This question builds off of this question.
In the HHL algorithm, how do you efficiently do the $\tilde{\lambda}_k$-controlled rotations on the ancilla qubit? It seems to me that since you don't know the eigenvalues a priori, you would have to control on every single $\lambda$ within your eigenvalue bounds $[\lambda_{\text{min}},\lambda_{\text{max}}]$ (since every $\lambda$ requires a different rotation angle), requiring a potentially exponential number of controlled rotations.
I kind of get how you can avoid an exponential number of controls in Shor's algorithm, because we can split up the modular exponentiation $a^x\pmod N$ so that we can deal with each bit of $x$ separately, $a^{2^{k-1}x_{k-1}}a^{2^{k-2}x_{k-2}}...a^{2^0 x_0} \pmod N$, so you only need as many controls as the number of bits of $x$. But I'm not sure how you can do something similar in the case of HHL, because $\tilde{\lambda}_k$ is not only in the denominator, but nested inside an arcsin, e.g. \begin{align} \mathrm{Controlled\ Rotation}=\sum_{m \in [\lambda_{\text{min}},\lambda_{\text{max}}]}\underbrace{|m\rangle\langle m|}_{\text{control reg.}} \otimes \underbrace{R_y\left(\sin^{-1}(2C/m) \right)}_{\text{anc reg.}} \end{align} where the number of terms in the sum is exponential in the number of bits of precision in $\lambda$. Is there a way to do this more efficiently, and if not, wouldn't this severely eat into the exponential speedup of the algorithm?