Help with a (simple) log-linearisation

As a solution to an economic model, I end up with the following equation: $$L=\frac{\alpha f^{\epsilon}s\left [ 1-Ak(1-\alpha+\alpha f^{\epsilon}s)^{\frac{1}{\epsilon}} f^{-1} s^\frac {-1}{\epsilon}\right ]}{\left [ 1-\alpha(1+f^{\epsilon}s) \right ]}$$

Is there any way to simplify it further and make it more readable, e.g. by differentiation or log-linearisation? How would it become in the latter case using the approximation that $$ln(1+x)\approx x$$?

I thank in advance anyone who takes a minute to help me.

Let's first make some substitution. Define $$g = f^\varepsilon s$$. This gives: \begin{align*} L &= \frac{\alpha g[1 - Ak(1 - \alpha + \alpha g)^{1/\varepsilon} g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\ &= \frac{\alpha g[1 - AK(1 - \alpha(1 - g))^{1/\varepsilon}g^{-1/\varepsilon}]}{1 - \alpha(1 + g)},\\ &= \frac{\alpha \left[1 - Ak\left(\frac{1- \alpha(1 - g)}{g}\right)^{1/\varepsilon}\right]}{\frac{1 - \alpha(1+g)}{\alpha g}} \end{align*} The two terms on $$\alpha$$ and $$g$$ in the denominator and numerator are almost the same except from a sign reverseal. So maybe this is a typo. I will continue by assuming that it is.

Assume that the signs on $$g$$ were flipped, then define $$h = \frac{1 + \alpha(1- g)}{g}$$

This allows the condensed form: $$L = \frac{\alpha[1 - Ak h^{1/\varepsilon}]}{h} \tag{1}$$

Then: $$\ln L = \ln \alpha + \ln(1 - A k h^{1/\varepsilon}) - \ln h$$

And: $$\frac{\partial \ln L}{\partial t} = \frac{\partial \ln(1 - A k h^{1/\varepsilon})}{\partial t} - \frac{\partial \ln h}{\partial t}$$

The first derivative on the right hand side equals: \begin{align*} \frac{\partial (1 - A k h^{1/\varepsilon})}{\partial t} &= -\frac{A k h^{1/\varepsilon} \dfrac{\dot k}{k} + \frac{1}{\varepsilon} A k h^{1/\varepsilon}\dfrac{\dot h}{h}}{1 - A k h^{1/\varepsilon}},\\ &=- \frac{A h^{1/\varepsilon}k}{1 - A k h^{1/\varepsilon}}\frac{\dot k}{k}- \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon} }{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h} \end{align*} So: $$\frac{\dot L}{L} = - \frac{A k h^{1/\varepsilon}}{1 - Ak h^{1/\varepsilon}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{A k h^{1/\varepsilon}}{1 - A k h^{1/\varepsilon}} \frac{\dot h}{h} - \frac{\dot h}{h}$$ Now we know from $$(1)$$ that: $$1 - A k h^{1/\varepsilon} = \frac{L h}{\alpha},$$ So: $$\frac{\dot L}{L} = - \dfrac{1-\dfrac{L h}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot k}{k} - \frac{1}{\varepsilon} \frac{1-\dfrac{Lh}{\alpha}}{\dfrac{Lh}{\alpha}} \frac{\dot h}{h}- \frac{\dot h}{h},\\ = - \left(\frac{\alpha}{Lh} - 1\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1\right)\frac{\dot h}{h} - \frac{\dot h}{h},\\ = \left(1- \frac{\alpha}{Lh}\right)\frac{\dot k}{k} - \frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{\dot h}{h}$$ Next, $$h = \dfrac{1 + \alpha(1 -g)}{g}$$ so taking logs gives: $$\ln h = \ln (1 + \alpha(1 - g)) - \ln g$$ taking the derivative with respect to time gives: $$\frac{\dot h}{h} = \frac{-\alpha g}{1 + \alpha(1 - g)}\frac{\dot g}{g} - \frac{\dot g}{g},\\ = - \frac{1 + \alpha}{1 + \alpha( 1 - g)}\frac{\dot g}{g},\\ = -\frac{(1+\alpha)}{hg} \frac{\dot g}{g},\\ = -\frac{hg + \alpha g}{hg} \frac{\dot g}{g},\\ = \frac{h + \alpha}{h} \frac{\dot g}{g}$$ Finally, $$g = f^\varepsilon s$$ so: $$\frac{\dot g}{g} = \varepsilon \frac{\dot f}{f} + \frac{\dot s}{s}$$ Together this gives: \begin{align*} \frac{\dot L}{L} &= \left(1 - \frac{\alpha}{Lh}\right) \frac{\dot k}{k} ,\\ &+ \left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot f}{f},\\ &+\frac{1}{\varepsilon}\left(\frac{\alpha}{Lh} - 1 + \varepsilon\right)\frac{h + \alpha}{h}\frac{\dot s}{s} \end{align*}

What are you trying to achieve? if your aim is simply log linearizing the equation, the best you can get is

$$ln(L) = ln(a) + {\epsilon}ln(f)+ ln(s)+ ln[1-Ak(1-\alpha+\alpha f^{\epsilon}s)^{\frac{1}{\epsilon}} f^{-1} s^\frac {-1}{\epsilon}] - ln[ 1-\alpha(1+f^{\epsilon}s)]$$