The integral does not exist. It is infinity if n is an even positive integer.

Note that if you use r:=5/2 than maple produces a symbolic answer but it will be incorrect because a continuous antiderivative is not found (too many symbolic parameters). For example in this case Q(6) returns 0 (obviously wrong).

A necessary and sufficient condition is given by

istetra:=proc(a,b,c,d,e,f)
is(`and`(       #((
a>0,b>0,c>0,d>0,e>0,f>0,
2*max(a,b,c)<a+b+c,
2*max(a,e,f)<a+e+f,
2*max(d,b,f)<d+b+f,
2*max(d,e,c)<d+e+c,
LinearAlgebra:-Determinant(
<0,a^2,b^2,c^2,1; a^2,0,f^2,e^2,1; b^2,f^2,0,d^2,1; c^2,e^2,d^2,0,1; 1,1,1,1,0>) >0
))
end;

In tomleslie's answer the triangle inequalities for faces are missing.

A more elegant version is:

IsTetra:=proc(d::Matrix(4, shape=symmetric))
Matrix(3, (i,j)-> d[4,i]^2+d[4,j]^2-d[i,j]^2);
LinearAlgebra:-IsDefinite(%, query = 'positive_definite')
end proc:

Example.

T:=Matrix(4, {(1,2)=3, (1,3)=4, (2,3)=5, (1,4)=4, (2,4)=4, (3,4)=4}, shape='symmetric'):
IsTetra(T);
        true

Try:

K:= f -> `if`(type(f,piecewise), select(type, [op(f)], And(algebraic,Not(undefined))), f);

f:=a*x^2 + b*x + c:
max(solve(f = 0, x)) assuming a>0, b^2-4*a*c >= 0;
         

max(solve(f = 0, x)) assuming a<0, b^2-4*a*c >= 0;
     

Note that the largest root is not necessarily positive. It will if a*c<0 or (a*c>0 and a*b<0).

Use assumptions or formal sums

sum(x^n,n=0..infinity) assuming abs(x)<1;
sum(x^n,n=0..infinity,  formal);

Download int.mw

It is not possible to use plots:-animate because HeatMap is implemented to display a background picture.
Explore can be used, but I am not sure whether it works in Maple 2016. In Maple 2017+ it's ok.

restart;
p := k -> Matrix(3, 3, (i,j) -> (i+j+k) mod 3):
f:=proc(k) Threads:-Sleep(0.2): Statistics:-HeatMap(p(k)) end:
Explore(  f(k), k=0..8, animate,loop, autorun);

When a variable is implicitely declared local, a warning message appears.
The rules for this are simple, see ?local

In

foo:= proc() 
  local x;
  plot(sin(x),x=-Pi..Pi); 
end proc:

local is indeed needed, otherwise x:=10  at top level will produce an error.

Alternatively you may use
plot(sin('x'),'x'=-Pi..Pi);

All the floating point computations are done with 1 significant digit. So, sqrt(5) is approximated to 2.  and  (2. -1.)/2.  is 0.5.

See also:

restart;
Digits:=1;
                          Digits := 1
a:=[seq(i/10., i=1..9)];
       a := [0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]
b:=[seq(i/10., i=9..1,-1)];
       b := [0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1]
add(a) = add(b);
                            6. = 4.
evalf[2](add(a)) = evalf[2](add(b));
                           4.5 = 4.5

Maple (at least the recent versions) can compute directly the double integral (without passing to polar coordinates):

int(sqrt(x^2+y^2), x = 0 .. B/2, y = 0 .. b/2) assuming b>0, B>0;

S:=simplify(eval(SA/t, [x=y*t, Log[t]=1/L])):
collect(S,y, expand);

Now you have a polynomial in the variables (L,y) with rational coefficients and you want to guess its coefficients or a generating function. Why do you think that a simple generating function does exist? A more realistic approach would be to come back to the original problem (which produced the expression) and try there.

.

n >= ceil(fsolve(1/(n+1)! = 0.00001));
                             8 <= n
n >= ceil(fsolve(1/(n+1)! * exp(0.1) = 0.00001));
                             8 <= n

Seems to be a bug. Workaround:

ex:=D[2](eta)(t,x)+D[2](phi)(t,x,0)+D[1](phi)(t,x,0);

inds:=[indets(ex)[]]:  cinds:=map(convert, inds, Diff):
subs(inds=~cinds, ex);

Download Interpolate.mw

a:=sqrt(6)/6 + sin(5/6) + exp(6/7) + 567:

f := u ->  `/`(subs(6=b, [op(u)])[]):
subsindets(subs(6=b,a), fraction, f);