An integral involving the inverse of $f(x)=\mathrm{log}x-\mathrm{log}\mathrm{cos}x+x\mathrm{tan}x$

Let the function $f:(0,\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \frac{\pi}{2}})\to \mathbb{R}$ be defined as

$f(x)=\mathrm{log}x-\mathrm{log}\mathrm{cos}x+x\mathrm{tan}x.$

Let its inverse be denoted as ${f}^{(-1)}:\mathbb{R}\to (0,\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \frac{\pi}{2}})$, it satisfies

$\mathrm{\forall}z\in \mathbb{R},\text{}\text{}f({f}^{(-1)}(z))=z.$

Consider the integral

$I={\int}_{0}^{1}z\phantom{\rule{thinmathspace}{0ex}}{e}^{z}\phantom{\rule{thinmathspace}{0ex}}{f}^{(-1)}(z)\phantom{\rule{thinmathspace}{0ex}}dz.$

Is it possible to evaluate the integral $I$ in a closed form?

Let the function $f:(0,\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \frac{\pi}{2}})\to \mathbb{R}$ be defined as

$f(x)=\mathrm{log}x-\mathrm{log}\mathrm{cos}x+x\mathrm{tan}x.$

Let its inverse be denoted as ${f}^{(-1)}:\mathbb{R}\to (0,\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \frac{\pi}{2}})$, it satisfies

$\mathrm{\forall}z\in \mathbb{R},\text{}\text{}f({f}^{(-1)}(z))=z.$

Consider the integral

$I={\int}_{0}^{1}z\phantom{\rule{thinmathspace}{0ex}}{e}^{z}\phantom{\rule{thinmathspace}{0ex}}{f}^{(-1)}(z)\phantom{\rule{thinmathspace}{0ex}}dz.$

Is it possible to evaluate the integral $I$ in a closed form?