I have functions $B(x), F(x), S(x)$ that is zero for $x<0$. And I had successfully represent the below equation as convolution and multiplication.

$\int_0^t {dt_1} \int_0^t {dt_2} B(t - t_2)F(t_2 - t_1)S(t - t_1)F(t_1)$

$=\int_0^t {dt_1}F(t_1) S(t - t_1)\int_{t_1}^t {dt_2} B(t - t_2)F(t_2 - t_1)$

$=\int_0^t {dt_1}F(t_1) S(t - t_1)\int_0^{t-t_1} {dt'_2} B(t - t_1 - t'_2)F(t'_2)$

$=\int_0^t {dt'_1}F(t - t'_1) S(t'_1)\int_0^{t'_1} {dt'_2} B(t'_1 - t'_2)F(t'_2)$

$=\int_0^t {dt'_1}F(t - t'_1) S(t'_1)\{B(t'_1)*F(t'_1)\}$

$=F(t) * [S(t)\{B(t)*F(t)\}]$

where $F(t)*G(t)=\int_0^tF(t-t')G(t')dt'$. The first equality holds because $F(t_2-t_1)=0$ for $t_2<t_1$. The second equality results by substitution $t'_2=t_2-t_1$. The third equality results by substitution $t'_1=t-t_1$.

It doesn't look like it can be represented as convolution at first line but actually it turns out that it is represented by combination of convolution and multiplication of functions.

But I failed to represent the below equation as combination of convolution and multiplication. Is it possible? or not? If it is not possible, how can one prove that it is not possible?

$\int_0^t {dt_1} \int_0^t {dt_2} \int_0^t {dt_3} B(t - t_3)F(t_3 - t_2)S(t - t_2)F(t_2 - t_1)F(t_1)$

For me, it doesn't look like it is possible. But like the equation in the first line which doesn't seem to be represented as convolution at first glance, It may be possible. If anyone succeed, please let me know. Thank you