" 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:"
Hello;
I myself do not edit or modify the input files. The Rubi input files are
maintained by Albert Rich.
This way there is only only source. If I modify or edit these files, then
things will get out of sync.
I will forward these integrals you found to Albert Rich and let him know
that these have known
anti-derivatives so he can update the files on his end for next release.
But first I would
need to know the list of the known antiderivatives for these to send them
as well.
For example the first integral in your list is integral number 399 in file
4_Trig_functions/4.1_Sine/4.1.7-d_trig-^m-a+b-c_sin-^n-^p.txt
And the Rubi input file says
{Cos[c + d*x]^4/(a + b*Sin[c + d*x]^3)^2, x, 0, Unintegrable[Cos[c +
d*x]^4/(a + b*Sin[c + d*x]^3)^2, x]}
It is the 4th field above which says it is not integrable. Otherwise it
will have the optimal
antiderivative there.
Mathematica gives solution but in terms of rootsof
In[2]:= Integrate[Cos[c+d*x]^4/(a+b*Sin[c+d*x]^3)^2,x]
Out[2]= (1/(18 a b d))(-I RootSum[-I b+3 I b #1^2+8 a #1^3-3 I b #1^4+I b
#1^6&,(2 b ArcTan[Sin[c+d x]/(Cos[c+d x]-#1)]-I b Log[1-2 Cos[c+d x]
#1+#1^2]+4 I a ArcTan[Sin[c+d x]/(Cos[c+d x]-#1)] #1+2 a Log[1-2 Cos[c+d x]
#1+#1^2] #1+12 b ArcTan[Sin[c+d x]/(Cos[c+d x]-#1)] #1^2-6 I b Log[1-2
Cos[c+d x] #1+#1^2] #1^2-4 I a ArcTan[Sin[c+d x]/(Cos[c+d x]-#1)] #1^3-2 a
Log[1-2 Cos[c+d x] #1+#1^2] #1^3+2 b ArcTan[Sin[c+d x]/(Cos[c+d x]-#1)]
#1^4-I b Log[1-2 Cos[c+d x] #1+#1^2] #1^4)/(b #1-4 I a #1^2-2 b #1^3+b
#1^5)&]+(24 Cos[c+d x] (a+b Sin[c+d x]))/(4 a+3 b Sin[c+d x]-b Sin[3 (c+d
x)]))
And Rubi does not solve it
In[4]:= Int[Cos[c+d*x]^4/(a+b*Sin[c+d*x]^3)^2,x]
Out[4]= Int[Cos[c+d x]^4/(a+b Sin[c+d x]^3)^2,x]
Maple gives solution is also in terms of rootof
sol:=int(cos(c + d*x)^4/(a + b*sin(c + d*x)^3)^2,x):
lprint(sol)
-2/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*tan(1/2*d*x+1/2*c)^3*
b+3*tan(1/2*d*x+1/2*c)^2*a+a)/a*tan(1/2*d*x+1/2*c)^5+2/3/d/(tan(1/2*d*x+1/2*c)^
6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*tan(1/2*d*x+1/2*c)^3*b+3*tan(1/2*d*x+1/2*c)^2*a+
a)/b*tan(1/2*d*x+1/2*c)^4+8/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*
a+8*tan(1/2*d*x+1/2*c)^3*b+3*tan(1/2*d*x+1/2*c)^2*a+a)/a*tan(1/2*d*x+1/2*c)^3+4
/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*tan(1/2*d*x+1/2*c)^3*b+
3*tan(1/2*d*x+1/2*c)^2*a+a)/b*tan(1/2*d*x+1/2*c)^2+2/3/d/(tan(1/2*d*x+1/2*c)^6*
a+3*tan(1/2*d*x+1/2*c)^4*a+8*tan(1/2*d*x+1/2*c)^3*b+3*tan(1/2*d*x+1/2*c)^2*a+a)
/a*tan(1/2*d*x+1/2*c)+2/3/d/(tan(1/2*d*x+1/2*c)^6*a+3*tan(1/2*d*x+1/2*c)^4*a+8*
tan(1/2*d*x+1/2*c)^3*b+3*tan(1/2*d*x+1/2*c)^2*a+a)/b+2/9/d/a/b*sum((_R^4*b+_R^3
*a+_R*a+b)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R =
RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))
Looking at Fricas I see it timed out on this:
sqlite> select fricas_anti from main where rowid=37649;
Timed out
But this must be because of the use of the flag set to true.
When I tried it now without setting the flag:
ii:=integrate(cos(d*x+c)^4/(a+b*sin(d*x+c)^3)^2,x);
Fricas gave a very large output also using Root object. But you said
antiderivative is
in terms of elementary functions?
I did not look at the rest of the list you showed.
Regards,
--Nasser
On Saturday, September 24, 2022 at 9:09:24 AM UTC-5 Waldek Hebisch wrote:
> 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 <(833)%20831-0374>_7230517291]
> [3.0, 0.8501565276_2501057619]
> [4.0, 0.8362068965_5172413793 <(517)%20241-3793>]
> [5.0, 0.8186005976 <(818)%20600-5976>_0956175299]
> [6.0, 0.8175120430 <(817)%20512-0430>_7169169736 <(716)%20916-9736>]
> [7.0, 0.7909229984_7016828149 <(701)%20682-8149>]
> [8.0, 0.7562585343_6504324078 <(650)%20432-4078>]
> [9.0, 0.7237969676_994067238]
> [10.0, 0.6470033034_4502123643 <(450)%20212-3643>]
> [11.0, 0.6482444733_4200260078]
> [12.0, 0.6384388807_0692194404]
> [13.0, 0.6023007395 <(602)%20300-7395>_2341824158]
> [14.0, 0.6090225563_9097744361 <(909)%20774-4361>]
> [15.0, 0.5511288180_6108897742 <(610)%20889-7742>]
> [16.0, 0.6092307692 <(609)%20230-7692>_3076923077]
> [17.0, 0.5546719681_9085487078 <(908)%20548-7078>]
> [18.0, 0.5558510638_2978723404]
> [19.0, 0.5223880597 <(522)%20388-0597>_0149253731]
> [20.0, 0.5607843137_2549019608 <(254)%20901-9608>]
>
>
> 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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