Getting from the ring of periods to the field of periods (which constant become periods in this case)?

The Ring of periods is a fascinating concept in number theory. However, it's rather restrictive, since many popular constants (such as $e$, $\gamma$) are not periods.

Periods are integrals of algebraic functions over algebraic domains. Periods form a ring, i.e. sum and product of periods is a period.

However, they do not form a field, as far as I know, i.e. the quotient of periods is not necessarily a period (only if the denomiator is an algebraic number, as far as I understand).

What happens if we also consider any quotient of periods a period? Will some additional important constants join the happy family of periods?

One example I have (and the motivation for this question) is the Gauss hypergeometric function, which has an interesting integral definition:

$${\displaystyle \mathrm {B} (b,c-b)\,_{2}F_{1}(a,b;c;z)=\int _{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,dx\qquad \Re (c)>\Re (b)>0,}$$

provided $|z| < 1$ or $|z| = 1$ and both sides converge

Using the integral definition of the Beta function we can write:

$$_2F_{1}(a,b;c;z)=\frac{\int _{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,dx}{\int_{0}^{1} x^{b-1} (1-x)^{c-b-1}dx}$$

Thus for any $a,b,c \in \mathbb{Q}$ (provided the integrals converge) and $z$ algebraic the hypergeometric function will belong to the field of periods.

The other example is $1/ \pi$, since it's still not known if it's a period (according to Wikipedia).

A more interesting example is from this answer:

$$\int_0^{\infty} \dfrac{\tanh(x)\,\tanh(x s)}{x^2}\,dx = \frac{4s}{\pi^2} \int_0^1 \ln\left(\frac{1-x}{1+x}\right) \ln\left(\frac{1-x^s}{1+x^s}\right) \,\frac{dx}{x}$$

This function belongs to the field of periods for any rational $s$, because $\pi^2$ is a period, and logarithms can be represented by integrals of rational functions (or algebraic functions if $s$ is not whole).

Any references on this topic (specifically the field of periods, not the ring of periods) will be appreciated as well.
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