\(\int \frac {e+f x}{(1+\sqrt {3}-x) \sqrt {1-x^3}} \, dx\) [174]

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 = 29, antiderivative size = 187 \[ \int \frac {e+f x}{\left (1+\sqrt {3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {\left (e+f+\sqrt {3} f\right ) \arctan \left (\frac {\sqrt {3+2 \sqrt {3}} (1-x)}{\sqrt {1-x^3}}\right )}{\sqrt {3 \left (3+2 \sqrt {3}\right )}}-\frac {\sqrt {2+\sqrt {3}} \left (e+\left (1-\sqrt {3}\right ) f\right ) (1-x) \sqrt {\frac {1+x+x^2}{\left (1+\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}-x}{1+\sqrt {3}-x}\right ),-7-4 \sqrt {3}\right )}{3^{3/4} \sqrt {\frac {1-x}{\left (1+\sqrt {3}-x\right )^2}} \sqrt {1-x^3}} \] Output:

-(e+f+3^(1/2)*f)*arctan((3+2*3^(1/2))^(1/2)*(1-x)/(-x^3+1)^(1/2))/(9+6*3^( 
1/2))^(1/2)-1/3*(1/2*6^(1/2)+1/2*2^(1/2))*(e+(1-3^(1/2))*f)*(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^(1/4)/((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 = 20.61 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.56 \[ \int \frac {e+f x}{\left (1+\sqrt {3}-x\right ) \sqrt {1-x^3}} \, dx=\frac {2 \sqrt {\frac {2}{3}} \sqrt {-\frac {i (-1+x)}{3 i+\sqrt {3}}} \left (-3 i f \sqrt {-i+\sqrt {3}-2 i x} \left (-i \left ((2+i)+\sqrt {3}\right )+\left ((2-i)+\sqrt {3}\right ) x\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {i+\sqrt {3}+2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )+2 \left (\sqrt {3} e+\left (3+\sqrt {3}\right ) f\right ) \sqrt {i+\sqrt {3}+2 i x} \sqrt {1+x+x^2} \operatorname {EllipticPi}\left (\frac {2 \sqrt {3}}{3 i+(1+2 i) \sqrt {3}},\arcsin \left (\frac {\sqrt {i+\sqrt {3}+2 i x}}{\sqrt {2} \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{\left (3 i+(1+2 i) \sqrt {3}\right ) \sqrt {i+\sqrt {3}+2 i x} \sqrt {1-x^3}} \] Input:

Integrate[(e + f*x)/((1 + Sqrt[3] - x)*Sqrt[1 - x^3]),x]
 

Output:

(2*Sqrt[2/3]*Sqrt[((-I)*(-1 + x))/(3*I + Sqrt[3])]*((-3*I)*f*Sqrt[-I + Sqr 
t[3] - (2*I)*x]*((-I)*((2 + I) + Sqrt[3]) + ((2 - I) + Sqrt[3])*x)*Ellipti 
cF[ArcSin[Sqrt[I + Sqrt[3] + (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I 
 + Sqrt[3])] + 2*(Sqrt[3]*e + (3 + Sqrt[3])*f)*Sqrt[I + Sqrt[3] + (2*I)*x] 
*Sqrt[1 + x + x^2]*EllipticPi[(2*Sqrt[3])/(3*I + (1 + 2*I)*Sqrt[3]), ArcSi 
n[Sqrt[I + Sqrt[3] + (2*I)*x]/(Sqrt[2]*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[ 
3])]))/((3*I + (1 + 2*I)*Sqrt[3])*Sqrt[I + Sqrt[3] + (2*I)*x]*Sqrt[1 - x^3 
])
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2566, 27, 759, 2565, 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.

