Ytukyg

I am having problems with one integral:

$$ int frac{1}{beta_1cdot e^{(gamma+m)cdot t} + e^{gamma cdot t} -beta_2cdot e^{(2gamma + m)cdot t}} , dt$$

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
$$begin{align}
&I= int frac{mathrm dt}{beta_1e^{(gamma,+,m),t} + e^{gamma, t} -beta_2e^{(2gamma,+,m),t}}
= int frac{e^{-gamma,t},mathrm dt}{1-beta_2e^{(gamma,+,m) ,t}+beta_1e^{m,t}}.tag1
end{align}$$

If the ratio $$r=dfrac mgamma$$ can be considered as integer, then function under the integral can be presented as the polynomials ratio,

$$begin{align}
&I= int frac{-e^{-(gamma,+,m),t}e^{-gamma,t},mathrm dt}{beta_2-beta_1e^{-gamma,t}-e^{-(gamma,+,m),t}} = begin{vmatrix}
x=e^{-gamma,t}\
dx=-gamma,e^{-gamma,t}\
end{vmatrix}
=intdfrac{gamma,x^{r+1}mathrm dx}{beta_2-beta_1x - x^{r+1}}.tag2
end{align}$$

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
$$z=w,e^{-mt/2},quad w=dfrac1{sqrt{beta_1}},quad y=b,e^{-(m-2gamma)/2},quad b=dfrac{beta_2}{2sqrt{beta_1}}.tag4$$
Then can be used expression for the generating function of second-order Chebyshev polynomials in the form of
$$g(z,y) = dfrac1{beta_1}sumlimits_{n=0}^infty U_n(y)z^n,tag5$$
where
$$begin{align}
&U_0(y)=1 = u_{00},\
&U_1(y)=2y = u_{11}y,\
&U_2(y)=4y^2-1 = u_{22}y^2-u_{20},\
&U_3(y)=8y^3-4y = u_{33}y^3 - u_{31}y,\
&U_4(y)=16y^4-12y^2+1=u_{44}y^4-u_{42}y^2+u_{40},\
&U_5(y)=32y^5-32y^3+y = u_{55}y^5-u_{53}y^3+u_{51}y,\
&U_6(y)=64y^6-80y^4+24y^2-1 = u_{66}y^6-u_{64}y^4+u_{62}y^2-u_{60},\
&U_{n}(y) = 2yU_{n-1}(y)-U_{n-2}(y),\
&U_n(y) = sumlimits_{k=0}^{left[frac n2right]}(-1)^k,u_{n,n-2k},y^{n-2k},\
&u_{n,i} = 2 u_{n-1,i-1} - u_{n-2,i},
end{align}tag6$$
$$ {u_{nn}} =
begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 2 & 0 & 0 & 0 & 0 & 0 \
-1 & 0 & 4 & 0 & 0 & 0 & 0 \
0 & -4 & 0 & 8 & 0 & 0 & 0 \
1 & 0 & -12 & 0 & 16 & 0 & 0 \
0 & 1 & 0 & -32 & 0 & 32 & 0 \
-1 & 0 & 24 & 0 & -80 & 0 & 64 \
end{pmatrix}.tag7$$
Therefore, the function under the integral can be presented as easily integrated series of
$$I = dfrac1beta_1int e^{-(gamma+m)t}sumlimits_{n=0}^infty U_n(b,e^{-(m-2gamma)/2})w^n,e^{-nmt/2},mathrm dt,tag8$$
wherein the exponent rates in the every term are negative iff $mge 2gamma.$

Let us calculate the integral.
begin{align}
&I =dfrac1beta_1int e^{-(gamma+m)t}sumlimits_{n=0}^infty w^n,e^{-nmt/2}sumlimits_{k=0}^{left[frac n2right]}(-1)^k,u_{n,n-2k},left(b,e^{-(m-2gamma)/2}right)^{n-2k},mathrm dt\
&=dfrac1beta_1sumlimits_{n=0}^infty sumlimits_{k=0}^{left[frac n2right]}int (-1)^k(wb)^n b^{-2k},u_{n,n-2k},e^{(2k+1-n)gamma+(k-n-1)m},mathrm dt\
&=dfrac1beta_1sumlimits_{n=0}^infty left(dfrac{beta_2}{2beta_1}right)^n sumlimits_{k=0}^{left[frac n2right]}(-1)^k ,u_{n,n-2k} left(dfrac{4beta_1}{beta_2^2}right)^kint,e^{(2k+1-n)gamma+(k-n-1)m},mathrm dt,\
end{align}
$$boxed{I=dfrac1beta_1sumlimits_{n=0}^infty left(dfrac{beta_2}{2beta_1}right)^n sumlimits_{k=0}^{left[frac n2right]}(-1)^k dfrac{u_{n,n-2k}}{(2k+1-n)gamma+(k-n-1)m} left(dfrac{4beta_1}{beta_2^2}right)^k,e^{(2k+1-n)gamma+(k-n-1)m},mathrm dt}.$$