Block diagram reduction: G1–G4 with H1, H2, H3

Name the signals: $A$ = output of $S_1$, $B$ = output of $G_3$. From the figure, $C = B + G_4A$.

Step 1: Move the take-off point of $H_1$ from $C$ back to $B$

Since $C = B + G_4A$:

$$H_1C = H_1B + G_4H_1A$$

So $H_1$ is now taken from $B$ (a loop around $G_3$), and an extra branch $G_4H_1$ goes from $A$ to $S_3$ with a minus sign.

Step 2: Move the extra input at $S_3$ back before $G_2$

A signal entering after $G_2$ can be moved before it by dividing by $G_2$: the branch becomes $G_4H_1/G_2$ entering $S_2$ (minus).

Step 3: Reduce the loop $G_3$, $H_1$

$$\frac{G_3}{1 + G_3H_1}$$

Step 4: Reduce the loop with $H_2$

Forward: $G_2\cdot\dfrac{G_3}{1+G_3H_1}$, feedback $H_2$:

$$\frac{B}{E_2} = \frac{G_2G_3}{1 + G_3H_1 + G_2G_3H_2}$$

where $E_2 = G_1A - \dfrac{G_4H_1}{G_2}A$ is the input of $G_2$. Hence

$$\frac{B}{A} = \frac{G_3(G_1G_2 - G_4H_1)}{1 + G_3H_1 + G_2G_3H_2}$$

Step 5: Add the parallel path $G_4$

$$\begin{aligned} \frac{C}{A} &= \frac{G_3(G_1G_2 - G_4H_1) + G_4(1 + G_3H_1 + G_2G_3H_2)}{1 + G_3H_1 + G_2G_3H_2} \\ &= \frac{G_1G_2G_3 + G_4 + G_2G_3G_4H_2}{1 + G_3H_1 + G_2G_3H_2} \end{aligned}$$

R ->(S1)--A--> [ N / D' ] -----+---> C
     ^-                        |
     +--------- H3 ------------+
 N  = G1G2G3 + G4 + G2G3G4H2
 D' = 1 + G3H1 + G2G3H2

Step 6: Close the outer loop with $H_3$

$$\frac{C}{R} = \frac{N/D'}{1 + H_3N/D'} = \frac{N}{D' + H_3N}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4 + G_2G_3G_4H_2}{1 + G_3H_1 + G_2G_3H_2 + G_1G_2G_3H_3 + G_4H_3 + G_2G_3G_4H_2H_3}$$

Check (Mason): paths $P_1 = G_1G_2G_3$ ($\Delta_1 = 1$), $P_2 = G_4$ ($\Delta_2 = 1 + G_2G_3H_2$, since the $H_2$ loop does not touch it). Loops $-G_2G_3H_2$, $-G_3H_1$, $-G_1G_2G_3H_3$, $-G_4H_3$; the pair ($-G_2G_3H_2$, $-G_4H_3$) is non-touching. Same result.