> As I wrote in other mail I would like to generate 'rootSum'
> instead of expanding roots. In principle after such change
> definite integrator would work with 'rootSum' directly
> skiping Rioboo, tantrick, etc.
I don't understand. Why should integrate(1/(x^8-1),x) and
oldk1331 wrote:
>
> Compare the result of
> integrate(1/(x^8-1),x)
> integrate(8*sqrt(2)/(x^8-1),x)
> integrate(a/(x^8-1),x)
>
> The first is more complex than the second one.
> Why? Because:
> (1) -> factor(x^4+1)
>
> 4
>(1) x + 1
>
