Formulation of Bioche's rules in familiar notation I was trying to find an interesting problem for my physics students involving a nontrivial flux integral, and I came up with one that produced the integral
$$\int \frac{dx}{1+\beta\cos x}$$
($\beta^2<1$). I resorted to the computer algebra system Maxima in order to integrate it, and that worked, but I wanted to understand what it had done. Playing around and searching on the web showed that this is an example that can naturally be approached using Bioche's rules, but the only description of those I could find was a French language wikipedia article. I'm finding the article hard to understand, I think not so much because of my weak French as what seems to be some archaic notation or old-fashioned ways of thinking about what we would today call a function. The WP article seems to say this (my attempted translation):
In the following $f(t)$ is a rational expression in $\sin t$ and $\cos t$. Then in order to calculate $\int f(t)dt$, one forms the integrand $\omega(t)=f(t)dt$. Then:


*

*If $\omega(-t)=\omega(t)$, a good change of variables is $u=\cos t$.

*If $\omega(\pi-t)=\omega(t)$, a good change of variables is $u=\sin t$.

*If $\omega(\pi+t)=\omega(t)$, a good change of variables is $u=\tan t$.

*If two of the preceding relations both hold, a good change of variables is $u=\cos 2t$.

*In all other cases, use $u=\tan(t/2)$.
I'm having a hard time interpreting the distinction between $f$ and $\omega$. Presumably rule 1 is equivalent to saying that $f$ is even. In 2, is the idea to do the substitution $t\rightarrow \pi-t$, which also implies $dt\rightarrow-dt$? This would seem equivalent to $f(\pi-t)=-f(t)$...?
Is there a reason not to just express the rules in terms of $f$?
My faith in my own translation/understanding is not reinforced when I try to apply the rules to my own example. It seems that 1 holds, because $f$ is even. Then the substitution $u=\cos x$ transforms my integral into
$$-\int\frac{du}{\sqrt{1-u^2}(1+\beta u)},$$
but this doesn't actually seem any better. It seems like the most general substitution $u=\tan(x/2)$ is required.
What is wrong with my analysis/translation/understanding?
 A: In fact yes, the $dt$ is important in the Bioche's rules.


*

*
*

*means $w(-t)=f(-t)d(-t)=w(t)=f(t)dt\\\implies f(-t)=-f(t)$
Then with the change $u=\cos x$

$f(x)dx=g(\cos x)(−\sin x dx)=g(u)du$



*

*

*means $w(\pi-t)=f(\pi-t)d(\pi-t)=w(t)=f(t)dt\\\implies f(\pi-t)=-f(t)$
Then with the change $u=\sin x$

$f(x)dx=g(\sin x)(\cos x dx)=g(u)du$



*

*

*means $w(\pi+t)=f(\pi+t)d(\pi+t)=w(t)=f(t)dt\\\implies f(\pi+t)=f(t)$
Then with the change $u=\tan x$

$f(x)dx=g(\tan x)(dx/\cos^2 x)=g(u)du$


You can see that performing the change $t\mapsto -t,\ t\mapsto \pi-t,\ t\mapsto \pi+t$ in the expression in yellow sees both $g$ and the $dx$ group in parenthesis unchanged, justifying this particular substitution.

So 1. when the integrand behaves as $\sin$ do a substitution in $\cos$
So 2. when the integrand behaves as $\cos$ do a substitution in $\sin$
So 3. when the integrand behaves as $\tan$ do a substitution in $\tan$
In your case, none is working because you have either $\dfrac{-1}{1+\beta}$ or $\dfrac{1}{1-\beta}$ thus you need to fall back to a change in $\tan(\theta/2)$.
A: Covering the same ground as @zwim while adding a bit more information, for integrals of the form
$$\int f(\sin x, \cos x) \, dx$$
where $f$ is a rational function of sine and cosine, I like to refer to $\omega (x) = f(\sin x, \cos x) \, dx$ as a differential form and it is this differential form that must remain invariant under one of the three substitutions: $x \mapsto -x, x \mapsto \pi - x, x \mapsto \pi + x$ if the rules of Bioche are to apply.
For the substitution where the differrntial form is invariant one sets $t = \phi (x)$, where $\phi (x)$ is the function $\cos x$, $\sin x$, or $\tan x$ that also remains invariant under the same substitution.
That is:
1. Set $t = \cos x$ when $x \mapsto -x$ leaves the differential form invariant since $\cos (-x) = \cos x$.
2. Set $t = \sin x$ when $x \mapsto \pi -x$ leaves the differential form invariant since $\sin (\pi -x) = \sin x$.
3. Set $t = \tan x$ when $x \mapsto \pi + x$ leaves the differential form invariant since $\tan (\pi + x) = \tan x$.
In the event that more than one of the initial substitutions leaves the differential form unchanged, the differential form will be unchanged under all three substitutions. In this case any one of the substitutions $t = \cos x$, $t = \sin x$, or $t = \tan x$ may be used but it is usually more efficient to:
4. Set $t = \cos 2x$ since in all three cases $x \mapsto -x, \pi - x$, and $\pi + x$ leave $\cos 2x$ unchanged.
Finally, if none of the initial three substitutions leave the differential form unchanged then as a last resort:
5. Set $t = \tan \frac{x}{2}$.

Comment
At least in the English speaking world, the Bioche rules do not seem to be widely know. I guess one possible reason for this is an integral consisting of a rational function of sine and cosine can always be found using the rationalising substitution of $t = \tan \frac{x}{2}$. However, when used, one often ends up with an integral requiring cumbersome partial fraction decompositions which can often be avoided by applying the rules of Bioche in cases where it works.
Three written sources (in English) that refer to the Bioche's rules can be found in:

*

*Handbook of Integration by Zwillinger on page 108 (though the name "Bioche's rules" is not used here).


*How to Integrate It: A practical guide to finding elementary integrals by Stewart on pages 190$-$197.


*The article Integrating rational functions of sine and cosine using the rules of Bioche found here.
