Let $\tau\in\mathbb C$ such that $\mathrm{Im}\left(\tau\right)>0$. Define $q=e^{2\pi i\tau}$. Then define the Weber modular functions as
$$ \mathfrak f\left(\tau\right)=q^{-\frac1{48}}\prod_{n=1}^\infty\left(1+q^{n-\frac12}\right)=e^{-\frac{i\pi}{24}}\frac{\eta\left(\frac{\tau+1}2\right)}{\eta\left(\tau\right)}=\frac{\eta\left(\tau\right)^2}{\eta\left(\frac\tau2\right)\eta\left(2\tau\right)}\\ \mathfrak f_1\left(\tau\right)=q^{-\frac1{48}}\prod_{n=1}^\infty\left(1-q^{n-\frac12}\right)=\frac{\eta\left(\frac\tau2\right)}{\eta\left(\tau\right)}\\ \mathfrak f_2\left(\tau\right)=\sqrt2q^{\frac1{24}}\prod_{n=1}^\infty\left(1+q^n\right)=\frac{\sqrt2\eta\left(2\tau\right)}{\eta\left(\tau\right)} $$
Can $$ S\left(\tau\right)=\mathfrak f\left(\tau\right)^2+\mathfrak f_1\left(\tau\right)^2+\mathfrak f_2\left(\tau\right)^2 $$ be written more compactly? Am I missing any helpful identites? This came up in a physics problem I was solving.
Extra Question
I managed to show that $S\left(\tau+8\right)=e^{-\frac{2\pi i}3}S\left(\tau\right)$, so I realised that this integral is interesting too: $$ g\left(a\right)=\frac18\int_0^8\left\lvert S\left(b+ia\right)\right\rvert^2\mathrm db $$ But I have no idea how to solve it. Any help would be appreciated.
I am not sure this answers your question:
As it is easily checked the sum $\,S\,$ is not an eta quotient, as user 'ccorn' alluded to briefly in a comment. Let us define $\,q:=e^{\pi i \tau},\,$ $$ S_1(\tau) :=\! \frac{\theta_3(0,q^2)}{\eta(\tau)} \!=\! \frac{\eta(4\tau)^5}{\eta(\tau)\, \eta(2\tau)^2\, \eta(8\tau)^2}. \tag{1} $$ It is the G.F. of OEIS sequence A226622. Now we get $$ 2 S_1(\tau) \!=\! \mathfrak f(\tau)^2 \!+\! \mathfrak f_1(\tau)^2 \!=\! q^{-1/24}(2\!+\!2q\!+\!8q^2\!+ ...). $$
Also let us define $$ S_2(\tau) :=\! \frac{\theta_3(0,q)}{\eta(8\tau)} \!=\! \frac{\eta(2\tau)^5}{\eta(\tau)^2 \eta(4\tau)^2\eta(8\tau)}. \tag{2} $$ This is an eta-quotient by definition and $$ S_2(\tau/8) \!=\! \mathfrak f(\tau)^2 \!+\! \mathfrak f_2(\tau)^2 \!=\! q^{-1/24}( 1 \!+\! 2q \!+\! 2q^4 \!+ ...).$$
Thus we get that $$ S(\tau) = 2S_1(\tau)+\mathfrak f_2(\tau)^2= S_2(\tau/8) + \mathfrak f_1(\tau)^2. \tag{3} $$