\(\int \frac {x^2}{\sqrt {-x-x^2+x^3} (-1+x^4)} \, dx\) [1910]

Optimal result

Integrand size = 27, antiderivative size = 133 \[ \int \frac {x^2}{\sqrt {-x-x^2+x^3} \left (-1+x^4\right )} \, dx=-\frac {1}{2} \arctan \left (\frac {\sqrt {-x-x^2+x^3}}{-1-x+x^2}\right )+\frac {1}{4} \sqrt {\frac {1}{5}+\frac {2 i}{5}} \arctan \left (\frac {\sqrt {1-2 i} \sqrt {-x-x^2+x^3}}{-1-x+x^2}\right )+\frac {1}{4} \sqrt {\frac {1}{5}-\frac {2 i}{5}} \arctan \left (\frac {\sqrt {1+2 i} \sqrt {-x-x^2+x^3}}{-1-x+x^2}\right ) \] Output:

-1/2*arctan((x^3-x^2-x)^(1/2)/(x^2-x-1))+1/20*(5+10*I)^(1/2)*arctan((1-2*I 
)^(1/2)*(x^3-x^2-x)^(1/2)/(x^2-x-1))+1/20*(5-10*I)^(1/2)*arctan((1+2*I)^(1 
/2)*(x^3-x^2-x)^(1/2)/(x^2-x-1))
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02 \[ \int \frac {x^2}{\sqrt {-x-x^2+x^3} \left (-1+x^4\right )} \, dx=\frac {\sqrt {x} \sqrt {-1-x+x^2} \left (-2 \sqrt {5} \arctan \left (\frac {\sqrt {x}}{\sqrt {-1-x+x^2}}\right )+\sqrt {1+2 i} \arctan \left (\frac {\sqrt {1-2 i} \sqrt {x}}{\sqrt {-1-x+x^2}}\right )+\sqrt {1-2 i} \arctan \left (\frac {\sqrt {1+2 i} \sqrt {x}}{\sqrt {-1-x+x^2}}\right )\right )}{4 \sqrt {5} \sqrt {x \left (-1-x+x^2\right )}} \] Input:

Integrate[x^2/(Sqrt[-x - x^2 + x^3]*(-1 + x^4)),x]
 

Output:

(Sqrt[x]*Sqrt[-1 - x + x^2]*(-2*Sqrt[5]*ArcTan[Sqrt[x]/Sqrt[-1 - x + x^2]] 
 + Sqrt[1 + 2*I]*ArcTan[(Sqrt[1 - 2*I]*Sqrt[x])/Sqrt[-1 - x + x^2]] + Sqrt 
[1 - 2*I]*ArcTan[(Sqrt[1 + 2*I]*Sqrt[x])/Sqrt[-1 - x + x^2]]))/(4*Sqrt[5]* 
Sqrt[x*(-1 - x + x^2)])
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 1.79 (sec) , antiderivative size = 985, normalized size of antiderivative = 7.41, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2467, 25, 2035, 7276, 2009}

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.

\(\displaystyle \int \frac {x^2}{\sqrt {x^3-x^2-x} \left (x^4-1\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt {x} \sqrt {x^2-x-1} \int -\frac {x^{3/2}}{\sqrt {x^2-x-1} \left (1-x^4\right )}dx}{\sqrt {x^3-x^2-x}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {x} \sqrt {x^2-x-1} \int \frac {x^{3/2}}{\sqrt {x^2-x-1} \left (1-x^4\right )}dx}{\sqrt {x^3-x^2-x}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2-x-1} \int \frac {x^2}{\sqrt {x^2-x-1} \left (1-x^4\right )}d\sqrt {x}}{\sqrt {x^3-x^2-x}}\)