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

\(\Big \downarrow \) 2566

\(\displaystyle \frac {1}{12} \left (\frac {e+f}{\sqrt {3}}+f\right ) \int -\frac {6 \left (-x-\sqrt {3}+1\right )}{\left (-x+\sqrt {3}+1\right ) \sqrt {1-x^3}}dx-\frac {1}{2} \left (f-\frac {e+f}{\sqrt {3}}\right ) \int \frac {1}{\sqrt {1-x^3}}dx\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2565

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

\(\Big \downarrow \) 216

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

Input:

Int[(e + f*x)/((1 + Sqrt[3] - x)*Sqrt[1 - x^3]),x]
 

Output:

-(((f + (e + f)/Sqrt[3])*ArcTan[(Sqrt[3 + 2*Sqrt[3]]*(1 - x))/Sqrt[1 - x^3 
]])/Sqrt[3 + 2*Sqrt[3]]) + (Sqrt[2 + Sqrt[3]]*(f - (e + f)/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 + 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 2565
Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ 
Symbol] :> With[{k = Simplify[(d*e + 2*c*f)/(c*f)]}, Simp[(1 + k)*(e/d)   S 
ubst[Int[1/(1 + (3 + 2*k)*a*x^2), x], x, (1 + (1 + k)*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^2*c 
^6 - 20*a*b*c^3*d^3 - 8*a^2*d^6, 0] && EqQ[6*a*d^4*e - c*f*(b*c^3 - 22*a*d^ 
3), 0]
 

rule 2566
Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x 
_Symbol] :> Simp[-(6*a*d^4*e - c*f*(b*c^3 - 22*a*d^3))/(c*d*(b*c^3 - 28*a*d 
^3))   Int[1/Sqrt[a + b*x^3], x], x] + Simp[(d*e - c*f)/(c*d*(b*c^3 - 28*a* 
d^3))   Int[(c*(b*c^3 - 22*a*d^3) + 6*a*d^4*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^2*c^6 
 - 20*a*b*c^3*d^3 - 8*a^2*d^6, 0] && NeQ[6*a*d^4*e - c*f*(b*c^3 - 22*a*d^3) 
, 0]
 
Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.39

method result size
default \(\frac {2 i \left (e +f +\sqrt {3}\, f \right ) \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}-\sqrt {3}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}\, \left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}-\sqrt {3}\right )}+\frac {2 i f \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}}\) \(259\)
elliptic \(\frac {2 i f \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}}-\frac {2 i \left (-e -f -\sqrt {3}\, f \right ) \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}-\sqrt {3}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}\, \left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}-\sqrt {3}\right )}\) \(264\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 735, normalized size of antiderivative = 3.93 \[ \int \frac {e+f x}{\left (1+\sqrt {3}-x\right ) \sqrt {1-x^3}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/3*(sqrt(3)*(I*e + I*f) - 3*I*f)*weierstrassPInverse(0, 4, x) + 1/12*sq 
rt(3*e^2 - 6*e*f - 2*sqrt(3)*(e^2 - e*f + f^2))*log(-((e^2 + 2*e*f - 2*f^2 
)*x^8 + 16*(e^2 + 2*e*f - 2*f^2)*x^7 + 112*(e^2 + 2*e*f - 2*f^2)*x^6 + 16* 
(e^2 + 2*e*f - 2*f^2)*x^5 + 112*(e^2 + 2*e*f - 2*f^2)*x^4 - 224*(e^2 + 2*e 
*f - 2*f^2)*x^3 + 64*(e^2 + 2*e*f - 2*f^2)*x^2 + 4*((2*e - f)*x^6 + 18*(e 
- f)*x^5 + 6*(7*e - 2*f)*x^4 + 8*(e - 5*f)*x^3 + 36*f*x^2 - 24*(e + f)*x + 
 sqrt(3)*((e - f)*x^6 + 6*(2*e - f)*x^5 + 6*(3*e - 4*f)*x^4 + 8*(2*e + f)* 
x^3 - 12*(e + f)*x^2 + 24*f*x - 8*e - 16*f) + 8*e + 32*f)*sqrt(-x^3 + 1)*s 
qrt(3*e^2 - 6*e*f - 2*sqrt(3)*(e^2 - e*f + f^2)) + 112*e^2 + 224*e*f - 224 
*f^2 - 128*(e^2 + 2*e*f - 2*f^2)*x + 16*sqrt(3)*((e^2 + 2*e*f - 2*f^2)*x^7 
 + 2*(e^2 + 2*e*f - 2*f^2)*x^6 + 6*(e^2 + 2*e*f - 2*f^2)*x^5 - 5*(e^2 + 2* 
e*f - 2*f^2)*x^4 + 2*(e^2 + 2*e*f - 2*f^2)*x^3 - 6*(e^2 + 2*e*f - 2*f^2)*x 
^2 - 4*e^2 - 8*e*f + 8*f^2 + 4*(e^2 + 2*e*f - 2*f^2)*x))/(x^8 - 8*x^7 + 16 
*x^6 + 16*x^5 - 56*x^4 - 32*x^3 + 64*x^2 + 64*x + 16)), -1/3*(sqrt(3)*(I*e 
 + I*f) - 3*I*f)*weierstrassPInverse(0, 4, x) + 1/6*sqrt(-3*e^2 + 6*e*f + 
2*sqrt(3)*(e^2 - e*f + f^2))*arctan(1/6*(3*f*x^2 - 6*(e - f)*x - sqrt(3)*( 
(e + f)*x^2 + 2*(2*e - f)*x - 2*e + 4*f) + 6*e)*sqrt(-x^3 + 1)*sqrt(-3*e^2 
 + 6*e*f + 2*sqrt(3)*(e^2 - e*f + f^2))/((e^2 + 2*e*f - 2*f^2)*x^3 - e^2 - 
 2*e*f + 2*f^2))]
 

