\(\int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx\) [491]

Optimal result

Integrand size = 23, antiderivative size = 282 \[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=-\frac {1+x^3}{x \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {1+x^3}{\sqrt {1+x} \left (1+\sqrt {3}+x\right ) \sqrt {1-x+x^2}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} E\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}}+\frac {\sqrt {2} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.66 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=-\frac {\sqrt {1+x} \sqrt {1-x+x^2}}{x}+\frac {(1+x)^{3/2} \left (\frac {12 \sqrt {-\frac {i}{3 i+\sqrt {3}}} \left (1-x+x^2\right )}{(1+x)^2}+\frac {3 \sqrt {2} \left (1-i \sqrt {3}\right ) \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} E\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {1+x}}+\frac {i \sqrt {2} \left (3 i+\sqrt {3}\right ) \sqrt {\frac {3 i+\sqrt {3}-\frac {6 i}{1+x}}{3 i+\sqrt {3}}} \sqrt {\frac {-3 i+\sqrt {3}+\frac {6 i}{1+x}}{-3 i+\sqrt {3}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {1+x}}\right )}{12 \sqrt {-\frac {i}{3 i+\sqrt {3}}} \sqrt {1-x+x^2}} \] Input:

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

Output:

-((Sqrt[1 + x]*Sqrt[1 - x + x^2])/x) + ((1 + x)^(3/2)*((12*Sqrt[(-I)/(3*I 
+ Sqrt[3])]*(1 - x + x^2))/(1 + x)^2 + (3*Sqrt[2]*(1 - I*Sqrt[3])*Sqrt[(3* 
I + Sqrt[3] - (6*I)/(1 + x))/(3*I + Sqrt[3])]*Sqrt[(-3*I + Sqrt[3] + (6*I) 
/(1 + x))/(-3*I + Sqrt[3])]*EllipticE[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3] 
)]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[1 + x] + (I*Sqrt[2 
]*(3*I + Sqrt[3])*Sqrt[(3*I + Sqrt[3] - (6*I)/(1 + x))/(3*I + Sqrt[3])]*Sq 
rt[(-3*I + Sqrt[3] + (6*I)/(1 + x))/(-3*I + Sqrt[3])]*EllipticF[I*ArcSinh[ 
Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[3]) 
])/Sqrt[1 + x]))/(12*Sqrt[(-I)/(3*I + Sqrt[3])]*Sqrt[1 - x + x^2])
 

Rubi [A] (warning: unable to verify)

Time = 0.66 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1210, 847, 832, 759, 2416}

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.

\(\displaystyle \int \frac {1}{x^2 \sqrt {x+1} \sqrt {x^2-x+1}} \, dx\)

\(\Big \downarrow \) 1210

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 832

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

\(\Big \downarrow \) 759

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

\(\Big \downarrow \) 2416

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

Input:

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

Output:

(Sqrt[1 + x^3]*(-(Sqrt[1 + x^3]/x) + ((2*Sqrt[1 + x^3])/(1 + Sqrt[3] + x) 
- (3^(1/4)*Sqrt[2 - Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^ 
2]*EllipticE[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]]) 
/(Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3]) - (2*(1 - Sqrt[3])*Sqrt 
[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[Ar 
cSin[(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]))/2))/(Sqrt[1 + x]*Sqrt[1 - x + 
 x^2])
 

Defintions of rubi rules used

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 832

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 - Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && PosQ[a]
 

rule 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 1210

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + 
c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p])   Int[(f + g*x)^n*(a*d + c* 
e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 
 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
 

rule 2416

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S 
imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt 
[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) 
*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 

Maple [A] (verified)

Time = 2.17 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.76

method	result	size
elliptic	\(\frac {\sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (-\frac {\sqrt {x^{3}+1}}{x}+\frac {\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{\sqrt {x^{3}+1}}\right )}{\sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(215\)
risch	\(-\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x}+\frac {\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right ) \sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}}{\sqrt {x^{3}+1}\, \sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(222\)
default	\(\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (i \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) \sqrt {3}\, x -6 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticE}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x +3 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x -2 x^{3}-2\right )}{2 x \left (x^{3}+1\right )}\)	\(363\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.11 \[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=-\frac {x {\rm weierstrassZeta}\left (0, -4, {\rm weierstrassPInverse}\left (0, -4, x\right )\right ) + \sqrt {x^{2} - x + 1} \sqrt {x + 1}}{x} \] Input:

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

Output:

-(x*weierstrassZeta(0, -4, weierstrassPInverse(0, -4, x)) + sqrt(x^2 - x + 
 1)*sqrt(x + 1))/x
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {1}{x^{2} \sqrt {x + 1} \sqrt {x^{2} - x + 1}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} - x + 1} \sqrt {x + 1} x^{2}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int { \frac {1}{\sqrt {x^{2} - x + 1} \sqrt {x + 1} x^{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {1}{x^2\,\sqrt {x+1}\,\sqrt {x^2-x+1}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^2 \sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{5}+x^{2}}d x \] Input:

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

Output:

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