\(\Big \downarrow \) 7276

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2-x-1} \int \left (-\frac {1}{4 (x-1) \sqrt {x^2-x-1}}+\frac {1}{4 (x+1) \sqrt {x^2-x-1}}-\frac {1}{2 \left (x^2+1\right ) \sqrt {x^2-x-1}}\right )d\sqrt {x}}{\sqrt {x^3-x^2-x}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2-x-1} \left (-\frac {\arctan \left (\frac {\sqrt {1-2 i} \sqrt {x}}{\sqrt {x^2-x-1}}\right )}{8 \sqrt {1-2 i}}-\frac {\arctan \left (\frac {\sqrt {1+2 i} \sqrt {x}}{\sqrt {x^2-x-1}}\right )}{8 \sqrt {1+2 i}}-\frac {\left (1+\sqrt {5}\right ) \sqrt {2 x+\sqrt {5}-1} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{4 \sqrt {2} \left (3+\sqrt {5}\right ) \sqrt {x^2-x-1}}-\frac {\left (1+\sqrt {5}\right ) \sqrt {2 x+\sqrt {5}-1} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{4 \sqrt {2} \left (1-\sqrt {5}\right ) \sqrt {x^2-x-1}}+\frac {\sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x+2}{\left (1-\sqrt {5}\right ) x+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{4 \sqrt [4]{5} \left (3+\sqrt {5}\right ) \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x+2}} \sqrt {x^2-x-1}}+\frac {\sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x+2}{\left (1-\sqrt {5}\right ) x+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{4 \sqrt [4]{5} \left (1-\sqrt {5}\right ) \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x+2}} \sqrt {x^2-x-1}}-\frac {\sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x+2}{\left (1-\sqrt {5}\right ) x+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} \sqrt {x}}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x\right )-2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{8 \sqrt [4]{5} \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x+2}} \sqrt {x^2-x-1}}+\frac {\left (2+\sqrt {5}\right ) \sqrt {2 x+\sqrt {5}-1} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (-1-\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{2 \sqrt {2} \left (3+\sqrt {5}\right ) \sqrt {x^2-x-1}}-\frac {\sqrt {2 x+\sqrt {5}-1} \sqrt {1-\frac {2 x}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (1+\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} \sqrt {x}\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{2 \sqrt {2} \left (1-\sqrt {5}\right ) \sqrt {x^2-x-1}}\right )}{\sqrt {x^3-x^2-x}}\)

Input:

Int[x^2/(Sqrt[-x - x^2 + x^3]*(-1 + x^4)),x]
 

Output:

(-2*Sqrt[x]*Sqrt[-1 - x + x^2]*(-1/8*ArcTan[(Sqrt[1 - 2*I]*Sqrt[x])/Sqrt[- 
1 - x + x^2]]/Sqrt[1 - 2*I] - ArcTan[(Sqrt[1 + 2*I]*Sqrt[x])/Sqrt[-1 - x + 
 x^2]]/(8*Sqrt[1 + 2*I]) - ((1 + Sqrt[5])*Sqrt[-1 + Sqrt[5] + 2*x]*Sqrt[1 
- (2*x)/(1 + Sqrt[5])]*EllipticF[ArcSin[Sqrt[2/(1 + Sqrt[5])]*Sqrt[x]], (- 
3 - Sqrt[5])/2])/(4*Sqrt[2]*(1 - Sqrt[5])*Sqrt[-1 - x + x^2]) - ((1 + Sqrt 
[5])*Sqrt[-1 + Sqrt[5] + 2*x]*Sqrt[1 - (2*x)/(1 + Sqrt[5])]*EllipticF[ArcS 
in[Sqrt[2/(1 + Sqrt[5])]*Sqrt[x]], (-3 - Sqrt[5])/2])/(4*Sqrt[2]*(3 + Sqrt 
[5])*Sqrt[-1 - x + x^2]) - (Sqrt[-2 - (1 - Sqrt[5])*x]*Sqrt[(2 + (1 + Sqrt 
[5])*x)/(2 + (1 - Sqrt[5])*x)]*EllipticF[ArcSin[(Sqrt[2]*5^(1/4)*Sqrt[x])/ 
Sqrt[-2 - (1 - Sqrt[5])*x]], (5 - Sqrt[5])/10])/(8*5^(1/4)*Sqrt[(2 + (1 - 
Sqrt[5])*x)^(-1)]*Sqrt[-1 - x + x^2]) + (Sqrt[-2 - (1 - Sqrt[5])*x]*Sqrt[( 
2 + (1 + Sqrt[5])*x)/(2 + (1 - Sqrt[5])*x)]*EllipticF[ArcSin[(Sqrt[2]*5^(1 
/4)*Sqrt[x])/Sqrt[-2 - (1 - Sqrt[5])*x]], (5 - Sqrt[5])/10])/(4*5^(1/4)*(1 
 - Sqrt[5])*Sqrt[(2 + (1 - Sqrt[5])*x)^(-1)]*Sqrt[-1 - x + x^2]) + (Sqrt[- 
2 - (1 - Sqrt[5])*x]*Sqrt[(2 + (1 + Sqrt[5])*x)/(2 + (1 - Sqrt[5])*x)]*Ell 
ipticF[ArcSin[(Sqrt[2]*5^(1/4)*Sqrt[x])/Sqrt[-2 - (1 - Sqrt[5])*x]], (5 - 
Sqrt[5])/10])/(4*5^(1/4)*(3 + Sqrt[5])*Sqrt[(2 + (1 - Sqrt[5])*x)^(-1)]*Sq 
rt[-1 - x + x^2]) + ((2 + Sqrt[5])*Sqrt[-1 + Sqrt[5] + 2*x]*Sqrt[1 - (2*x) 
/(1 + Sqrt[5])]*EllipticPi[(-1 - Sqrt[5])/2, ArcSin[Sqrt[2/(1 + Sqrt[5])]* 
Sqrt[x]], (-3 - Sqrt[5])/2])/(2*Sqrt[2]*(3 + Sqrt[5])*Sqrt[-1 - x + x^2...
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(219\) vs. \(2(105)=210\).

