An integral involving the inverse of $f(x)=\log x-\log\cos x+x\tan x$ Let the function $f:\left(0,\,\displaystyle\frac\pi2\right)\to\mathbb{R}$ be defined as
$$f(x)=\log x-\log\cos x+x\tan x.$$
Let its inverse be denoted as $f^{(-1)}:\mathbb{R}\to\left(0,\,\displaystyle\frac\pi2\right)$, it satisfies
$$\forall z\in\mathbb{R},\ \ f(f^{(-1)}(z))=z.$$
Consider the integral
$$I=\int_0^1z\,e^z\,f^{(-1)}(z)\,dz.$$
Is it possible to evaluate the integral $I$ in a closed form?
Known special functions (that appear at MathWorld or DLMF) are okay.
 A: I know it has been some time now but we can transform this question into something more manageable. We'll just use the substitution $z = f(x)$ with $dz = f'(x) dx$. The bounds of integration change to the points $x$ where $f(x)$ is respectively $0$ or $1$. Call these points $x_0$ and $x_1$. Numerically they are 
$$x_0 = 0.576412723\ldots$$
$$x_1 = 0.807626251\ldots$$
And so we have
$$I = \int_{x_0}^{x_1} f(x) e^{f(x)}f^{(-1)}(f(x)) f'(x) dx = \int_{x_0}^{x_1} x f'(x)f(x)e^{f(x)} dx$$
Now we perform integration by parts with 
$$u = x f(x),\quad du = (f(x) + x f'(x)) dx$$
$$dv = f'(x) e^{f(x)} dx ,\quad v = e^{f(x)}$$
and the integral becomes
$$I = x_1f(x_1)e^{f(x_1)}-x_0f(x_0)e^{f(x_0)}-\int_{x_0}^{x_1}f(x)e^{f(x)} dx - \int_{x_0}^{x_1} x f'(x)e^{f(x)} dx$$
Since $f(x_0) = 0$ and $f(x_1) = 1$ we have
$$I = x_1 -\int_{x_0}^{x_1}f(x)e^{f(x)} dx - \int_{x_0}^{x_1} x f'(x)e^{f(x)} dx$$
Now we perform integration by parts on the latter integral with
$$u = x ,\quad du = dx$$
$$dv = f'(x) e^{f(x)} dx ,\quad v = e^{f(x)}$$
and we get
$$I = x_1 - x_1e^{f(x_1)}+x_0e^{f(x_0)}- \int_{x_0}^{x_1}f(x)e^{f(x)} dx+\int_{x_0}^{x_1}e^{f(x)} dx$$
It turns out we can perform the integration in the first integral since
$$e^{f(x)} = \frac{x}{\cos x} e^{x\tan x}$$
and
$$\int e^{f(x)} dx = e^{x \tan x}\cos x$$
After cleaning up a bit
$$I = x_0+ e^{x_1 \tan x_1}\cos x_1-e^{x_0 \tan x_0}\cos x_0 - \int_{x_0}^{x_1} f(x)e^{f(x)} dx$$
and this is where I am getting stuck. Mathematica is having a problem with it too. On $[x_0,x_1]$ we have
$$f(x) e^{f(x)} = \frac{x e^{x\tan(x)}}{\cos x}\left(\ln\left(\frac{x}{\cos x}\right)+x\tan x\right)$$
We also have another form
$$f(x) e^{f(x)} = \left(\frac{x^2 \sin x}{\cos^2x}+\frac{x}{\cos x}\ln\left(\frac{x}{\cos x}\right)\right)e^{x\tan x}$$
which suggest that we might have better luck looking at the following two integrals
$$J = \int_{x_0}^{x_1} \frac{x^2 \sin x}{\cos^2x}e^{x\tan x} dx$$
and
$$K = \int_{x_0}^{x_1} \ln\left(\frac{x}{\cos x}\right)e^{x\tan x} dx$$
The other problem perhaps, is that $x_0$ and $x_1$ are given numerically. I have no idea if one could find a closed for expression for them or even perhaps something like a series representation.
EDIT1: I'm not sure if this could be called a reduction but using the equation defining $f(x)$ we can say that
$$\cos x_1 e^{x_1 \tan x_1} = x_1 e^{-1+2 x_1 \tan x_1}$$
$$\cos x_10 e^{x_0 \tan x_0} = x_0 e^{2 x_0 \tan x_0}$$
