I'm trying to find the continued fraction expansion for the square root of a quartic polynomial, e.g. $\sqrt{x^4+10x^2-96x-71}$. I am more interested in the method than the answer, since it appears that one of the criteria for $$\int\frac{x\,\mathrm{d}x}{\sqrt{x^4+ax^2+bx+c}}$$ to have an elementary closed-form solution is for the continued fraction expansion of the denominator to have a quadratic term after the leading coefficient.
I'm trying to prove that $$\sqrt{x^4+10x^2-96x-71}=x^2+5-\cfrac{48}{x-1+\cfrac{6}{x+5+\cfrac{24}{x-4-\cfrac{3}{x-4+\cfrac{24}{x+5+\cfrac{12}{x-1+\cfrac{48}{x^2+5+\ddots}}}}}}}$$ I can get everything through the constant term: $$\sqrt{x^4+10x^2-96x-71}-(x^2+5)=-\frac{96(x+1)}{x^2+5+\sqrt{x^4+10x^2-96x-71}}$$ It's clear to me how to get 48 by substituting this into itself, but I'm not sure how to get the correct denominator since straight substitution gets me $$\sqrt{x^4+10x^2-96x-71}=x^2+5-\cfrac{96(x+1)}{2(x^2+5)-\cfrac{96(x+1)}{2(x^2+5)-\ddots}} $$
Can anyone help me derive the continued fraction expansion above, so I can apply it to other similar functions?
Thanks in advance.