On Thu, Sep 22, 2022 at 04:37:21AM -0700, 'Nasser M. Abbasi' via FriCAS -
computer algebra system wrote:
> Fyi,
>
> CAS integrations tests SQL database is now fully build.
>
Thanks for the database. I looked at result in terms of Rubi
steps and FriCAS performace is decreasing with number of Rubi
steps (first is number of steps, second is fraction solved
by FriCAS):
[1.0, 0.8345051379_1238507301]
[2.0, 0.8338310374_7230517291]
[3.0, 0.8501565276_2501057619]
[4.0, 0.8362068965_5172413793]
[5.0, 0.8186005976_0956175299]
[6.0, 0.8175120430_7169169736]
[7.0, 0.7909229984_7016828149]
[8.0, 0.7562585343_6504324078]
[9.0, 0.7237969676_994067238]
[10.0, 0.6470033034_4502123643]
[11.0, 0.6482444733_4200260078]
[12.0, 0.6384388807_0692194404]
[13.0, 0.6023007395_2341824158]
[14.0, 0.6090225563_9097744361]
[15.0, 0.5511288180_6108897742]
[16.0, 0.6092307692_3076923077]
[17.0, 0.5546719681_9085487078]
[18.0, 0.5558510638_2978723404]
[19.0, 0.5223880597_0149253731]
[20.0, 0.5607843137_2549019608]
BTW: I am not sure how you maintain "has_known_anti". The following
9 have has_known_anti = 0, but belong to well-known class having
elementary integrals:
integrate(cos(d*x+c)^4/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")
integrate(cos(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")
integrate(1/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")
integrate(sec(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")
integrate(sec(d*x+c)^4/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")
integrate(sinh(d*x+c)^3/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")
integrate(sinh(d*x+c)/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")
integrate(csch(d*x+c)/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")
integrate(csch(d*x+c)^3/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")
FriCAS, Maple and almost surely MMa can do them. I suspect that
Maple or MMa answer is nicer, so you can use it as optimal one.
There is buch of others, one would have to check if they are correct.
But the above are easy ones.
--
Waldek Hebisch
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