (warning: unable to verify)

Time = 0.62 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {847, 832, 759, 2416}

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.

\(\displaystyle \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {1}{2} \int \frac {x}{\sqrt {1-x^3}}dx-\frac {\sqrt {1-x^3}}{x}\)

\(\Big \downarrow \) 832

\(\displaystyle \frac {1}{2} \left (\int \frac {-x-\sqrt {3}+1}{\sqrt {1-x^3}}dx-\left (1-\sqrt {3}\right ) \int \frac {1}{\sqrt {1-x^3}}dx\right )-\frac {\sqrt {1-x^3}}{x}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {1}{2} \left (\int \frac {-x-\sqrt {3}+1}{\sqrt {1-x^3}}dx+\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}}\right )-\frac {\sqrt {1-x^3}}{x}\)

\(\Big \downarrow \) 2416

\(\displaystyle \frac {1}{2} \left (\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x+\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {-x-\sqrt {3}+1}{-x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{\sqrt {\frac {1-x}{\left (-x+\sqrt {3}+1\right )^2}} \sqrt {1-x^3}}-\frac {2 \sqrt {1-x^3}}{-x+\sqrt {3}+1}\right )-\frac {\sqrt {1-x^3}}{x}\)

Input:

Int[1/(x^2*Sqrt[1 - x^3]),x]
 

Output:

-(Sqrt[1 - x^3]/x) + ((-2*Sqrt[1 - x^3])/(1 + Sqrt[3] - x) + (3^(1/4)*Sqrt 
[2 - Sqrt[3]]*(1 - x)*Sqrt[(1 + x + x^2)/(1 + Sqrt[3] - x)^2]*EllipticE[Ar 
cSin[(1 - Sqrt[3] - x)/(1 + Sqrt[3] - x)], -7 - 4*Sqrt[3]])/(Sqrt[(1 - x)/ 
(1 + Sqrt[3] - x)^2]*Sqrt[1 - x^3]) + (2*(1 - Sqrt[3])*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]])/(3^(1/4)*Sqrt[(1 - x)/(1 + Sq 
rt[3] - x)^2]*Sqrt[1 - x^3]))/2
 

Defintions of rubi rules used

rule 759
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]
 

rule 832
Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && PosQ[a]
 

rule 847
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]
 

rule 2416
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S 
imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt 
[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) 
*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 
Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.63 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.06

method result size
meijerg \(-\frac {\operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {1}{2}\right ], \left [\frac {2}{3}\right ], x^{3}\right )}{x}\) \(15\)
default \(-\frac {\sqrt {-x^{3}+1}}{x}+\frac {i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \left (\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )+\operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )\right )}{3 \sqrt {-x^{3}+1}}\) \(173\)
elliptic \(-\frac {\sqrt {-x^{3}+1}}{x}+\frac {i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \left (\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )+\operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )\right )}{3 \sqrt {-x^{3}+1}}\) \(173\)
risch \(\frac {x^{3}-1}{x \sqrt {-x^{3}+1}}+\frac {i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {-1+x}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \left (\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )+\operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )\right )}{3 \sqrt {-x^{3}+1}}\) \(177\)

Input:

int(1/x^2/(-x^3+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/x*hypergeom([-1/3,1/2],[2/3],x^3)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.10 \[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=\frac {-i \, x {\rm weierstrassZeta}\left (0, 4, {\rm weierstrassPInverse}\left (0, 4, x\right )\right ) - \sqrt {-x^{3} + 1}}{x} \] Input:

integrate(1/x^2/(-x^3+1)^(1/2),x, algorithm="fricas")
 

Output:

(-I*x*weierstrassZeta(0, 4, weierstrassPInverse(0, 4, x)) - sqrt(-x^3 + 1) 
)/x
 

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.12 \[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=\frac {\Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{2 i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} \] Input:

integrate(1/x**2/(-x**3+1)**(1/2),x)
 

Output:

gamma(-1/3)*hyper((-1/3, 1/2), (2/3,), x**3*exp_polar(2*I*pi))/(3*x*gamma( 
2/3))
 

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=\int { \frac {1}{\sqrt {-x^{3} + 1} x^{2}} \,d x } \] Input:

integrate(1/x^2/(-x^3+1)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(-x^3 + 1)*x^2), x)
 

Giac [F]

\[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=\int { \frac {1}{\sqrt {-x^{3} + 1} x^{2}} \,d x } \] Input:

integrate(1/x^2/(-x^3+1)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(-x^3 + 1)*x^2), x)
 

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=-\frac {\sqrt {1-x^3}}{x}+\frac {\left (\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\left (-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {E}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3-1}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{\sqrt {1-x^3}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:

int(1/(x^2*(1 - x^3)^(1/2)),x)
 

Output:

((((3^(1/2)*1i)/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)) - ((3^(1/2)*1i)/2 
- 3/2)*ellipticE(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)))*((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))/((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)) 
- (1 - x^3)^(1/2)/x
 

Reduce [F]

\[ \int \frac {1}{x^2 \sqrt {1-x^3}} \, dx=-\left (\int \frac {\sqrt {-x^{3}+1}}{x^{5}-x^{2}}d x \right ) \] Input:

int(1/x^2/(-x^3+1)^(1/2),x)
 

Output:

 - int(sqrt( - x**3 + 1)/(x**5 - x**2),x)