Sympy [F]

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

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

Output:

-Integral(e/(x*sqrt(1 - x**3) - sqrt(3)*sqrt(1 - x**3) - sqrt(1 - x**3)), 
x) - Integral(f*x/(x*sqrt(1 - x**3) - sqrt(3)*sqrt(1 - x**3) - sqrt(1 - x* 
*3)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate((f*x+e)/(1+3^(1/2)-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]%%%} / %%%{%%{[2,4]:[1,0,-3]%%},[2]%%%} Error: Bad Ar 
gument Va
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e+f x}{\left (1+\sqrt {3}-x\right ) \sqrt {1-x^3}} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {e+f x}{\left (1+\sqrt {3}-x\right ) \sqrt {1-x^3}} \, dx=\frac {-\sqrt {3}\, \left (\int \frac {\sqrt {-x^{3}+1}}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right ) e +\sqrt {3}\, \left (\int \frac {\sqrt {-x^{3}+1}\, x^{2}}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right ) f +\sqrt {3}\, \left (\int \frac {\sqrt {-x^{3}+1}\, x}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right ) e -\sqrt {3}\, \left (\int \frac {\sqrt {-x^{3}+1}\, x}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right ) f +3 \left (\int \frac {\sqrt {-x^{3}+1}}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right ) e +3 \left (\int \frac {\sqrt {-x^{3}+1}\, x}{x^{5}-2 x^{4}-2 x^{3}-x^{2}+2 x +2}d x \right ) f}{\sqrt {3}} \] Input:

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

Output:

( - sqrt(3)*int(sqrt( - x**3 + 1)/(x**5 - 2*x**4 - 2*x**3 - x**2 + 2*x + 2 
),x)*e + sqrt(3)*int((sqrt( - x**3 + 1)*x**2)/(x**5 - 2*x**4 - 2*x**3 - x* 
*2 + 2*x + 2),x)*f + sqrt(3)*int((sqrt( - x**3 + 1)*x)/(x**5 - 2*x**4 - 2* 
x**3 - x**2 + 2*x + 2),x)*e - sqrt(3)*int((sqrt( - x**3 + 1)*x)/(x**5 - 2* 
x**4 - 2*x**3 - x**2 + 2*x + 2),x)*f + 3*int(sqrt( - x**3 + 1)/(x**5 - 2*x 
**4 - 2*x**3 - x**2 + 2*x + 2),x)*e + 3*int((sqrt( - x**3 + 1)*x)/(x**5 - 
2*x**4 - 2*x**3 - x**2 + 2*x + 2),x)*f)/sqrt(3)