\(\int \frac {1}{(2^{2/3}-x) \sqrt {1-x^3}} \, dx\) [119]

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 = 23, antiderivative size = 160 \[ \int \frac {1}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} x\right )}{\sqrt {1-x^3}}\right )}{3 \sqrt {3}}-\frac {2 \sqrt [3]{2} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}} \] Output:

-2/9*arctan(3^(1/2)*(1-2^(1/3)*x)/(-x^3+1)^(1/2))*3^(1/2)-2/9*2^(1/3)*(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 = 10.18 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {4 i \sqrt {2} \sqrt {-\frac {i (-1+x)}{3 i+\sqrt {3}}} \sqrt {1+x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{i+2 i 2^{2/3}+\sqrt {3}},\arcsin \left (\frac {\sqrt {i+\sqrt {3}+2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{\left (1+2\ 2^{2/3}-i \sqrt {3}\right ) \sqrt {1-x^3}} \] Input:

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

Output:

((-4*I)*Sqrt[2]*Sqrt[((-I)*(-1 + x))/(3*I + Sqrt[3])]*Sqrt[1 + x + x^2]*El 
lipticPi[(2*Sqrt[3])/(I + (2*I)*2^(2/3) + Sqrt[3]), ArcSin[Sqrt[I + Sqrt[3 
] + (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])])/((1 + 2*2^( 
2/3) - I*Sqrt[3])*Sqrt[1 - x^3])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2559, 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.

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

\(\Big \downarrow \) 2559

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2562

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(-2*ArcTan[(Sqrt[3]*(1 - 2^(1/3)*x))/Sqrt[1 - x^3]])/(3*Sqrt[3]) - (2*2^(1 
/3)*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 2559
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Simp[2/ 
(3*c)   Int[1/Sqrt[a + b*x^3], x], x] + Simp[1/(3*c)   Int[(c - 2*d*x)/((c 
+ d*x)*Sqrt[a + b*x^3]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 - 4* 
a*d^3, 0]
 

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]
 
Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.89

method result size
default \(\frac {2 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}\, \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}\right )}\) \(143\)
elliptic \(\frac {2 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}\, \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}\right )}\) \(143\)

Input:

int(1/(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)/(-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 = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {1}{9} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (5 \, x^{3} - 2^{\frac {2}{3}} {\left (x^{5} - x^{2}\right )} - 2^{\frac {1}{3}} {\left (7 \, x^{4} - 4 \, x\right )} - 2\right )} \sqrt {-x^{3} + 1}}{6 \, {\left (2 \, x^{6} - 3 \, x^{3} + 1\right )}}\right ) - \frac {2}{3} i \cdot 2^{\frac {1}{3}} {\rm weierstrassPInverse}\left (0, 4, x\right ) \] Input:

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

Output:

-1/9*sqrt(3)*arctan(-1/6*sqrt(3)*(5*x^3 - 2^(2/3)*(x^5 - x^2) - 2^(1/3)*(7 
*x^4 - 4*x) - 2)*sqrt(-x^3 + 1)/(2*x^6 - 3*x^3 + 1)) - 2/3*I*2^(1/3)*weier 
strassPInverse(0, 4, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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