Time = 3.85 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.65

method	result	size
default	\(-\frac {\sqrt {5}\, \ln \left (\frac {x \sqrt {5}-\sqrt {x \left (x^{2}-x -1\right )}\, \sqrt {-2+2 \sqrt {5}}+x^{2}-x -1}{x}\right )-\sqrt {5}\, \ln \left (\frac {x \sqrt {5}+\sqrt {x \left (x^{2}-x -1\right )}\, \sqrt {-2+2 \sqrt {5}}+x^{2}-x -1}{x}\right )-10 \arctan \left (\frac {\sqrt {x \left (x^{2}-x -1\right )}}{x}\right ) \sqrt {2+2 \sqrt {5}}+\left (5+\sqrt {5}\right ) \left (\arctan \left (\frac {\sqrt {-2+2 \sqrt {5}}\, x +2 \sqrt {x \left (x^{2}-x -1\right )}}{x \sqrt {2+2 \sqrt {5}}}\right )-\arctan \left (\frac {\sqrt {-2+2 \sqrt {5}}\, x -2 \sqrt {x \left (x^{2}-x -1\right )}}{x \sqrt {2+2 \sqrt {5}}}\right )\right )}{20 \sqrt {2+2 \sqrt {5}}}\)	\(220\)
pseudoelliptic	\(-\frac {\sqrt {5}\, \ln \left (\frac {x \sqrt {5}-\sqrt {x \left (x^{2}-x -1\right )}\, \sqrt {-2+2 \sqrt {5}}+x^{2}-x -1}{x}\right )-\sqrt {5}\, \ln \left (\frac {x \sqrt {5}+\sqrt {x \left (x^{2}-x -1\right )}\, \sqrt {-2+2 \sqrt {5}}+x^{2}-x -1}{x}\right )-10 \arctan \left (\frac {\sqrt {x \left (x^{2}-x -1\right )}}{x}\right ) \sqrt {2+2 \sqrt {5}}+\left (5+\sqrt {5}\right ) \left (\arctan \left (\frac {\sqrt {-2+2 \sqrt {5}}\, x +2 \sqrt {x \left (x^{2}-x -1\right )}}{x \sqrt {2+2 \sqrt {5}}}\right )-\arctan \left (\frac {\sqrt {-2+2 \sqrt {5}}\, x -2 \sqrt {x \left (x^{2}-x -1\right )}}{x \sqrt {2+2 \sqrt {5}}}\right )\right )}{20 \sqrt {2+2 \sqrt {5}}}\)	\(220\)
trager	\(\text {Expression too large to display}\)	\(911\)
elliptic	\(\text {Expression too large to display}\)	\(1436\)

Input:

int(x^2/(x^3-x^2-x)^(1/2)/(x^4-1),x,method=_RETURNVERBOSE)
 

Output:

