A gamma-ratio continued fraction formula with four gamma-function pairs

Let $x,l,n,\eta\in\mathbb{C}$, and define $P$ by

$$P=\frac{\Gamma\left(\frac{x+l+n+\eta+1}{4}\right)\Gamma\left(\frac{x+l-n-\eta+1}{4}\right)}{\Gamma\left(\frac{x+l+n-\eta+3}{4}\right)\Gamma\left(\frac{x+l-n+\eta+3}{4}\right)}\frac{\Gamma\left(\frac{x-l+n-\eta+3}{4}\right)\Gamma\left(\frac{x-l-n+\eta+3}{4}\right)}{\Gamma\left(\frac{x-l+n+\eta+1}{4}\right)\Gamma\left(\frac{x-l-n-\eta+1}{4}\right)}.$$

Suppose that $x$ is complex with $\operatorname{Re}(x)>0$, or that either $n$ or $\eta$ is an odd integer, or that $l$ is an even integer. The four-pair gamma-ratio continued-fraction conjecture. Then

$$\frac{1-P}{1+P}=\frac{l}{x-n\eta+\mathop{\mathrm{K}}_{m=1}^{\infty}\left(\frac{a_m}{b_m(x)}\right)},$$

where

$$a_{2m}=(2m)^2-l^2,\qquad a_{2m-1}=\frac{\left((2m-1)^2-n^2\right)\left((2m-1)^2-\eta^2\right)}{(2m-1)^2},$$ $$b_{2m}(x)=x-\frac{n\eta}{2m+1},\qquad b_{2m-1}(x)=x+\frac{n\eta}{2m-1}.$$

This is an open problem in the paper's study of continued-fraction expansions for ratios of gamma functions; no proof or disproof is supplied in the given material.