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

3.1.2.1 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((1-2^(1/3)*x)*3^(1/2)/(-x^3+1)^(1/2))*3^(1/2)-2/9*2^(1/3)*(1-x 
)*EllipticF((1-x-3^(1/2))/(1-x+3^(1/2)),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^ 
(1/2))*((x^2+x+1)/(1-x+3^(1/2))^2)^(1/2)*3^(3/4)/(-x^3+1)^(1/2)/((1-x)/(1- 
x+3^(1/2))^2)^(1/2)
 

3.1.2.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.10 (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])
 

3.1.2.3 Rubi [A] (verified)

Time = 0.43 (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])
 

3.1.2.3.1 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]
 

3.1.2.4 Maple [A] (verified)

Time = 2.90 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.89

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

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)/(-2 
^(2/3)-1/2+1/2*I*3^(1/2))*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)/(-2^(2/3)-1/2+1/2*I*3^(1/2)),(I*3^(1/2)/(-3/2+1/2 
*I*3^(1/2)))^(1/2))
 

3.1.2.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (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)
 

3.1.2.6 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)
 

3.1.2.7 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)
 

3.1.2.8 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
 

3.1.2.9 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)