\(\int \frac {x}{(2^{2/3}-x) \sqrt {1-x^3}} \, dx\) [112]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 160 \[ \int \frac {x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} x\right )}{\sqrt {1-x^3}}\right )}{3 \sqrt {3}}+\frac {2 \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}} \] Output:

-2/9*2^(2/3)*arctan(3^(1/2)*(1-2^(1/3)*x)/(-x^3+1)^(1/2))*3^(1/2)+2/9*(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)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.56 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.31 \[ \int \frac {x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=\frac {2 \sqrt {\frac {1-x}{1+\sqrt [3]{-1}}} \left (-\frac {\left (\sqrt [3]{-1}+x\right ) \sqrt {\frac {\sqrt [3]{-1}+(-1)^{2/3} x}{1+\sqrt [3]{-1}}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right ),\sqrt [3]{-1}\right )}{\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}}+\frac {i 2^{2/3} \sqrt {1+x+x^2} \operatorname {EllipticPi}\left (\frac {i \sqrt {3}}{\sqrt [3]{-1}+2^{2/3}},\arcsin \left (\sqrt {\frac {1-(-1)^{2/3} x}{1+\sqrt [3]{-1}}}\right ),\sqrt [3]{-1}\right )}{\sqrt [3]{-1}+2^{2/3}}\right )}{\sqrt {1-x^3}} \] Input:

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

Output:

(2*Sqrt[(1 - x)/(1 + (-1)^(1/3))]*(-((((-1)^(1/3) + x)*Sqrt[((-1)^(1/3) + 
(-1)^(2/3)*x)/(1 + (-1)^(1/3))]*EllipticF[ArcSin[Sqrt[(1 - (-1)^(2/3)*x)/( 
1 + (-1)^(1/3))]], (-1)^(1/3)])/Sqrt[(1 - (-1)^(2/3)*x)/(1 + (-1)^(1/3))]) 
 + (I*2^(2/3)*Sqrt[1 + x + x^2]*EllipticPi[(I*Sqrt[3])/((-1)^(1/3) + 2^(2/ 
3)), ArcSin[Sqrt[(1 - (-1)^(2/3)*x)/(1 + (-1)^(1/3))]], (-1)^(1/3)])/((-1) 
^(1/3) + 2^(2/3))))/Sqrt[1 - x^3]
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2564, 27, 759, 2562, 216}

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.

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

\(\Big \downarrow \) 2564

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2562

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(-2*2^(2/3)*ArcTan[(Sqrt[3]*(1 - 2^(1/3)*x))/Sqrt[1 - x^3]])/(3*Sqrt[3]) + 
 (2*Sqrt[2 + Sqrt[3]]*(1 - x)*Sqrt[(1 + x + x^2)/(1 + Sqrt[3] - x)^2]*Elli 
pticF[ArcSin[(1 - Sqrt[3] - x)/(1 + Sqrt[3] - x)], -7 - 4*Sqrt[3]])/(3*3^( 
1/4)*Sqrt[(1 - x)/(1 + Sqrt[3] - x)^2]*Sqrt[1 - x^3])
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 2562
Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ 
Symbol] :> Simp[2*(e/d)   Subst[Int[1/(1 + 3*a*x^2), x], x, (1 + 2*d*(x/c)) 
/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] 
&& EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]
 

rule 2564
Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x 
_Symbol] :> Simp[(2*d*e + c*f)/(3*c*d)   Int[1/Sqrt[a + b*x^3], x], x] + Si 
mp[(d*e - c*f)/(3*c*d)   Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x], x 
] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && (EqQ[b*c^3 - 4*a* 
d^3, 0] || EqQ[b*c^3 + 8*a*d^3, 0]) && NeQ[2*d*e + c*f, 0]
 
Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.58

method result size
default \(\frac {2 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}}+\frac {2 i 2^{\frac {2}{3}} \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}\, \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}\right )}\) \(253\)
elliptic \(\frac {2 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}}+\frac {2 i 2^{\frac {2}{3}} \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}\, \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}\right )}\) \(253\)

Input:

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

Output:

2/3*I*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2)*((x-1)/(-3/2+1/2*I*3 
^(1/2)))^(1/2)*(-I*(x+1/2+1/2*I*3^(1/2))*3^(1/2))^(1/2)/(-x^3+1)^(1/2)*Ell 
ipticF(1/3*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),(I*3^(1/2)/(-3/ 
2+1/2*I*3^(1/2)))^(1/2))+2/3*I*2^(2/3)*3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^ 
(1/2))^(1/2)*((x-1)/(-3/2+1/2*I*3^(1/2)))^(1/2)*(-I*(x+1/2+1/2*I*3^(1/2))* 
3^(1/2))^(1/2)/(-x^3+1)^(1/2)/(-1/2+1/2*I*3^(1/2)-2^(2/3))*EllipticPi(1/3* 
3^(1/2)*(I*(x+1/2-1/2*I*3^(1/2))*3^(1/2))^(1/2),I*3^(1/2)/(-1/2+1/2*I*3^(1 
/2)-2^(2/3)),(I*3^(1/2)/(-3/2+1/2*I*3^(1/2)))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.52 \[ \int \frac {x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {1}{3} \cdot 2^{\frac {1}{6}} \sqrt {\frac {2}{3}} \arctan \left (\frac {2^{\frac {1}{6}} \sqrt {\frac {2}{3}} {\left (2 \, x^{5} - 2 \, x^{2} + 2^{\frac {2}{3}} {\left (7 \, x^{4} - 4 \, x\right )} - 2^{\frac {1}{3}} {\left (5 \, x^{3} - 2\right )}\right )} \sqrt {-x^{3} + 1}}{4 \, {\left (2 \, x^{6} - 3 \, x^{3} + 1\right )}}\right ) + \frac {2}{3} i \, {\rm weierstrassPInverse}\left (0, 4, x\right ) \] Input:

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

Output:

-1/3*2^(1/6)*sqrt(2/3)*arctan(1/4*2^(1/6)*sqrt(2/3)*(2*x^5 - 2*x^2 + 2^(2/ 
3)*(7*x^4 - 4*x) - 2^(1/3)*(5*x^3 - 2))*sqrt(-x^3 + 1)/(2*x^6 - 3*x^3 + 1) 
) + 2/3*I*weierstrassPInverse(0, 4, x)
 

Sympy [F]

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

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

Output:

-Integral(x/(x*sqrt(1 - x**3) - 2**(2/3)*sqrt(1 - x**3)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[2]%%%} / %%%{%%{[2,0]:[1,0,0,-2]%%},[2]%%%} Error: Bad 
Argument
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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