Product of time ordered exponential

The following expectation of a time ordered exponential is easy to work with:
$$ leftlangle n right| T leftlbrace e^{-i int_{t_1}^{t_2} V_A(t’)dt’ }rightrbrace T leftlbrace e^{-i int_{t_0}^{t_1} V_B(t’)dt’ }rightrbrace left| n rightrangle $$
with $t_2>t_1>t_0$ and where $T$ is the time-ordering operator and $V_A, V_B$ two operators acting on the space spanned by $left| n rightrangle$. In this case it can easily be transformed into one time ordered exponential $Tleftlbrace e^{-i int_{t_0}^{t_2} V(t’)dt’ }rightrbrace $ with $V(t) = V_A(t)$ for $t_1<t<t_2$ and $V(t) = V_B(t)$ elsewhere. Then I can expand it and deal with the operators changing in function of the time variable. For example, the second order term is given by
$$ frac{(-i)^2}{2!}int_{t_0}^{t_2}int_{t_0}^{t_2}T{V(t)V(t’)}dtdt’,$$
which can be divided into 4 areas, then time ordered and then integrated. The problem is when the operator arise at the same time this technique cannot be used (I guess). For example, this expectation value

$$ leftlangle n right| T leftlbrace e^{iint_{0}^{t} V_A(t-t’)dt’ }rightrbrace T leftlbrace e^{iint_{0}^{t} V_B(t’)dt’ }rightrbrace Tleftlbrace e^{-iint_{0}^{t} V_A(t’-t)dt’ }rightrbrace T leftlbrace e^{-iint_{0}^{t} V_A(-t’)dt’ }rightrbraceleft| n rightrangle $$

which arise in this paper https://arxiv.org/abs/cond-mat/0609105 eqn.6 is more complex. The two left time ordered exponential arise after the two on the right. But the two pair are operator in the same time area $[0,t]$ for the left and $[-t,0]$ for the right.

I have tried many way to regroup them to be able to expand them in a way similar to the previous expectation value but I was not able. Many questions touch the same subject especially this one Product of time ordered products (exponentials) where why my approach fail is clearly explained. But I could not find a solution.