Evaluation of $ \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx$ 
Evaluation of $\displaystyle \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int\frac{x^2-1}{\sqrt{x^4+4x^3+7x^2+4x+1}}dx = \int\frac{x^2-1}{\sqrt{(x+1)^4+x^2}}dx$
So $\displaystyle I = \int\frac{x^2-1}{x\cdot \sqrt{\left(\frac{(x+1)^2}{x}\right)^2+1}}dx\;,$ ow Put $\displaystyle \frac{(x+1)^2}{x}=t\;,$
Then $\displaystyle \frac{2x(x+1)-(x+1)^2}{x^2}dx = dt\Rightarrow \left(\frac{x^2-1}{x^2}\right)dx = dt$
Can we solve the Integral in terms of elementry function.
If yes then plz explain here, 
Thanks

It looks to me that an elliptic integral is unavoidable. 
By replacing $\sqrt{x}+\frac{1}{\sqrt{x}}$ with $y$ the problem boils down to computing:
$$ \int \left(y^3-2y+y^2\sqrt{y^2-4}\right)\,\frac{dy}{\sqrt{1+y^4}}=\frac{\sqrt{1+y^4}}{2}+\text{arcsinh}(y^2)+\frac{1}{2}\int\sqrt{\frac{z^2-4z}{z^2+1}}\,dz $$
where in the last step I set $z=y^2$.
