On Mon, Dec 04, 2023 at 01:52:39AM +0100, Waldek Hebisch wrote:
> Third version in the attachemtnt, this time handling rather
> general quartics where we can find real parts of roots. It
> uses precomputed resolvent. After normalizing so that sum
> of roots is zero resolvent is an even polynomial, so we
> get equation of degree 3 for square of real part. If
> this has one posivite real root which can be obtained via
> factoring, then we are in business, otherwise this fails.
The routine fails on integrals like
integrate((z^2+z)^(1/2)/(1+z^2)^2, z)
where the poly is a^4 - (1/64)*a^2 + 1/8192 and
resolvent is 64*u^3 + (1/2)*u^2 - (1/1024)*u which
factors as 64* u*(u^2 + (1/128)*u - 1/65536. Here linear
factor must be discarded and quadratic factor has one
positive root. We should use this positive root, but
currently the routine gives up.
--
Waldek Hebisch
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