@ecterrab Maybe I miss something (I use Physics very seldom) but the Wirtinger derivatives seem to be used only at a formal level:

diff(conjugate(z)+z*conjugate(z)^2, conjugate(z));  # ok

but, diff(Re(z), conjugate(z)) and diff(abs(z), conjugate(z)) are both 0.
(probably a convert(..., conjugate) is mandatory).

Edit. Actually it is not complicated to compute the Wirtinger derivatives; more work would be to integrate them in the system.

diffW := proc(f::algebraic, z::{And(symbol,Not(constant)), specfunc(And(symbol,Not(constant)),conjugate)})
  local g:= convert(f,conjugate),zb;
  g:=subs(conjugate(op(z))=zb,g);
  subs(zb=conjugate(op(z)),diff(g,`if`(type(z,symbol),z,zb)))
end proc:

@Axel Vogt Yes, the derivative is real. As mentioned, the modified `diff/abs` implements d/dx abs(f(x+I*y)). But as well known, if f is analytic, d/dz f(z) = d/dx f(x+I*y), where z = x + I*y. The complex derivative d/dz abs(z) does not exit at any point (not even at real points). The same for abs(GAMMA(2*t + I)): the complex derivative does not exist.

Unfortunately, the Wirtinger derivatives d/dz, d/dzbar  are not implemented in Maple (or other CASes AFAIK).

@Pascal4QM I know that diff(abs(i + sin(z)), z)  works even without Physics, because evalc simplifies the expression.
But for abs(GAMMA(2*t + i))  it fails with Physics too.
The problem is that `diff/abs` is implemented as if abs were analytic, and it is not so. 

Note that if `diff/abs` is defined as in my post (after loading Physics), then the correct result is obtained in Physics too (but without converting to conjugate, because `diff/conjugate`  must be changed too).

@robertocooper 

abs(1,z)  is by definition  diff(abs(z),z). Note abs is not differentiable in C\R.
evalc should not be used here because u(...) is not real.
Anyway, the provided solution is not verified. 

@mmcdara Yes, there is even a "chaotic" billiard.
For large rational m,n, the curve appears to be dense due to the physical thickness of the line.

plot([sin(127/11*t), sin(63/5*t), t=0..200*Pi], view=[(0.50 .. 0.51) $ 2]);

(but the plotting algorithm is not very accurate for such graphs).

Note that 
{ (sin(127/11*t), sin(63/5*t)) | t in R } =
{ (sin(127*5*t), sin(63*11*t)) | t in [0, 2*Pi] } 
so, it is a compact set (hence not dense).

@mmcdara  Actually the curve is dense (for t in R) iff m and n are incommensurable i.e. m/n irrational (e.g. m = sqrt(2), n = 1).

@Zeineb They have the same values, so one of the graphs will cover the other.
To see both, you may use the transparency, e,g.

plot([(x+1)^2, x^2+2*x+1], x=-2..1, 
color=[red,green], transparency=[0,0.5], thickness=[4,15]);

@AHSAN You should be aware that an antiderivative has very seldom a closed form. E.g.
int((4*x + 3*x^2*exp(x^3))/(a*x^2 + exp(x^3))^2, x)

is elementary only for a=2 (probably).

@AHSAN You have changed the expression!

V:=simplify(v); # v is your original expression

@Ugurgozutok If you know that g divides f, simply compute the remainder rem(f, g, x) (or rem(f, g, y)) and equate its coefficients with 0.

They are called Lissajous curves (see ?Lissajous and the wiki article) and are usually plotted via

plot([sin(4*t), sin(5*t), t=0..2*Pi]);

However, when x0 is a pole, the order corresponds to the number of terms:

series(1/x^7*exp(x),x=0,3);

        1/x^7 + 1/x^6 + 1/2/x^5 + O(1/x^4)  

@tomleslie Without this "main" condition the problem is trivial: a and x are arbitrary (x<>a) and
m/n = a^2*x^2/(x-a)^2+x^2 

@tomleslie You forgot about the main condition: four integer roots.

restart;
eval(a^2*x^2/(x-a)^2+x^2 = m/n, {a = 21, m = 15616744, n = 73}):
solve(%);

    22, 462, -221 - sqrt(58123), -221 + sqrt(58123)

(isolve is useless here)

@Maple_lover1 Now, for a=1, the two formulae agree for B = r*Pi/k * 2*n/(n-1) but w2 is set to 0 when the argument of sin is not in the interval (-Pi,Pi). Note however that in Maple w2 will be complex when sin(...) < 0 (for n>3).