Re: [Reduce-algebra-developers] sometimes reduce process gets killed on its own

[Arthur N. <ac...@ca...>]

On Sun, 7 Jan 2024, Arthur Norman wrote:
>> redcsl -w
>> Reduce (CSL, rev 6657), 10-Dec-2023 ...
>> 
>> 1: int( 
>> x*(1+x)^(2/3)*(1-x)^(1/2)/(-(1-x)^(5/6)*(1+x)^(1/3)+(1-x)^(2/3)*(1+x)^(1/2)),x);
>> Killed
>> After about 10-15 minutes. I am running this on Linux Manjaro inside
>> virtual box with 64 GB RAM for the VBox, and on fast PC.
>
This message is mainly for other Reduce maintainers. When I arrange to be 
able to get a backtrace from this (which took be a little bit of fiddling, 
but redefining systemgchook to raise an error dis the trick) I see a 
fragment of the traceback saying as follows where there has been an 
attempt to make progress via trig functions, where abominably large 
integers are in play and I end up in a hugely deep recursion of !*multf. 
This after only around 30 seconds of CPU time. The expression I see looks 
as if it includes a cos(!!intvar2)^1364. Ha ha ha.

Via "on trint" I can see it has started by shifting so that instead of 
both x+1 and x-1 it has y and y-2 or some such and there is anther 
conversion => 1-w^2. We get

Start of Integration; integrand is

                          2     2/3                      2
            2*( - !intvar1  + 2)   *!intvar1*( - !intvar1  + 1)
--------------------------------------------------------------------------
          1/3          1/3             2     1/3                   2
  !intvar1   *(!intvar1   *( - !intvar1  + 2)    - sqrt( - !intvar1  + 2))


and that prompts a conversion to

Start of Integration; integrand is

                 1/3  2/3              2                                  2
(4*cos(!intvar2)   *2   *cos(!intvar2) *sin(!intvar2)*( - 2*cos(!intvar2) 
+ 1))

          1/3               1/3
/(sqrt(2)   *(sin(!intvar2)   *cos(!intvar2)*sqrt(2)

                         1/3                  2     1/3              2/3 
1/3
                - sqrt(2)   *( - cos(!intvar2)  + 1)   *cos(!intvar2)   *2 
))

Determination of the differential field descriptor
gives the functions:
((expt (cos !!intvar2) (quotient 1 3)) (expt (plus (minus (expt (cos 
!!intvar2) 2)) 1) (quotient 1 3)) (expt (sin !!intvar2) (quotient 1 3)) 
(sin !!intvar2) (cos !!intvar2) !!intvar2)

after which things fairly repidly go ugly!

Any ideas?
Arthur


> 
-2006582604045247462173839524414111582012306138161916297721207009532444822043
> ))))))
> Inside: transcendentalcase
> Arg1: ((((#1=(cos !!intvar2) . 4) ((#2=(expt #1# (quotient 1 3)) . 1) 
((#3=(e
>  2 (quotient 1 3)) . 2) ((#4=(sin !!intvar2) . 1) . -8)))) ((#1# . 2) 
((#2# .
>  ((#3# . 2) ((#4# . 1) . 4))))) ((#1# . 1) (((expt #4# (quotient 1 3)) . 
1) (
> =(expt #5=(sqrt 2) (quotient 1 3)) . 1) ((#5# . 1) . 1)))) ((#2# . 2) 
(((expt
> plus (minus (expt #1# 2)) 1) (quotient 1 3)) . 1) ((#6# . 2) ((#3# . 1) 
. -1)
> )
> Arg2: !!intvar2
> Arg3: nil
> Arg4: ((expt #1=(cos !!intvar2) (quotient 1 3)) (expt (plus (minus (expt 
#1#
>  1) (quotient 1 3)) (expt #2=(sin !!intvar2) (quotient 1 3)) #2# #1# 
!!intvar
> Arg5: ((expt 2 (quotient 1 3)) (expt #1=(sqrt 2) (quotient 1 3)) #1#)
> Inside: integratesq
> Arg1: ((((#1=(cos !!intvar2) . 4) ((#2=(expt #1# (quotient 1 3)) . 1) 
((#3=(e
>  2 (quotient 1 3)) . 2) ((#4=(sin !!intvar2) . 1) . -8)))) ((#1# . 2) 
((#2# .
>  ((#3# . 2) ((#4# . 1) . 4))))) ((#1# . 1) (((expt #4# (quotient 1 3)) . 
1) (
> =(expt #5=(sqrt 2) (quotient 1 3)) . 1) ((#5# . 1) . 1)))) ((#2# . 2) 
(((expt
> plus (minus (expt #1# 2)) 1) (quotient 1 3)) . 1) ((#6# . 2) ((#3# . 1) 
. -1)
> )
> Arg2: !!intvar2
> Arg3: nil
> Arg4: nil
> Calling transcendentalcase from integratesq
> Arg1: ((((#1=(expt (plus (minus (expt !!intvar1 6)) 2) (quotient 1 3)) . 
2) (
> !!intvar1 . 10) . -6) ((!!intvar1 . 4) . 6))) ((#1# . 1) ((!!intvar1 . 
1) . 1
> (((sqrt (plus (minus (expt !!intvar1 6)) 2)) . 1) . -1))
> Arg2: !!intvar1