How to apply Risch's algorithm to $\int\frac{x}{\sqrt{x^4-2x^3+3x^2+4x+1}}\,\mathrm{d}x$?

Question

It is known that, for $y^2=x^4-2x^3+3x^2+4x+1$, $$\int\frac{x\,\mathrm{d}x}{y}=\frac13\mathrm{arctanh}\frac{x^2-x+1}{y}+\frac13\mathrm{arctanh}\frac{x^2-x-1}{y}+\frac16\mathrm{arctanh}\frac{x^2-x-1/2}{y}+C.$$

To get this result, I tried to apply Risch's algorithm as described in Manuel Bronstein's Symbolic Integration Tutorial, as follows. The integrand is $xy/D$ where $y^2=D=x^4-2x^3+3x^2+4x+1$ is square-free, so the indefinite integral has only a logarithm part, and we then get to section 2.5. There a procedure is described, which in step 3 computes the polynomial $$R(t)=\mathrm{Resultant}_x(\mathrm{Resultant}_y(xy-t\frac{\mathrm{d}D}{\mathrm{d}x},y^2-D),D).$$ From this formula, we have $R(t)=186624t^8$ for the given integrand. This has no nonzero roots, so by step 6 the integral is not elementary. This contradicts the antiderivative result given above. What is wrong?

Answer 1

The residue polynomial calculated from the double resultant only picks up residues at finite places. In this case, since the form $\frac{dx}{y}$ is holomorphic at all finite places on the curve $y^2=x^4-2x^3+3x^2+4x+1$, the resultant fails to pick up anything.

In reality, the residues live at infinity: the hyperelliptic curve $y^2=P(x)$ has two unramified points over infinity if the degree of $P$ is even, and in the case deg $P=4$, we know that $\frac{dx}{y}$ extends to a holomorphic, nonzero form there. But with $x=\infty$, $x \left(\frac{dx}{y}\right)$ picks up a simple pole at infinity.

Consider the change of variables $x=1+\frac{1}{z}$. This sends $\infty$ to $0$, and we get the new integrand $-(1+z)w/(7z^5+8z^4+3z^3+2z^2+z) dz$, with $w = z^2y$ defined by $w^2=7z+8z+3z+2z+1$. We compute the residue polynomial $9144576 z^{10}-9144576 z^8$, which has roots $0$ and $\pm 1$. We recover the two roots $\pm 1$.

The roots of the residue polynomial are nonzero, rational multiples of the residues. Noting that the integrand can also be written as $-(1+z)/z * (dz/w)$, and $dz/w$ is holomorphic everywhere, the only residues occur when z=0. At the two locations above $z=0$, which are $(0,1)$ and (0,-1). The residue at $(0,1)$ is $-1$, and the residue at $(0,-1)$ is $1$.

Continuing along the tutorial, in step $4$ we compute $s=2, q_1=1, k=1,$ $\alpha_1=r_{11} * 1 = 1, \alpha_2 = r_{21} * 1 = -1$. In step 5 we are asked to form the divisor $\delta_j = \sum_{i=1}^{s} r_{ij} \sum_l r_l P_l = \sum_{i=1}^{2} r_{ij} \sum_l P_l$ for each $1 \leq j \leq k$; so

$$\delta_1 = [0,-1]-[0,1]$$

Thus, we seek a function with a degree $k$ zero at $[0,-1]$ and a degree $k$ pole at $[0,1]$ for some nonzero integer $k$, and no zeros or poles anywhere else. This turns out to be highly nontrivial by hand, but if passed into friCAS, results in the function $u=\frac{(2z^4-2z^3+6z^2-2z+2)w+z^6-12z^5-8z^3-6z^2-2}{z^6}$ having the prescribed zeros and poles with $k=6$, whereby the integral is $\frac{\log(u)}{6}$, which after change of variables results in the final answer.