On 12/19/23 01:37, Waldek Hebisch wrote:
On Mon, Dec 18, 2023 at 06:12:01PM +0800, Qian Yun wrote:
I'm opening a new thread for this newly added rsimp.spad.
I have tested with first 10k of Nasser's integrals, the recent
changes to integrator is positive:
It affects the result of a few percent integrals, almost all are
positive improvement.
Only a few regressions, in 3 classes:
1. not invertiable:
integrate(x^6/(2*x^5+3)^3,x)
This is caused by troubles with other routines. Namely 'complex_roots'
gets polynomial witgh dependent roots which causes failure in 'factor'.
To reproduce do:
a := (-1)^(1/5)/(108)^(1/5)
p1 := x^4 + a*x^3/250 + a^2*x^2/62500 + a^3*x/15625000 + a^4/3906250000
factor(p1::UP('x, EXPR(INT))::SUP(EXPR(INT)))
If one replaces a by
a := subst(b^(1/5), b = -1/108)
then there is error later, again coused by dependent roots. This
time 'factor' returns empty list of factors for polynomial of
degree 2...
I have a workaround, but ATM I am not sure how to fix this properly.
Do you think this part is problematic?
"alpha := rootSimp zeroOf p"
for p is (x^5+1), alpha is (-1)^(1/5) instead of -1.
(unlike radicalSovle)
With this patch I can get better result.
=====
diff --git a/src/algebra/irexpand.spad b/src/algebra/irexpand.spad
index 4195f3b7..760f14bd 100644
--- a/src/algebra/irexpand.spad
+++ b/src/algebra/irexpand.spad
@@ -185,7 +185,12 @@ IntegrationResultToFunction(R, F) : Exports ==
Implementation where
d = 4 => quartic(p, lg.logand, x)
odd? d and
((r := retractIfCan(reductum p)@Union(F, "failed")) case F) =>
- pairsum([cmplex(alpha := rootSimp zeroOf p, lg.logand)],
+ sn := sign(r)
+ if sn case Integer and (sn::Integer) > 0 then
+ alpha := - rootSimp nthRoot(r, d)
+ else
+ alpha := rootSimp zeroOf p
+ pairsum([cmplex(alpha, lg.logand)],
lg2func([lg.scalar,
(p exquo (monomial(1, 1)$UP - alpha::UP))::UP,
lg.logand], x))
=====
2. third "impossible" in rsimp.spad:
integrate(1/(x^4+4*x^2+1),x)
3. fourth "impossible" in rsimp.spad:
This error happens when resolvent contains zero factor, it seems.
integrate(log(x^2+(-x^2+1)^(1/2)),x)
integrate(log(1+x*(x^2+1)^(1/2)),x)
integrate(atan(x*(-x^2+1)^(1/2)),x)
integrate(1/(x^4+3*x^2-1),x)
integrate(1/(x^4-3*x^2-1),x)
integrate(1/(x^2-1)^(1/2)/(x^(1/2)+(x^2-1)^(1/2))^2,x)
integrate((x^(1/2)-(x^2-1)^(1/2))^2/(-x^2+x+1)^2/(x^2-1)^(1/2),x)
All of the above are fixed by adding special case for biquadratic.
I missed possibility that irreducible quartic may have 4 purely imaginary
roots, so real part may be 0. Case of general quartic needs rechecking,
Just to clarify, you revert some class of biquadratic to old behavior,
right?
Otherwise, your patch works.
- Qian
while in our case real part may be 0 implies biquadratic, so will
no longer happen but it is possible that I also missed something
else.
Patch for biquadratic in the attachment.
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