Transfer Function of PI Controller
For a start you might as well get rid of R2 and R3 - all they do is reduce the setpoint at which the error is calculated and do not at all affect the AC transfer function of the Op-amp. If you reduce this further to imagining Vref at 0V, the AC transfer function again is unaltered.

This leaves you with a simple inverting op-amp configuration to solve: -

\$\dfrac{V_{ERR}}{V_{(+IN)}} = 1 + \dfrac{X_{feedback}}{R_1}\$

So, calculate what Xf is in terms of jw or s, then do some math and you have your transfer function. If you want to then accomadate R2 and R3 then it's a simple case of multiplying the answer by: -

\$\dfrac{R3}{R2+R3}\$

The actual DC value of Vref is unimportant unless this is also being moved around rapidly.

If you want a gain of 8 concentrate on R4 and R1 as the important components. Clearly at DC the circuit is an integratorbut at some point the AC gain should be largely defined by those two resistors with C1 being much less significant. If C1 isn't insignificant for mid-band AC gain then you have to assume that C1 is much smaller than C2 and therefore can be realized as being in parallel with R4.

From my experience I would usually say that C2 as the main integration capacitor and C1 is a high frequency roll-off component that barely affects mid-band gain.