How to integrate $ \int \frac{x}{\sqrt{x^4+10x^2-96x-71}}dx$?

Question

I read about $ \int \dfrac{x}{\sqrt{x^4+10x^2-96x-71}}dx$ on the Wikipedia Risch algorithm page. They gave an answer but I don't understand how they got it.

Answer 1

The problem is $\int\frac{x\,\mathrm{d}x}{y}$ where $y^2=x^4+10x^2-96x-71$ is a polynomial with integer coefficients and no double roots. There is a rarely-known “Chebyshev substitution” which decides if integrals like this one are elementary functions.

His criterion is the following: the $\mathbb{Z}[x]$-continued fraction expansion of $y$ contains a quadratic coefficient after the constant (e.g. $y=[c_0;c_1,\ldots,c_n,c_{n+1},\ldots]$ where $c_0,c_{n+1}$ are quadratic and $c_1,\ldots,c_n$ are linear) iff the integral is an elementary function. In this case “$w=$ the numerator of $[c_0;c_1,\ldots,c_n]$” is a good substitution.

So we first work out the continued fraction expansion of $y$: $$y=x^2+5-\frac{48}{x-1}\mathbin{\raise{-11mu}{+}}\frac{6}{x+5}\mathbin{\raise{-11mu}{+}}\frac{24}{x-4}\mathbin{\raise{-11mu}{-}}\frac{3}{x-4}\mathbin{\raise{-11mu}{+}}\frac{24}{x+5}\mathbin{\raise{-11mu}{+}}\frac{12}{x-1}\mathbin{\raise{-11mu}{-}}\frac{48}{x^2+5+y}.$$ Now $c_1,\ldots,c_6$ are linear, but $c_7$ is quadratic, so the integral is an elementary function (or, a pseudoelliptic integral). We have $$ [c_0;c_1,\ldots,c_6]=\frac{x^8+20x^6-128x^5+54x^4-1408x^3+3124x^2+10001}{x^6+15x^4-80x^3+27x^2-528x+781}=\frac{w}{z}.$$ For these polynomials we observe that $w^2-y^2z^2=2^{16}\cdot3^7$ and $w'=8xz$, so $$\int\frac{x\,\mathrm{d}x}{y}=\int\frac{xz\,\mathrm{d}x}{yz}=\frac18\int\frac{\mathrm{d}w}{\sqrt{w^2-2^{16}\cdot3^7}}=\frac18\log\left(w+\sqrt{w^2-2^{16}\cdot3^7}\right)+C.$$

This method is found in Chebyshev's paper “On the integration of differentials that contain a square root of a third or fourth degree polynomial”, available in French here.