My impression is that the only problem is the limited capability of assume, as the next very simple examples show.

assume(p::integer, p>1, p<4);
assume(x<3,x>2);
about(p);
   Originally p, renamed p~:
     is assumed to be: AndProp(integer,RealRange(2,3))

[floor(x)-2, floor(p/2)-1, sin(floor(p/2)*Pi), cos(floor(p/2)*Pi)+1, sin(p-2)*(p-3)];



(should be all 0).

@Preben Alsholm 

I don't think that equality is the real issue here because <= is for sure a valid condition and p<=0,p>=0 gives p=0 (and about(p) confirms this).

@Preben Alsholm 

It seems that the assume database is not always inspected (correctly?):

[p, sin(p), sin(Pi*p)] assuming p=0;
  
    [p, sin(p),0]

@Markiyan Hirnyk 

int(x^(-ln(x)), x = 0 .. b);

also works (without assumptions)!

@JacquesC 

But most of these examples can be obtained in Maple 2015 as "one-liners", e.g.

Change(int(x^(-ln(x)),x=0..b), x=ln(t)) assuming b > 0;

Edit: I wanted x=exp(t), but even this strange change of variable works!

@Kitonum 

They are so easy to compose ... E.g.

J := Int((2+ln(x))/(sqrt((ln(x)+1)*x)*sqrt(1-x-x*ln(x))), x);

ListTools:-Search is not faster; it actually calls member. Just type:

showstat(ListTools:-Search);

The error can be easily solved for Maple 2015 (probably in 2016 it's ok):

with(IntegrationTools):
f:=t^2*exp(-t)/ln(t):
J:=Int(f,t=0..infinity):
Change(J,t=exp(s)):
Split(%,[-eps,eps]):
J1,J2:=op(1,%),op(3,%):
J1a:=subs(t=s, Change(J1,s=-t,t)) assuming eps>0:
Jeps:=normal(combine(J1a+J2)):
J:=eval(Jeps,eps=0);

evalf(J);
    2.153547692

But the following theoretical problem appears.

If f is a function defined in an interval (a,b) having a singularity  s in (a,b),
h is a C^2 diffeomorphism from (a,b) to (c,d) and h is used as a change of variables
int(f, a..b) ---->  int(F, c..d)

then the following situations are possible:

1. f is integrable in the sense of Cauchy principal value but F is not.
2. F is integrable in the sense of Cauchy principal value but f is not.

@Axel Vogt 

Just consider

Int(a/x + 1, x=0..1)

for a-->0.

Edit. The changes of variables are in the intervals [0,1-a] (where the integral is not improper)
and [1+a, oo) where the integrability is obvious. (0<a<1).

I also vote up yours because it's general (it appears in Fihtengolts, you probably know this excellent textbook).

@Markiyan Hirnyk

@Markiyan Hirnyk 

1. I know those theorems (I teach measure theory BTW).

2. I asked Maple to compute the limit of the integral (not the integral of the limit). So, in principle, you should ask Maple to base its computations.
[I admit that Maple probably did not check the conditions].

3. It's not necessary to have uniform convergence. It's not difficult to use the Lebesgue dominated convergence for the integrand of J1+J2 (which, BTW, is not your expression).

@Declan 

f := (i,j) -> `if`(j=1, (i+4)^2, (i+4)^3):

@Declan 

The function (Maple procedure) f defines the (i,j) element of the matrix (i = row, j = column).

If j=1 then f(i,j) = i^2, otherwise (j=2 in our case), f(i,j) = i^3.

@Declan 

Why don't you use the clear and short preudocode algorithm at

https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test

Note also that you should not use FactorInteger (see the pseudocode above).
(BTW, FactorInteger is in the MmaTranslator; otherwise ifactors does the job).

Also note that your fea(x, e, n)  exists in Maple as  x &^ e mod n.

@JessyOw 

Have you tried the provided hint? Then the answer should be clear: you can't.