Output voltage function (circuit-analysis)
Ok, so op amps 3, 4 and 5 form a PID block. Considering their respective output voltages as Vp, Vi and Vd:

$$ V_p = -\frac{R_9}{R_{10}}V_{in} $$

$$ V_i = -\frac{1}{R_{11}C_{1}}\int{V_{in}dt} $$

$$ V_d = -R_{12}C_2\frac{dV_{in}}{dt} $$

Op amp 2 and resistors R3-8 are used to sum these voltages. Using superposition, one can solve this block's output (Vx). Let's say Vi and Vd are 0; R5, R7 and R8 are then in parallel and the voltage on the op amp '+' will be Vp/4.

$$ V_{x(V_p)} = \left(\frac{R_3}{R_4} + 1\right)\frac{V_p}{4} = 0.75V_p$$

Therefore, with superposition, Vx is 0.75(Vp + Vi + Vd). The last op amp is just a regular negative gain circuit, and Vout = -Vx. Final answer:

$$ V_{out} = 0.75\left(\frac{R_9}{R_{10}}V_{in} + \frac{1}{R_{11}C_{1}}\int{V_{in}dt} + R_{12}C_2\frac{dV_{in}}{dt}\right)$$