@MDD 

So, you want "disjoint" monomials.

restart;
vars:=[x,y,z]:
m1,m2,m3:= (a-b)*x*y*z^2, c*x^3, 3*b*y*z^2:
DisjMon:= (m1,m2,vars::list(name)) -> not depends(gcd(m1,m2), vars):
DisjMon(m1,m2,vars), DisjMon(m2,m3,vars), DisjMon(m1,m3,vars);

                                  false, true, false

@MDD 

EqMon tests if two terms have the same monomials, with possible distinct coefficients.
Define exactly what you want to test. Your example is not relevant and you have a typo.
 

@9009134 

Here is the the [2,3] integral.

Digits:=15:
ti:=time[real]():
F23:=(3/4*(501230769273.847*cos(theta)^2*r^2+5012307692738.48*r*cos(theta)+12530769231846.2)*sin(theta)*r^4*cos(theta)*cos(3*phi)/(5+r*cos(theta))^4-3/4*(.835384615384617e-1*cos(theta)^2*r+278461538504.616*cos(theta)^2*r^3+.139230769230770*cos(theta)+2784615385046.16*cos(theta)*r^2+6961538462615.36*r)*r^2*sin(theta)*cos(3*phi)/(5+r*cos(theta))^3-.710076923076924*sin(theta)*r^2*cos(theta)*cos(3*phi)/(5+r*cos(theta))^2-.229730769230770*r*sin(theta)*cos(3*phi)/(5+r*cos(theta))+.187961538461539*r^3*sin(theta)*cos(3*phi)/(5+r*cos(theta))^3+.187961538461539*sin(theta)*r^4*cos(theta)*cos(3*phi)/(5+r*cos(theta))^4-.417692307692310*r*sin(theta)*cos(3*phi)/(5+r*cos(theta))^2-.835384615384620*sin(theta)*r^2*cos(theta)*cos(3*phi)/(5+r*cos(theta))^3)*r^4*cos(theta)*cos(3*phi):
temp:=int(F23, phi=0..2*Pi):
L:=[op(expand(temp))]:
add(op(1,J)*int( J/op(1,J), [theta=0..2*Pi,r=2/10..1], numeric), J=L);
time=time[real]()-ti;
#                        1.09206001690410*10^(-6)
#                          time = 0.427

N.B. It would be nice to isolate yourself the problem before posting.

@acer 

I thought that these local variables have another status.
So, they differ from the regular ones, only by being "quasi" inaccessible.

@Carl Love  My question was not related to your original answer.

Thank you both for the clarification.

@Carl Love 

lambda is an escaped local variable, as in:

restart;
f:=proc()
local LOC:=13;
'LOC'
end proc:
f();
#                              LOC
%;
#                               13

My question: isn't LOC removed from the system after the first gc()?
How long is LOC going to survive?

@MDD 

You must provide a formal definition of "appear" for a question like

Does b-1 "appear" in G = [a*n*z^3-1, -b*x+3]   ?

Changing b-1 to c ==>

Does c "appear" in H = [a*n*z^3-1, -(1+c)*x+3]   ?

Does it?

@Carl Love 

A new kind of maths indeed :-)
But vote up for `diff/Zeta`.

@Mariusz Iwaniuk 

If f(n) = the number of positive divisors of n, I don't think you can define f'(3). Nobody can.

Do you really want to differentiate with respect to a natural number? Maybe you want fracdiff.

@Rouben Rostamian  

Yes, the existing numerical method is not stable, but a solution exists (even C^oo).
Must we change a sign and solve another problem, just because it has a stable numerical method implemented?
Maybe the method will work with some adjustments: in our example, adding 'abstol'=0.1  ==> OK for t=0..3.
Or, maybe a change of variables will help.

@Rouben Rostamian  

I don't agree entirely with your argument. The backward heat equation could work quite nicely sometimes.

For example,

sol:=pdsolve(diff(u(x,t),t) = -diff(u(x,t),x,x), {u(x,0)=sin(x), u(0,t)=0, u(Pi,t)=0}, numeric):
sol:-plot3d(u(x,t), x=0..Pi, t=0..1);

It fully agrees with the exact solution u(x,t)=sin(x)*exp(t).

@sgils1 

Your problem is actually an integral (so, instead of dsolve, it's possible to simply use int).

Unfortunately, Maple does not find a continuous antiderivative in the symbolic case i.e. it is valid only locally.
(This is not a bug, it is documented; for example int(1/(2+cos(x)), x) is discontinuous and it is exactly the integral obtained by hand using the substitution tan(x/2)=t; note that for this example there is a simple continuous antiderivative, but in general a CAS does not try to find it).

For example:

Here, (3) and (4) are continuous, but (2) is discontinuous.
The interesting fact is that (3) is also valid for a<0, but Maple does not care.

For your simple example, there are workarounds (looking for discontinuities and adding constants if needed; or use assuming).

@torabi 

Not quite useful. I was hoping to see an example where both input and output are Maple expressions.
I see instead a problem which must be formalized and has lots of unessential parameters such as lambda, mu, J2 etc which seem to be constants.
I don't know what the circled triple integral means. What is V? Why that change of variables? It is not clear what quantities are constans. I don't understand the assumption (4) ; Aijk are unique? Are Jijk etc given? etc, etc.
In conclusion, I cannot help here.

@torabi 

It would be useful to see:
1. A short example (just the example, no subs)
2. A description of what you want to obtain
3. The expected result.

If 1-2-3 are clear enough, then it will be probably possible to find a general method for doing this.

@tomleslie 

Actually the integral is oo (diverges).