Near the curve the sqrt's are almost 0 and may take imaginary values. So, the sign changes are absent and the implicitplot algorithm does not work well in this case. That's why Acer's direct solution is probably the best!

(Of course, the algorithm could implement a sign check for Re and Im).

@acer 

Strangely,

plots:-implicitplot(Re(ee)=0, b=0..5, p=0..5,gridrefine=1, crossingrefine=7);
also works, but
plots:-implicitplot(abs(ee)=0, b=0..5, p=0..5,gridrefine=1, crossingrefine=7);
fails.

(Actually for abs it's not that strange because the needed sign changes are absent).

@Markiyan Hirnyk 

The parametrization covers the whole curve. This follows from the solve result wrt {b,p} .

@Markiyan Hirnyk 

It's simply a substitution; other substitutions are also possible.
What for? We want a parametrization, isn't it?

@mlog 

The procedures are not really necessary, but make the manipulations easier. On the other hand evalf(Int(...)) tends to work better for procedures. The symbolics are difficult for floats, so convert(...,rational)  is usually a good idea; here also because the parameters have large magnitudes.

@Markiyan Hirnyk 

The curve was parametrized. E.g. using:

f:=simplify(sqrt(b)*sqrt(1-4*p/b)-2*arctan(sqrt((9*p/b-22201/10000)/(9/4-9*p/b)))) assuming b>0:
par:=solve([2*indets(f, arctrig)[]=t,f=0],{b,p});

@Carl Love 

Yes, you are right of course, thank you; sum should be add in my formulation.

@Adam Ledger

It is not clear what you try to obtain. Why do you think that f(n) deserves to be considered?
If you want better asymptotic approximations for ithprime(n)  and Pi(x), there are many results, e.g.

nthprime:=n->
n*(ln(n) + ln(ln(n)) -1 + (ln(ln(n))-2)/ln(n) - ( ln(ln(n))^2 - 6*ln(ln(n)) + 11 )/(2*ln(n)^2)); #  + 0(n/ln(n)^2) 

#  Remark
abs(Pi(x)-Li(x)) <= sqrt(x)*ln(x)/(8*Pi);   # assuming RH, x>=2657

But (1) is trivial and has not much to do with Prime Number Theorem; it is true for any sequence p_n > n.

sqrt(n^2 - n) <= f(n) <= n,  so f(n)/n --> 1.

@rahinui 

For some reason (probably a bug) assuming in this context produces the error.
Of course here the assuming clause is useless, and removing it ==> the expected result.

BTW, restart(); does nothing. It should be restart;

@Mariusz Iwaniuk 

Actually a workaround is not needed because F works. The problem is about interface & typesetting.
(F is the first user procedure I met which refuses to be printed).

@samira moradi 

You cannot, there are infinitely many.

Why don't you simply keep the original expression?
It is obvious that it's impossible to retrive z from |z|. Even in Mathematica :-) for a more complicated z.

@Carl Love 

I still don't understand why should we rely on int to find f(x) when f(x) = F(x,y)/F(a,y).
It is very easy to produce F for which int fails.

Edit. BTW, the user _must_ know (and give) such an `a` in the domain.
For example F(x,y) = x * ( y + x*sqrt(x*y) + sqrt(x^3*y) )
can be split only for x<0, so e.g. a = -1 is needed.

@Carl Love 

int will not be able to compute the integral if F is not simple enough. E.g.

'expand( (randpoly(exp(x))+cos(x)) * (randpoly(exp(y))+sin(y)) )':
F:=%/%;