-1/20/(2+2*5^(1/2))^(1/2)*(5^(1/2)*ln((x*5^(1/2)-(x*(x^2-x-1))^(1/2)*(-2+2 
*5^(1/2))^(1/2)+x^2-x-1)/x)-5^(1/2)*ln((x*5^(1/2)+(x*(x^2-x-1))^(1/2)*(-2+ 
2*5^(1/2))^(1/2)+x^2-x-1)/x)-10*arctan((x*(x^2-x-1))^(1/2)/x)*(2+2*5^(1/2) 
)^(1/2)+(5+5^(1/2))*(arctan(((-2+2*5^(1/2))^(1/2)*x+2*(x*(x^2-x-1))^(1/2)) 
/x/(2+2*5^(1/2))^(1/2))-arctan(((-2+2*5^(1/2))^(1/2)*x-2*(x*(x^2-x-1))^(1/ 
2))/x/(2+2*5^(1/2))^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (97) = 194\).

Time = 0.16 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.16 \[ \int \frac {x^2}{\sqrt {-x-x^2+x^3} \left (-1+x^4\right )} \, dx=\frac {1}{4} \, \sqrt {\frac {1}{10} \, \sqrt {5} + \frac {1}{10}} \arctan \left (-\frac {{\left (5 \, x^{2} - \sqrt {5} {\left (x^{2} + 4 \, x - 1\right )} - 5\right )} \sqrt {\frac {1}{10} \, \sqrt {5} + \frac {1}{10}}}{4 \, \sqrt {x^{3} - x^{2} - x}}\right ) + \frac {1}{16} \, \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}} \log \left (\frac {x^{4} - 4 \, x^{3} + 6 \, x^{2} + 4 \, \sqrt {x^{3} - x^{2} - x} {\left (\sqrt {5} {\left (x^{2} - x - 1\right )} + 5 \, x\right )} \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}} + 4 \, \sqrt {5} {\left (x^{3} - x^{2} - x\right )} + 4 \, x + 1}{x^{4} + 2 \, x^{2} + 1}\right ) - \frac {1}{16} \, \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}} \log \left (\frac {x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, \sqrt {x^{3} - x^{2} - x} {\left (\sqrt {5} {\left (x^{2} - x - 1\right )} + 5 \, x\right )} \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}} + 4 \, \sqrt {5} {\left (x^{3} - x^{2} - x\right )} + 4 \, x + 1}{x^{4} + 2 \, x^{2} + 1}\right ) + \frac {1}{4} \, \arctan \left (\frac {x^{2} - 2 \, x - 1}{2 \, \sqrt {x^{3} - x^{2} - x}}\right ) \] Input:

integrate(x^2/(x^3-x^2-x)^(1/2)/(x^4-1),x, algorithm="fricas")
 

Output:

1/4*sqrt(1/10*sqrt(5) + 1/10)*arctan(-1/4*(5*x^2 - sqrt(5)*(x^2 + 4*x - 1) 
 - 5)*sqrt(1/10*sqrt(5) + 1/10)/sqrt(x^3 - x^2 - x)) + 1/16*sqrt(1/10*sqrt 
(5) - 1/10)*log((x^4 - 4*x^3 + 6*x^2 + 4*sqrt(x^3 - x^2 - x)*(sqrt(5)*(x^2 
 - x - 1) + 5*x)*sqrt(1/10*sqrt(5) - 1/10) + 4*sqrt(5)*(x^3 - x^2 - x) + 4 
*x + 1)/(x^4 + 2*x^2 + 1)) - 1/16*sqrt(1/10*sqrt(5) - 1/10)*log((x^4 - 4*x 
^3 + 6*x^2 - 4*sqrt(x^3 - x^2 - x)*(sqrt(5)*(x^2 - x - 1) + 5*x)*sqrt(1/10 
*sqrt(5) - 1/10) + 4*sqrt(5)*(x^3 - x^2 - x) + 4*x + 1)/(x^4 + 2*x^2 + 1)) 
 + 1/4*arctan(1/2*(x^2 - 2*x - 1)/sqrt(x^3 - x^2 - x))
 

Sympy [F]

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

integrate(x**2/(x**3-x**2-x)**(1/2)/(x**4-1),x)
 

Output:

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

Maxima [F]

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

integrate(x^2/(x^3-x^2-x)^(1/2)/(x^4-1),x, algorithm="maxima")
 

