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, $$I(t)=\frac{\pi ^3}{8} \frac{\pi t \cosh \pi t+2\pi t-3 \sinh \pi t}{\sinh^4(\pi t/2)}.$$ This expression for $I(t)$ can be obtained from formula 3.986.4 of Gradshteyn and Ryzhik by using substitutions $\pi x=r$ and $\beta=\pi t/2$, to get $$J(t):=\int_0^\infty\frac{1-\cos rt}{\sinh^2 r}\,dr =\frac{\pi t}2\,\coth\frac{\pi t}2-1$$ and then noting that $I(t)=-J'''(t)$.

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.

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, $$I(t)=\frac{\pi ^3}{8} \frac{\pi t \cosh \pi t+2\pi t-3 \sinh \pi t}{\sinh^4(\pi t/2)}.$$ This expression for $I(t)$ can be obtained from formula 3.986.4 of Gradshteyn and Ryzhik by using substitutions $\pi x=r$ and $\beta=\pi t/2$, to get $$J(t):=\int_0^\infty\frac{1-\cos rt}{\sinh^2 r}\,dr =\frac{\pi t}2\,\coth\frac{\pi t}2-1,$$ and then noting that $I(t)=-J'''(t)$.

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.

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.

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, $$I(t)=\frac{\pi ^3}{8} \frac{\pi t \cosh \pi t+2\pi t-3 \sinh \pi t}{\sinh^4(\pi t/2)}.$$ This expression for $I(t)$ can be obtained from formula 3.986.4 of Gradshteyn and Ryzhik by using substitutions $\pi x=r$ and $\beta=\pi t/2$, to get $$J(t):=\int_0^\infty\frac{1-\cos rt}{\sinh^2 r}\,dr =\frac{\pi t}2\,\coth\frac{\pi t}2-1$$ and then noting that $I(t)=-J'''(t)$.

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.