I have to integrate following function

$\int \frac{x(t)^n}{K + x(t)^n} dt $

with a quite complex function

$x(t) = a b x_0 \left[\frac{1}{\left( b - a \right)} \left(\frac{e^{-bt}}{\left( b - d y_2 \right)} - \frac{e^{-at}}{\left( a - d y_2 \right)} \right) + \frac{ e^{-d y_2t}}{\left( a - d y_2 \right)\left( b - d y_2 \right)} \right]$

Since I have no idea how to solve this integral I started with an easy version of this function $x(t) = e^{-at}$ and I left out the exponent $n$

$\int \frac{e^{-at}}{K + e^{-at}} dt $

via substitution $x = e^{-at}$ I get the following result:

$\int \frac{e^{-at}}{K + e^{-at}} dt = -\frac{1}{a}\log{(K + e^{-at} )}$

My next step is to solve the integral with the following function $x =e^{-bt} - e^{-at} $

Here the substitution $x =e^{-bt} - e^{-at} $ doesn't work out anymore because numerator and denominator don't cancel out anymore.

$\int \frac{e^{-bt} - e^{-at}}{K + e^{-bt} - e^{-at}} dt$

$\frac{dx}{dt} = -ae^{-at} + b e^{-bt} $

$\int \frac{e^{-bt} - e^{-at}}{K + x} \frac{dx}{-ae^{-at} + b e^{-bt}}$

Here I get stuck. How do I solve this integral? I tried to solve the complete integral with Mathematica but I got no solution.

  • $\begingroup$ Look at your title: "integrate Hill's equation" and you give an integral... Can you say where is Hill's equation in there ? $\endgroup$ – Jean Marie Apr 20 '17 at 10:17
  • $\begingroup$ If you were telling more about the context, we could help you more... The main point is that you are looking for an antiderivative (aka a "closed form for an indefinite integral") : what makes you think that such an antiderivative exist, especially for such a complicated expression ? I'm confident at 99.9% that it doesn't exist (the fact that mathematica doesn't find is an indication, although you must pay attention not to give a general exponent $n$, but particular exponents : replace $n$ by $3$, or $4$ or...). Are you sure you want an exact result ? $\endgroup$ – Jean Marie Apr 20 '17 at 10:23
  • $\begingroup$ I just realized that there are 2 very different Hill's equations (en.wikipedia.org/wiki/Hill_equation). You are talking about the one in biochemistry... that I didn't know (I know the other). $\endgroup$ – Jean Marie Apr 20 '17 at 10:43
  • $\begingroup$ Im talking about Hill's equation in a biological context. [(en.wikipedia.org/wiki/Hill_equation_(biochemistry))] $\endgroup$ – Alexander Tille Apr 20 '17 at 11:06
  • $\begingroup$ Yes I had understood... But how do you react to my advise not to look at an exact expression : I do think it's desparate, and moreover I don't see the interest in the framework of biology. $\endgroup$ – Jean Marie Apr 20 '17 at 11:08

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