Probably not a very elegant method, as it makes a great use of a CAS. However it seems quite general for this kind of series. Inversely, it can be used to create similar results.
The terms of the series, which are a rational function of the index, can be decomposed into a sum of rational terms
\begin{equation}
u_k=\sum_{j=1}^n\frac{\lambda_j}{k+a_j}
\end{equation}
(we suppose the order of the poles is 1).
When $\left|x\right|<1$, the series
\begin{equation}
f_j(x)=\sum_{k=0}^\infty \frac{x^k}{k+a_j}=x^{-a_j}\int_0^x \frac{t^{a_j}}{1-t}\,dt+\frac{1}{a_j}
\end{equation}
This can be verified by developping the $(1-t)^{-1}$ term in the integral.
Then, the series
\begin{align}
S(x)&=\sum_{k=0}^\infty u_kx^k\\
&=\sum_{k=0}^\infty\sum_{j=1}^n\frac{\lambda_jx^k}{k+a_j}\\
&=\sum_{j=1}^n\lambda_j\left[x^{-a_j}\int_0^x \frac{t^{a_j}}{1-t}\,dt+\frac{1}{a_j}\right]\\
&=\sum_{j=1}^n\frac{\lambda_j}{a_j}+\int_0^x \frac{\sum_{j=1}^n\lambda_jx^{-a_j}t^{a_j}}{1-t}\,dt
\end{align}
The proposed series corresponds to $\lim_{x\to1^{-}}S(x)$. Due to the denominator in the integral, in order that this limit exists, the condition
\begin{equation}
\sum_{j=1}^n\lambda_j=0
\end{equation}
must hold. Then
\begin{equation}
S(1)=\sum_{j=1}^n\frac{\lambda_j}{a_j}+\int_0^1 \frac{\sum_{j=1}^n\lambda_jt^{a_j}}{1-t}\,dt
\end{equation}
The remaining integral can be directly calculated.
In the proposed case, using a CAS,
\begin{align}
u_k&=\frac{48}{371}\frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\
&=-{\frac {181203}{3799040\,k+21844480}}+{\frac {418643}{759808\,k+
3229184}}-{\frac {293677}{759808\,k+2849280}}\\
&\,\quad+{\frac {743573}{3799040
\,k+12346880}}+{\frac {181203}{759808\,k+3988992}}-{\frac {1868267}{
3799040\,k+18045440}}\\
&\,\quad-{\frac {56237}{759808\,k+2089472}}+{\frac {56237
}{3799040\,k+8547840}}
\end{align}
after some calculations, one obtains
\begin{align}
\sum_{j=1}^n\frac{\lambda_j}{a_j}&=\frac{7516928}{124151182155}\\
\sum_{j=1}^n\lambda_jt^{a_j}&=-{\frac {181203}{3799040}{t}^{{\frac{23}{4}}}}+{\frac {418643}{759808}
{t}^{{\frac{17}{4}}}}-{\frac {293677}{759808}{t}^{{\frac{15}{4}}}}+{
\frac {743573}{3799040}{t}^{{\frac{13}{4}}}}\\
&\,\quad+{\frac {181203}{759808}{t
}^{{\frac{21}{4}}}}-{\frac {1868267}{3799040}{t}^{{\frac{19}{4}}}}-{
\frac {56237\,{t}^{11/4}}{759808}}+{\frac {56237\,{t}^{9/4}}{3799040}}\\
&=\frac{1}{3799040} \left( 181203t+56237 \right)t^{9/4}\left( 1-\sqrt{t} \right)^5
\end{align}
The above function vanishes at $t=1$, as expected. We have to evaluate
\begin{align}
S(1)&=\frac{7516928}{124151182155}+\frac{1}{3799040} \int_0^1 \frac{\left( 181203t+56237 \right)t^{9/4}\left( 1-\sqrt{t} \right)^5}{1-t}\,dt \\
&=\frac{7516928}{124151182155} +\frac{1}{949760}\int_0^1\frac{\left( 181203v^4+56237 \right)v^{12}\left( 1-v^2 \right)^5}{1+v^2}\,dv
\end{align}
To evaluate the integral, by devloping the numerator, we have to calculate terms as
\begin{equation}
I_n=\int_0^1\frac{v^{2n}}{1+v^2}\,dv
\end{equation}
A recurrence relation can be found easily:
\begin{equation}
I_n=\frac{1}{2n-1}-I_{n-1}
\end{equation}
from which we have (with $I_0=\pi/4$)
\begin{equation}
I_n=(-1)^{n-1}\sum_{p=0}^{n-1}\frac{(-1)^{p}}{2p+1}+(-1)^n\frac{\pi}{4}
\end{equation}
After (rather uninteresting) calculations, we get
\begin{equation}
\frac{1}{949760}\int_0^1\frac{\left( 181203v^4+56237 \right)v^{12}\left( 1-v^2 \right)^5}{1+v^2}\,dv=\pi-{\frac{780059253811}{248302364310}}
\end{equation}
Finally
\begin{equation}
S(1)=\pi-\frac{333}{106}
\end{equation}
as expected.
