Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$ Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$  but nothing happen.
I felt that it must be in Elliptical Integral of first kind, second kind or in third  kind.
Can someone explain me how to write in elliptical form

Maple expresses it as
$$ {\frac { \left( i\sqrt {3}+1 \right) \sqrt {4+2\,{x}^{2}-2\,i{x}
^{2}\sqrt {3}}\sqrt {4+2\,{x}^{2}+2\,i{x}^{2}\sqrt {3}} \left( i{\it 
EllipticF} \left(  \left( i\sqrt {3}+1 \right) x/2,(1+i\sqrt {3})/2
 \right) \sqrt {3}+3\,{\it EllipticF} \left( \left( i\sqrt {3}+1
 \right) x/2,(1+i\sqrt {3})/2 \right) -2\,{\it EllipticE} \left(
 \left( i\sqrt {3}+1 \right) x/2,(1+i\sqrt {3})/2 \right)  \right) }{16\;
\sqrt {x^4+x^2+1}}}
$$
So yes, it is expressed in terms of elliptic integrals of the first and second kinds (EllipticF is first kind, EllipticE is second kind).
