Trying to solve a transcendental differential equation

[Portuga]

Homework Statement

Solving a Brachistochrone curve with drag.

Relevant Equations

$\vec F = -k \vec v$, $v = \frac{C}{\sqrt{1 + \left( y^\prime \right)^2}}$

Well, I followed the strategy used by A.S. Parnovsky in his article (\url{http://info.ifpan.edu.pl/firststep/aw-works/fsV/parnovsky/parnovsky.pdf}) and found this differential equation: $$-\frac{g x}{C^{2}} = -\frac{\beta^{2} {y^{\prime}}^{2} \arctan\left({y^{\prime}}\right) + \beta {y^{\prime}}^{2} \log\left({y^{\prime}}^{2} + 1\right) - 2 \, \beta {y^{\prime}}^{2} \log\left(\beta - {y^{\prime}}\right) - \beta^{3} - \beta^{2} {y^{\prime}} + \beta^{2} \arctan\left({y^{\prime}}\right) - {y^{\prime}}^{2} \arctan\left({y^{\prime}}\right) + \beta \log\left({y^{\prime}}^{2} + 1\right) - 2 \, \beta \log\left(\beta - {y^{\prime}}\right) - \beta - {y^{\prime}} - \arctan\left({y^{\prime}}\right)}{2 \, {\left(\beta^{2} + 1\right)}^{2} {\left({y^{\prime}}^{2} + 1\right)}},$$ where $\beta, g$ and $C$ are constants.
The problem is: how to proceed? This is clearly an transcendental equation for $y^\prime$ and the autor didn't solve it explicitly. He purposed the use of a series expansion of reciprocal powers, but I was imagining if this was the most reasonable way to do it.

 

[FactChecker]

Closed-form solutions are few and far between. Anything that gets a little complicated (and this example is very complicated) can not be expected to have a simple closed-form solution. I suspect that the use of a series expansion is the most practical way to get a closed-form approximation to any desired accuracy.

 

[aheight]

but I was imagining if this was the most reasonable way to do it.

I believe the most reasonable way to handle this is to numerically solve the DE. However, before doing this, looks like he made a typo in eqn 35 in the denominator of integrand. Should it not be (where I use x for y'):
$$
\int \frac{x}{(1+x^2)^2(x-\beta)}dx
$$
And even when I just try to solve the integral in Mathematica, the answers do not agree:

expA=Integrate[x /((1+x^2)^2(x-\[Beta])),x]//Apart
expB=(x+\[Beta])/(2(1+x^2)(1+\[Beta]^2))+
(1-\[Beta]^2)/(2(1+\[Beta]^2)^2) (ArcTan[x]-Pi/2)+
\[Beta]/(2(1+\[Beta]^2)^2) Log[(x-\[Beta])^2/(1+x^2)]

expA===expB

expA/.{x->1,\[Beta]->0.5}
expB/.{x->1,\[Beta]->0.5}

False

0.155785

-0.221206

Would have to resolve these issues first .