# Express result of calculation in terms of Gamma functions only

I would like to to express the result of my integration just in terms of Gamma functions. The following integral is at hand:

$$\int_0^1dz\int_0^1dy(z(1-z))^{-\epsilon}(1-y)^{1-2\epsilon}y^{-1-\epsilon}(z(1-z)-2)$$

yielding the result with mathematica

$$\frac{\sqrt{\pi}2^{2\epsilon-3}(7\epsilon-11)\Gamma(2-2\epsilon)\Gamma(1-\epsilon)\Gamma(-\epsilon)}{\Gamma(2-3\epsilon)\Gamma(\frac{5}{2}-\epsilon)},$$

where one can rewrite the $$\sqrt{\pi}2^{2\epsilon-3}$$ in terms of Gamma functions. By employing

Series[%, {\[Epsilon], 0, 0}]


we obtain

$$\frac{11}{6\epsilon}+\frac{1}{6}\left(4+11\gamma+22\log(2)+11\psi^{(0)}\left(\frac{5}{2}\right)\right)+\mathcal{O}(\epsilon^1).$$

Is it possible to express the result of the integration in terms of Gamma functions or to express the last line as $$\frac{11}{6\epsilon}+\frac{50}{9}+\mathcal{O}(\epsilon^1)$$?

(int = Assuming[Re[ϵ] < 1, Integrate[
(z (1 - z))^-ϵ (1 - y)^(1 - 2 ϵ) y^(-1 - ϵ) (z (1 - z) - 2),
{z, 0, 1}, {y, 0, 1}] // Simplify]) // TraditionalForm


Series[int, {ϵ, 0, 0}] // FullSimplify


Series[int, {ϵ, 0, 1}] // FullSimplify


EDIT: Note that since

Gamma[1/2]


then

rule = Gamma[1/2] -> 8/15*Inactive[Gamma][7/2]


(int2 = int /. rule) // TraditionalForm


int == int2 // Activate

(* True *)