Let $wcolon (0,T)timesmathbb{T}^d to mathbb{R}^n$ be such that
$$ w_{L^infty(BMO)} := sup_{tin(0,T)}w(t,cdot)_{BMO} leq C $$
and $int_{mathbb{T}^d} w(t,x)mathrm{d}x = 0 $ for all $t$.
The corollary from the JohnNirenberg inequality states that
$$ int_{mathbb{T}^d} e^{pw(t,x)} mathrm{d}x leq C $$
for $pleq frac{c_2}{w_{L^infty(BMO)}}$.
Moreover, let $x(t,y)in C^1((0,T)timesmathbb{T}^d)$ be such that $x(t,cdot)$ is a diffemorphism for any $t$ and the Jacobian $J(t,y) = det D_y x(t,y)$ satisfies
$$ 0 < frac{1}{c} leq J(t,y) leq c $$
My problem: I’m trying to show that for a fixed $p>1$ we can choose sufficiently small $t$ such that
begin{equation}label{int}
int_{mathbb{T}^d} expleft(pint_0^t w(s,x(s,y))mathrm{d}sright)mathrm{d}y tag{*} end{equation}
is finite.
So far I tried multiple things that didn’t work out, but I believe they may be helpful:

If we had in the exponent just $int_0^t w(s,x)mathrm{d}s$, then as
$$ leftint_0^t w(s,cdot)mathrm{d}sright_{BMO} leq int_0^t w(s,cdot)_{BMO} mathrm{d}s leq tw_{L^infty(BMO)} leq Ct, $$
we would have the integrability for all $p leq frac{c_2}{Ct}$, so choosing sufficiently small $t$ we can take $p$ as big as we want. However, the function $w(t,x(t,cdot))$ may no longer be in $BMO$, so we cannot do it this way. 
The obvious way to deal with (*) is to use convexity of $exp$ and get rid of the integral in the exponent. Then
$$int_{mathbb{T}^d} expleft(pint_0^t w(s,x(s,y))mathrm{d}sright)mathrm{d}y leq int_{mathbb{T}^d}int_0^t e^{pw(s,x(s,y))} mathrm{d}smathrm{d} y $$
and then using Fubini and the bounds on $J(t,y)$, we get
$$ int_{mathbb{T}^d}int_0^t e^{pw(s,x(s,y))} mathrm{d}smathrm{d} y leq cint_0^t int_{mathbb{T}^d} e^{pw(s,x(s,y))} J(s,y) mathrm{d}ymathrm{d}s = cint_0^t int_{mathbb{T}^d} e^{pw(s,x)} mathrm{d}x mathrm{d}y. $$
Now we can apply the JohnNirenberg inequality, but we wouldn’t get arbitrary large $p$. I was wondering if there are some functional inequalities, which would allow me to put the dependence of small $t$ again in the exponent, but so far I didn’t find any.
 My third idea was also incorrect, but maybe it can be fixed somehow. It uses the fact that for nonnegative integrable functions there exist $xiin(0,t)$ such that $int_0^t f(s)mathrm{d}s leq tf(xi)$. In my case $win L^1((0,T)timesmathbb{T}^d)$ and I would have for almost all $y$
$$ int_0^t w(s,x(s,y))mathrm{d}s leq tw(xi,x(xi,y)). $$
Then
$$int_{mathbb{T}^d} expleft(pint_0^tw(s,x(s,y))mathrm{d}sright) mathrm{d} y leq int_{mathbb{T}^d} expleft(ptw(xi,x(xi,y))right) mathrm{d} y $$
and I could perform the change of variables in the same way as in 1. and eventually from JohnNirenberg obtain the integrability for $p leq frac{c_2}{Ct}$.
However, what I didn’t take into account is that my $xi$ depends on $y$, so instead I have
$$ int_o^t w(s,x(s,y))mathrm{d}s leq t w(xi(y), x(xi(y),y)) $$
and I don’t know if I can do something with it (after the change of variables I would still have the dependence of $x(t,y)$ inside $w$).
I would be really happy from the slightiest hints, as I got out of ideas. I’m also not 100% sure if this is even true, although it looks like it should work…