Series $\sum(n^2+dm^2)^{-s}$ and its relation to unique decomposability of $\mathbb Z[\sqrt{-d}]$ Denote $\sum_{n,m}'=\sum_{(n,m)\in\mathbb Z^2\setminus\{(0,0)\}}$. Let $d\in\mathbb Z^+$. Suppose $$S_d(s)=\sum_{n,m}'\frac1{(n^2+dm^2)^s},$$
$$S_{4d-1}^\ast(s)=\sum_{n,m}'\frac1{(n^2+nm+dm^2)^s}.$$
By integral convergence test, it converges iff $\Re s>1$.
The result when $d=1$ is well known and can be found on this site.
I succeeded to evaluate $S_2(s)$ and $S_3^\ast(s)$.
$$\begin{aligned}S_2(s)&=\sum_{n,m}'\frac{1}{(n^2+2m^2)^s}\\
&=\sum_{n,m}'\left|n+\sqrt{-2}m\right|^{-2s}\\
&=\#U(\mathbb Z[\sqrt{-2}])\prod_{\mathfrak p}\frac1{1-|\mathfrak p|^{-2s}}\\
&=2\cdot\frac1{1-2^{-s}}\cdot\prod_{p\equiv 1,3(\operatorname{mod} 8)}\frac1{(1-p^{-s})^2}\\
&=\frac2{1-2^{-s}}\zeta(s)L(s,\chi_8)\\&\text{ where }\chi_8(3)=1,\chi_8(5)=\chi_8(7)=-1
\end{aligned}$$
In this evaluation, it seems that the unique decomposibility of the ring $\mathbb Z[\sqrt{-d}]$ (resp.  $\mathbb Z\left[\frac{1+\sqrt{-d}}2\right]$)is quite important in reducing computation. Also, I noticed, by numerical calculation, that the closed form does not imply that $\mathbb Z[\sqrt{-d}]$ is a UFD. To show this, I made a table.
$$\begin{array}{|c|c|c|c|c|c|}\hline
d&1&2&\color{red}3&4&7\\\hline
\frac{S_d(3)\sqrt d}{\pi^3\zeta(3)}&\frac18&\frac3{32}&\color{red}{\frac{11}{108}}&\frac{29}{256}&\frac{50}{343}\\\hline
\frac{S_d^\ast(3)\sqrt d}{\pi^3\zeta(3)}&-&-&\frac{8}{27}&-&\frac{64}{343}\\\hline
\end{array}$$
The missing columns are the $d$'s that I failed to find a rational number to approximate the result.  

These two series are heavily related to the prime factorization of a number ring. As $d$ grows bigger, it is increasingly harder to get a closed form.
  (Main question) Is there a general method to obtain $S_d(s)$ and $S_d^\ast(s)$?
  (Broad question) Is there any evidence to suggest any family of $d$ that $S$ and $S^\ast$ does not have an obvious closed form containing Dirichlet L functions? 

Note that $$\begin{cases}S&d\ne4n-1\\S^\ast&d=4n-1\end{cases}$$
coincides with Dedekind zeta function $\zeta_{\mathbb Q[\sqrt{-d}]}(s)$ when the field has class number one. So for a finite family of $d$ this question is investigated.
 A: Suppose $d$ is squarefree and let 
$$
T_d(s)=\left\{
  \begin{array}{@{}ll@{}}
    S_d^*(s), & \text{if}\ -d\equiv 1 \mod 4 \\
    S_d(s), & \text{otherwise}
  \end{array}\right.
$$
When $-d\equiv 1\mod 4$, $S_d(s)$ is harder to understand, but $T_d(s)$ can always be written as a nice sum of L-functions using the independence of characters.
For simplicity, I'll assume that $d \neq 1,3$ so that the only units in $\mathbb{Q}[\sqrt{-d}]$ are $\pm 1$, but similar arguments could be made for $d=1,3$. Under this assumption, we evidently have
$$
\frac{1}{2}T_d(s) = \sum_{\mathfrak{a}\neq 0,\mathfrak{a}\textrm{ principal}} N(\mathfrak{a})^{-s}.
$$
Now let $C$ be the class group of the ring of integers in $\mathbb{Q}[\sqrt{-d}]$, and let $X$ be the set of characters of $C$, i.e. each $\chi\in X$ is a homomorphism $\chi:C \to \mathbb{C}$. By independence of characters, for any ideal $\mathfrak{a}$ we have
$$
\sum_{\chi\in X}\chi(\mathfrak{a})=\left\{
  \begin{array}{@{}ll@{}}
    |C|, & \text{if}\ \mathfrak{a}\textrm{ principal} \\
    0, & \textrm{otherwise}
  \end{array}\right.
$$
Now for each character $\chi\in X$, let
$$
L(s,\chi)=\sum_{\mathfrak{a}\neq 0} \chi(\mathfrak{a})N(\mathfrak{a})^{-s}
$$
be the Hecke L-function associated to $\chi$. Then we easily derive
\begin{align}
\frac{1}{2}T_d(s) &= \sum_{\mathfrak{a}\neq 0,\mathfrak{a}\textrm{ principal}} N(\mathfrak{a})^{-s}\\
&= \frac{1}{|C|}\sum_{\mathfrak{a}\neq 0} \sum_{\chi\in X}\chi(\mathfrak{a})N(\mathfrak{a})^{-s}\\
&= \frac{1}{|C|} \sum_{\chi\in X} L(s,\chi).
\end{align}
