integration - Asymptotic expansion of $int_0^1 frac{operatorname{K}(r x)}{sqrt{(1-r^2 x^2)(1-x^2)}} , mathrm{d} x $

Notation: For $\varphi \in [0,\frac{\pi}{2}]$ and $k \in [0,1)$ the definitions
$$ \operatorname{F}(\varphi,k) = \int \limits_0^\varphi \frac{\mathrm{d} \theta}{\sqrt{1-k^2 \sin^2(\theta)}} = \int \limits_0^{\sin(\varphi)} \frac{\mathrm{d} x}{\sqrt{(1 - k^2 x^2)(1-x^2)}} $$
and $\operatorname{K}(k) = \operatorname{F}(\frac{\pi}{2},k)$ will be used for the elliptic integrals of the first kind.

When answering this question, I came across the function
$$ \psi \colon [0,1) \to (0,\infty) \, , \, \psi(k) = \int \limits_0^1 \frac{\operatorname{K}(k x)}{\sqrt{(1-k^2 x^2)(1-x^2)}} \, \mathrm{d}x = \int \limits_0^{\pi/2} \frac{\operatorname{K}(k \sin(\theta))}{\sqrt{1-k^2 \sin^2 (\theta)}} \, \mathrm{d} \theta \, .$$

While $\psi(k) = \frac{\pi^2}{4} [1+ \frac{3}{8} k^2 + \mathcal{O}(k^4)]$ near $k=0$ is readily found using Maclaurin series, the expansion at $k=1$ is more elusive.

The naive attempt of replacing $\operatorname{K}(kx)$ by $\operatorname{K}(k)$ (since the largest contribution to the integral comes from the region near $x=1$) yields $\psi(k) \simeq \operatorname{K}^2 (k)$, which according to plots is not too far off but also not quite right. Integration by parts reproduces this term, but the remaining integral does not look very nice:
\begin{align}
\psi(k) &= \operatorname{K}^2 (k) - k \int \limits_0^1 \operatorname{K}'(k x) \operatorname{F}(\arcsin(x),k) \, \mathrm{d} x \\
&= \operatorname{K}^2 (k) - \int \limits_0^1 \left[\frac{\operatorname{E}(k x)}{1-k^2 x^2} - \operatorname{K}(kx)\right] \frac{\operatorname{F}(\arcsin(x),k)}{x} \, \mathrm{d} x \, .
\end{align}
The expansion $\operatorname{K}(k) = -\frac{1}{2} \log(\frac{1-k}{8}) + \mathcal{o}(1)$ is useful for the final steps, but I do not know how to extract all the leading terms, so:

How can we find the asymptotic expansion (ideally up to and including the constant terms) of $\psi(k)$ as $k \nearrow 1$ ?

Answer

Some heuristics:
From DMLF we get the asymptotic Expansion of $K(z)$ up toll all orders around $z=1_-$
$$
K(z)\sim-\frac12\log(1-z^2)+2\log(2)+O((1-z^2) \log(1-z^2))
$$

disregarding the terms which vanish at $z=1_{-}$ we obtain (Unluckily i don't see a way of making a precise estimate of the $O$-term, but it should definitly be $o(1)$)

$$
\psi(k)\sim -\int_0^1\frac{1}{2}\frac{\log(1-k^2x^2)}{\sqrt{1-x^2}{\sqrt{1-(kx)^2}}}dx-2 \log(2)\int_0^1\frac{1}{\sqrt{1-x^2}{\sqrt{1-(kx)^2}}}
$$

or (a proof for the first integral can be found here in section 7., the second is just the very defintion of $K(k)$)

$$
\psi(k)\sim -\frac12K(k)\log(1-k^2)-2\log(2)K(k)
$$

which fits Pretty well (the relative error for $k=0.999999999$ is about $2.1$% and for $k=0.9999999$ is about $2.5$%, so we are most likely off by a constant)

Using the asympototic Expansion from the beginning of the post again we could further simplifiy the above, but i leave it like that...