This is to complement the answer by Carlo Beenakker by showing that 
$$I(t)\le\pi^4 te^{-\pi t}\tag{1}\label{1}$$
for real $t\ge0$, where $I(t)$ is the integral in question. 

Indeed, according to Carlo Beenakker, 
$$I(t)=\frac{\pi ^3}{8}  \frac{\pi  t \cosh \pi  t+2\pi t-3 \sinh \pi  t}{\sinh^4(\pi t/2)}.$$ 
So, using the substitution $t=\dfrac{\ln(1+x)}\pi$, one rewrites 
inequality \eqref{1} as 
$$d(x):=\left(8 x^3+19 x^2+18 x+6\right) \ln(x+1)-3 x (x+1)^2 (x+2)\le0\tag{2}\label{2}$$
for real $x\ge0$. 

In turn, inequality \eqref{2} follows immediately because $d(0)=d'(0)=d''(0)=d'''(0)=0$ and 
$$d''''(x)=-\frac{2 x \left(36 x^3+120 x^2+139 x+58\right)}{(x+1)^4}\le0$$
for real $x\ge0$. $\quad\Box$

One may note that $I(t)\sim\pi^4 te^{-\pi t}$ as $t\to\infty$, so that the upper bound $\pi^4 te^{-\pi t}$ on $I(t)$ in \eqref{1} is exact.