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It is known that $$\int\frac{x}{\sqrt{x^4-2x^3+3x^2+4x+1}}\,\mathrm{d}x=$$ $$-\frac{1}{6}\log\Big((2x^4-10x^3+24x^2-28x+14)\sqrt{x^4-2x^3+3x^2+4x+1}-2x^6+12x^5-36x^4+56x^3-42x^2+13\Big)+C.$$

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?

Yizhen Chen
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  • I don't know the specifics of Risch's algorithm nor have time now to look into this, but possibly the work I did in my extensive answer to the following less specific question about a similar integral could be of use: How to integrate $ \int \frac{x}{\sqrt{x^4+10x^2-96x-71}}dx$? – Dave L. Renfro Aug 09 '20 at 17:47
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    @dave-l-renfro The integral can be reduced to elliptic functions using the standard procedure you mentioned. However, the integral is actually pseudo-elliptic, so I need a procedure that can identify pseudo-elliptic integrals. Thus I am asking how to apply Risch's algorithm which claims to work. – Yizhen Chen Aug 11 '20 at 07:53

0 Answers0