You can factor x^8+1 by adjoining a root of unity, e.g.
factor(x^8+1,a^2+1) gives a product of two quartics
factor(x^8+1,a^4+1) gives a product of four quadratics
In both cases, 1/... is integrable. You can then substitute back in the
value of a.
But the result is rather messy.
On 10/31/05, Lodder, Josje wrote:
>
> Is it possible with Maxima to factor x^8+1 in quadratic terms? In fact I
> tried to integrate 1/(x^8+1) but since Maxima couldn't do this (is that
> wright) I tried the factorization, Josje lodder
>