Output:

integrate(x^2/((x^4 - 1)*sqrt(x^3 - x^2 - x)), x)
                                                                                    
                                                                                    
 

Giac [F]

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

integrate(x^2/(x^3-x^2-x)^(1/2)/(x^4-1),x, algorithm="giac")
 

Output:

integrate(x^2/((x^4 - 1)*sqrt(x^3 - x^2 - x)), x)
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 533, normalized size of antiderivative = 4.01 \[ \int \frac {x^2}{\sqrt {-x-x^2+x^3} \left (-1+x^4\right )} \, dx =\text {Too large to display} \] Input:

int(x^2/((x^4 - 1)*(x^3 - x^2 - x)^(1/2)),x)
 

Output:

((5^(1/2)/2 + 1/2)*(x/(5^(1/2)/2 + 1/2))^(1/2)*((x + 5^(1/2)/2 - 1/2)/(5^( 
1/2)/2 - 1/2))^(1/2)*((5^(1/2)/2 - x + 1/2)/(5^(1/2)/2 + 1/2))^(1/2)*ellip 
ticPi(- (5^(1/2)*1i)/2 - 1i/2, asin((x/(5^(1/2)/2 + 1/2))^(1/2)), -(5^(1/2 
)/2 + 1/2)/(5^(1/2)/2 - 1/2)))/(2*(x^3 - x^2 - x*(5^(1/2)/2 - 1/2)*(5^(1/2 
)/2 + 1/2))^(1/2)) - ((5^(1/2)/2 + 1/2)*(x/(5^(1/2)/2 + 1/2))^(1/2)*((x + 
5^(1/2)/2 - 1/2)/(5^(1/2)/2 - 1/2))^(1/2)*((5^(1/2)/2 - x + 1/2)/(5^(1/2)/ 
2 + 1/2))^(1/2)*ellipticPi(5^(1/2)/2 + 1/2, asin((x/(5^(1/2)/2 + 1/2))^(1/ 
2)), -(5^(1/2)/2 + 1/2)/(5^(1/2)/2 - 1/2)))/(2*(x^3 - x^2 - x*(5^(1/2)/2 - 
 1/2)*(5^(1/2)/2 + 1/2))^(1/2)) - ((5^(1/2)/2 + 1/2)*(x/(5^(1/2)/2 + 1/2)) 
^(1/2)*((x + 5^(1/2)/2 - 1/2)/(5^(1/2)/2 - 1/2))^(1/2)*((5^(1/2)/2 - x + 1 
/2)/(5^(1/2)/2 + 1/2))^(1/2)*ellipticPi(- 5^(1/2)/2 - 1/2, asin((x/(5^(1/2 
)/2 + 1/2))^(1/2)), -(5^(1/2)/2 + 1/2)/(5^(1/2)/2 - 1/2)))/(2*(x^3 - x^2 - 
 x*(5^(1/2)/2 - 1/2)*(5^(1/2)/2 + 1/2))^(1/2)) + ((5^(1/2)/2 + 1/2)*(x/(5^ 
(1/2)/2 + 1/2))^(1/2)*((x + 5^(1/2)/2 - 1/2)/(5^(1/2)/2 - 1/2))^(1/2)*((5^ 
(1/2)/2 - x + 1/2)/(5^(1/2)/2 + 1/2))^(1/2)*ellipticPi((5^(1/2)*1i)/2 + 1i 
/2, asin((x/(5^(1/2)/2 + 1/2))^(1/2)), -(5^(1/2)/2 + 1/2)/(5^(1/2)/2 - 1/2 
)))/(2*(x^3 - x^2 - x*(5^(1/2)/2 - 1/2)*(5^(1/2)/2 + 1/2))^(1/2))
 

Reduce [F]

\[ \int \frac {x^2}{\sqrt {-x-x^2+x^3} \left (-1+x^4\right )} \, dx=\int \frac {\sqrt {x}\, \sqrt {x^{2}-x -1}\, x}{x^{6}-x^{5}-x^{4}-x^{2}+x +1}d x \] Input:

int(x^2/(x^3-x^2-x)^(1/2)/(x^4-1),x)
 

Output:

int((sqrt(x)*sqrt(x**2 - x - 1)*x)/(x**6 - x**5 - x**4 - x**2 + x + 1),x)