Transfer function with G3 feedforward, H1 and unity loop

Name the signals: $E$ = output of $S_1$, $A$ = output of $S_2$, $B$ = output of $G_2$.

Step 1: Combine the parallel paths from $A$ to $C$

$G_2$ (through $B$) and $G_3$ both go from $A$ to $S_3$:

$$\frac{C}{A} = G_2 + G_3$$

Step 2: Reduce the inner unity-feedback loop

$C$ is fed back to $S_2$ with unity gain:

$$\frac{C}{G_1E} = \frac{G_2 + G_3}{1 + G_2 + G_3}$$

Step 3: Move the take-off point of $H_1$ from $B$ to $C$

$B = G_2A$ and $A = \dfrac{C}{G_2 + G_3}$, so

$$H_1B = \frac{G_2H_1}{G_2 + G_3}\,C$$

R ->(S1)-> G1 -> [(G2+G3)/(1+G2+G3)] --+--> C
     ^-                                |
     +----- G2H1/(G2+G3) <-------------+

Step 4: Close the outer loop

$$\begin{aligned} G_f &= \frac{G_1(G_2 + G_3)}{1 + G_2 + G_3}, \qquad H = \frac{G_2H_1}{G_2 + G_3} \\ \frac{C}{R} &= \frac{G_f}{1 + G_fH} = \frac{G_1(G_2 + G_3)}{1 + G_2 + G_3 + G_1G_2H_1} \end{aligned}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_2 + G_3 + G_1G_2H_1}$$

Check (Mason): paths $G_1G_2$, $G_1G_3$; loops $-G_2$, $-G_3$, $-G_1G_2H_1$, all touching each other and both paths. Same result.