Sum of squares of Weber Modular Functions 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.
 A: I am not sure this answers your question:

Can
$$ S(\tau) = \mathfrak f(\tau)^2 + 
 \mathfrak f_1(\tau)^2 + \mathfrak f_2(\tau)^2 $$
be written more compactly?

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
$$ S(\tau) = 2S_1(\tau)+\mathfrak f_2(\tau)^2= 
S_2(\tau/8) + \mathfrak f_1(\tau)^2. \tag{3} $$
