I have examined integrals where rootSum is over degree 4, from a
subset of Nasser's test.
Other less common cases are:
%A^8+1 integrate(log(1/x^4+x^4),x)
%A^4-1/2048 integrate(1/(x^4-2),x)
%A^6-1/1492992 integrate(1/(x^6-2),x)
%A^6+1/1492992 integrate(1/(x^6+2),x)
%A^5-7776/3125 integrate(x^(1/2)/(-1/x^(1/3)+x^(1/2)),x)
(factors to %A^4+6/5*%A^3+36/25*%A^2+216/125*%A+1296/625)
%A^6+%A^3+1 integrate(((-3)*x^3+6)/(x^6+(-1)*x^3+1), x)
Of course, there are corresponding cases where coefficients are
POLY INT.
I'll do more testing to see if the generic resultant method can
handle them or special cases are required.
- Qian
On 11/25/23 04:45, Waldek Hebisch wrote:
On Tue, Nov 21, 2023 at 06:31:27PM +0800, Qian Yun wrote:
How do you suggest to proceed? Write a special case for biquadratic?
AFAICS we can quickly add case for x^4 + a, that when we can
determine sign of a. Case for biquadratic is more complicated
but should be doable with moderate effort. Both should be
rather efficient as they do not need more general factoring
nor complicated solve.
We could try to handle more cases with easy Galois group
by factoring resolvent (resultant of P and Q). This is
more complicated, but still moderately hard.
I think that handling more general cases requires more
research and will take some time. So it makes sense to
do easier cases first.
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