I am having problems with one integral:
$$ \int \frac{1}{\beta_1\cdot e^{(\gamma+m)\cdot t} + e^{\gamma \cdot t} -\beta_2\cdot e^{(2\gamma + m)\cdot t}} \, dt$$
The $\gamma$, $\beta_i$ and $m$ are all constants. I have tried several things in Mathematica but I haven't been able to solve it. Is there any change of variable that might cone in handy?
Assuming $$m>0,\quad \gamma>0,\quad \beta_1>0,\quad \beta_2>0,\quad t >0,$$ can be written $$\begin{align} &I= \int \frac{\mathrm dt}{\beta_1e^{(\gamma\,+\,m)\,t} + e^{\gamma\, t} -\beta_2e^{(2\gamma\,+\,m)\,t}} = \int \frac{e^{-\gamma\,t}\,\mathrm dt}{1-\beta_2e^{(\gamma\,+\,m) \,t}+\beta_1e^{m\,t}}.\tag1 \end{align}$$
If the ratio $$r=\dfrac m\gamma$$ can be considered as integer, then function under the integral can be presented as the polynomials ratio,
$$\begin{align} &I= \int \frac{-e^{-(\gamma\,+\,m)\,t}e^{-\gamma\,t}\,\mathrm dt}{\beta_2-\beta_1e^{-\gamma\,t}-e^{-(\gamma\,+\,m)\,t}} = \begin{vmatrix} x=e^{-\gamma\,t}\\ dx=-\gamma\,e^{-\gamma\,t}\\ \end{vmatrix} =\int\dfrac{\gamma\,x^{r+1}\mathrm dx}{\beta_2-\beta_1x - x^{r+1}}.\tag2 \end{align}$$
I.e. can be obtained closed form of the given integral in the elementary functions.
If this simplification does not satisfy, then the integral $(1)$ can be presented in the form of $$I = \int \frac{e^{-(\gamma+m)\,t}\,\mathrm dt}{\beta_1-\beta_2e^{\gamma\,t}+e^{-m\,t}}.\tag3$$
$$\beta_1-\beta_2e^{\gamma\,t}+e^{-m\,t} = \beta_1(1-2yz+z^2) = \beta_1\,g(z,y),\tag3$$ where $$z=w\,e^{-mt/2},\quad w=\dfrac1{\sqrt{\beta_1}},\quad y=b\,e^{-(m-2\gamma)/2},\quad b=\dfrac{\beta_2}{2\sqrt{\beta_1}}.\tag4$$ Then can be used expression for the generating function of second-order Chebyshev polynomials in the form of $$g(z,y) = \dfrac1{\beta_1}\sum\limits_{n=0}^\infty U_n(y)z^n,\tag5$$ where $$\begin{align} &U_0(y)=1 = u_{00},\\ &U_1(y)=2y = u_{11}y,\\ &U_2(y)=4y^2-1 = u_{22}y^2-u_{20},\\ &U_3(y)=8y^3-4y = u_{33}y^3 - u_{31}y,\\ &U_4(y)=16y^4-12y^2+1=u_{44}y^4-u_{42}y^2+u_{40},\\ &U_5(y)=32y^5-32y^3+y = u_{55}y^5-u_{53}y^3+u_{51}y,\\ &U_6(y)=64y^6-80y^4+24y^2-1 = u_{66}y^6-u_{64}y^4+u_{62}y^2-u_{60},\\ &U_{n}(y) = 2yU_{n-1}(y)-U_{n-2}(y),\\ &U_n(y) = \sum\limits_{k=0}^{\left[\frac n2\right]}(-1)^k\,u_{n,n-2k}\,y^{n-2k},\\ &u_{n,i} = 2 u_{n-1,i-1} - u_{n-2,i}, \end{align}\tag6$$ $$ \{u_{nn}\} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 4 & 0 & 0 & 0 & 0 \\ 0 & -4 & 0 & 8 & 0 & 0 & 0 \\ 1 & 0 & -12 & 0 & 16 & 0 & 0 \\ 0 & 1 & 0 & -32 & 0 & 32 & 0 \\ -1 & 0 & 24 & 0 & -80 & 0 & 64 \\ \end{pmatrix}.\tag7$$ Therefore, the function under the integral can be presented as easily integrated series of $$I = \dfrac1\beta_1\int e^{-(\gamma+m)t}\sum\limits_{n=0}^\infty U_n(b\,e^{-(m-2\gamma)/2})w^n\,e^{-nmt/2}\,\mathrm dt,\tag8$$ wherein the exponent rates in the every term are negative iff $m\ge 2\gamma.$
Let us calculate the integral. \begin{align} &I =\dfrac1\beta_1\int e^{-(\gamma+m)t}\sum\limits_{n=0}^\infty w^n\,e^{-nmt/2}\sum\limits_{k=0}^{\left[\frac n2\right]}(-1)^k\,u_{n,n-2k}\,\left(b\,e^{-(m-2\gamma)/2}\right)^{n-2k}\,\mathrm dt\\ &=\dfrac1\beta_1\sum\limits_{n=0}^\infty \sum\limits_{k=0}^{\left[\frac n2\right]}\int (-1)^k(wb)^n b^{-2k}\,u_{n,n-2k}\,e^{(2k+1-n)\gamma+(k-n-1)m}\,\mathrm dt\\ &=\dfrac1\beta_1\sum\limits_{n=0}^\infty \left(\dfrac{\beta_2}{2\beta_1}\right)^n \sum\limits_{k=0}^{\left[\frac n2\right]}(-1)^k \,u_{n,n-2k} \left(\dfrac{4\beta_1}{\beta_2^2}\right)^k\int\,e^{(2k+1-n)\gamma+(k-n-1)m}\,\mathrm dt,\\ \end{align} $$\boxed{I=\dfrac1\beta_1\sum\limits_{n=0}^\infty \left(\dfrac{\beta_2}{2\beta_1}\right)^n \sum\limits_{k=0}^{\left[\frac n2\right]}(-1)^k \dfrac{u_{n,n-2k}}{(2k+1-n)\gamma+(k-n-1)m} \left(\dfrac{4\beta_1}{\beta_2^2}\right)^k\,e^{(2k+1-n)\gamma+(k-n-1)m}\,\mathrm dt}.$$
