Well, I'll provide an output expression for your filter. First, let's recall the Fourier series of the absolute sine-wave.
The function \$x\mapsto f(x):=|\sin x|\$ is even and \$\pi\$-periodic; therefore \$f\$ has a Fourier series of the form:
$$f(x)=\frac{\text{a}_0}{2}+\sum_{\text{k}\space\geq\space1}\text{a}_\text{k}\cos\left(2\text{k}x\right)\tag1$$
With:
$$\text{a}_\text{k}={2\over\pi}\int\limits_0^\pi f(x)\cos\left(2\text{k}x\right)\space\text{d}x={2\over\pi}\int\limits_0^\pi \sin\left(x\right)\cos\left(2\text{k}x\right)\space\text{d}x\tag2$$
It follows that:
$$\eqalign{\text{a}_\text{k}&={1\over\pi}\int\limits_0^\pi\left(\sin\bigl((1+2\text{k})x\bigr)+\sin\bigl((1-2\text{k})x\bigr)\right)\space\text{d}x\cr\\&={1\over\pi}\left({\cos\bigl((2\text{k}-1)x\bigr)\over 2\text{k}-1}-{\cos\bigl((2\text{k}+1)x\bigr)\over 2\text{k}+1}\right)\Biggr|_0^\pi\cr\\&={2\over\pi}\left({1\over 2\text{k}+1}-{1\over 2\text{k}-1}\right)\cr\\&=-{4\over\pi(4\text{k}^2-1)}\cr}\tag3$$
Therefore we have:
$$\left|\sin\left(x\right)\right|={2\over\pi}-{4\over\pi}\sum_{\text{k}\space\geq\space1}{\cos(2\text{k}x)\over 4\text{k}^2-1}\tag4$$
The transfer function of your circuit is given by:
\begin{equation}
\begin{split}
\text{Y}\left(\text{s}\right)&=\frac{\displaystyle\text{R}\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\text{sL}+\left(\text{R}\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}}\right)}\\
\\
&=\frac{\displaystyle\frac{\displaystyle\text{R}\cdot\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\text{R}+\frac{\displaystyle1}{\displaystyle\text{sC}}}}{\displaystyle\text{sL}+\frac{\displaystyle\text{R}\cdot\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\text{R}+\frac{\displaystyle1}{\displaystyle\text{sC}}}}\\
\\
&=\frac{\displaystyle\frac{\displaystyle\text{R}}{\displaystyle1+\text{sCR}}}{\displaystyle\text{sL}+\frac{\displaystyle\text{R}}{\displaystyle1+\text{sCR}}}\\
\\
&=\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{sL}\left(1+\text{sCR}\right)}\\
\\
&=\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{Ls}+\text{CLRs}^2}
\end{split}\tag5
\end{equation}
Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.
Taking the inverse Laplace transform, we can see:
\begin{equation}
\begin{split}
y\left(x\right)&=\mathscr{L}_\text{s}^{-1}\left[\text{Y}\left(\text{s}\right)\right]_{\left(x\right)}\\
\\
&=\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{Ls}+\text{CLRs}^2}\right]_{\left(x\right)}\\
\\
&=\frac{2}{\sqrt{\text{L}\left(\text{L}-4\text{CR}^2\right)}}\cdot\exp\left(-\frac{x}{2\text{CR}}\right)\cdot\sinh\left(\frac{\sqrt{\text{L}\left(\text{L}-4\text{CR}^2\right)}}{2\text{CLR}}\cdot x\right)
\end{split}\tag6
\end{equation}
We can write the voltage after the full bridge rectifier as follows:
\begin{equation}
\begin{split}
\text{v}_\text{i}\left(t\right)&=\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left|\sin\left(2\pi\text{f}t\right)\right|\\
\\
&=\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left({2\over\pi}-{4\over\pi}\sum_{\text{k}\space\geq\space1}{\cos\left(4\pi\text{k}\text{f}t\right)\over 4\text{k}^2-1}\right)
\end{split}\tag7
\end{equation}
Where \$\hat{\text{u}}_\text{i}\space\left[\text{V}\right]\$ is the amplitude of the input voltage, \$\text{V}_\text{d}\space\left[\text{V}\right]\$ is the voltage drop across one diode.
Now, using the convolution property of the Laplace transform, we can write the output voltage:
\begin{equation}
\begin{split}
\text{V}_\text{o}\left(t\right)&=\int\limits_0^t\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\text{Y}\left(\text{s}\right)\right]_{\left(t-\tau\right)}\space\text{d}\tau\\
\\
&=\int\limits_0^t\text{v}_\text{i}\left(\tau\right)\cdot y\left(t-\tau\right)\space\text{d}\tau\\
\\
&=\int\limits_0^t\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left({2\over\pi}-{4\over\pi}\sum_{\text{k}\space\geq\space1}{\cos\left(4\pi\text{k}\text{f}\tau\right)\over 4\text{k}^2-1}\right)\cdot y\left(t-\tau\right)\space\text{d}\tau\\
\\
&=\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left({2\over\pi}\int\limits_0^t y\left(t-\tau\right)\space\text{d}\tau-\frac{4}{\pi}\sum_{\text{k}\space\geq\space1}\frac{1}{4\text{k}^2-1}\int\limits_0^t\cos\left(4\pi\text{k}\text{f}\tau\right)y\left(t-\tau\right)\space\text{d}\tau\right)
\end{split}\tag8
\end{equation}
