Evaluation of $\int_0^1\frac{\log x\,dx}{\sqrt{x(1-x)(1-cx)}}$ Assume $c$ is a small real number.

QUESTION. What is the value of this integral in terms of the complete elliptic function $K(k)$?
$$\int_0^1\frac{\log x}{\sqrt{x(1-x)(1-cx)}}\,dx.$$

I got as far as (give or take some silly errors) expressing the integral as
$$\int_{-\omega_1}^{\omega_1}\log\left(\wp(z)-\wp(\omega_2)\right)dz$$
where $\wp(\omega_1)=e_1, \wp(\omega_2)=e_3$ and $\wp(\omega_3)=e_2$ while the $e_j$'s are the (real) roots of the cubic equation associated to the Weierstrass elliptic function of the current problem.
UPDATE. I have found a solution to this problem using the Weierstrass functions. However, I welcome any sort of alternative approach.
 A: $\def\F{\operatorname F}$ Expand $\ln(x)$ as a limit:
$$\int_0^1\frac{\ln(x)}{\sqrt{x(1-x)(1-kx)}}dx=\lim_{s\to0}\frac1s \int_0^1 x^{s-\frac12}(1-x)^{-\frac12}(1-cx)^{-\frac12}dx-\frac1s\int_0^1 dx x^{-\frac12}(1-x)^{-\frac12}(1-cx)^{-\frac12}$$
Now recall the regularized Gauss hypergeometric function:
$$\Gamma(b)\Gamma(c-b)\,_2\tilde\F_1(a,b;c;z)= \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}\,_2\F_1(a,b;c;z) =\int_0^1 x^{b-1}(1-x)^{c-b-1}(1-xz)^{-a}dx$$
Therefore, we take the derivative using its limit definition:
$$\int_0^1\frac{\ln(x)}{\sqrt{x(1-x)(1-kx)}}dx=\lim_{s\to0}\frac1s\left(\sqrt\pi\Gamma\left(s+\frac12\right)\,_2\tilde\F_1\left(\frac12,s+\frac12;s+1;k\right)-2\text K(k)\right)=\pi\left.\frac{d\,_2\F_1\left(\frac12,b;1;k\right)}{db}\right|_{b=\frac12}+ \pi\left.\frac{d\,_2\F_1\left(\frac12,\frac12;c;k\right)}{dc}\right|_{c=1}-2\ln(4)\text K(k) $$
A: This is not an answer.
I did not see any way to introduce elliptic functions in this problem.
For the time being, I wrote
$$\frac{\log (x)}{\sqrt{x(1-x)(1-cx)}}=\sum_{n=0}^\infty (-1)^n \,\binom{-\frac{1}{2}}{n} \,c^n \,\,\frac{x^{n-\frac{1}{2}}\log (x)}{\sqrt{1-x}}$$ The antiderivative of the summand express in terms of hypergeometric function and if
$$I_n=\int_0^1 \frac{x^{n-\frac{1}{2}}\log (x)}{\sqrt{1-x}}\,dx$$
$$I_n=\sqrt{\pi }\,\,\frac{ \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}\left(\psi ^{(0)}\,\,\left(n+\frac{1}{2}\right)-\psi ^{(0)}(n+1)\right)$$
If $$a_n=(-1)^n\,\binom{-\frac{1}{2}}{n} \,c^n\,I_n \quad \implies \quad \frac {a_{n+1}}{a_n}=c\left(1-\frac{2}{n}+O\left(\frac{1}{n^2}\right)\right)$$ giving probably a quite fast convergence.
Computing the partial sums for $c=\frac 12$
$$\left(
\begin{array}{cc}
 p & \sum_{n=0}^p a_n \\
 0 & -4.35517 \\
 1 & -4.50687 \\
 2 & -4.53113 \\
 3 & -4.53699 \\
 4 & -4.53871 \\
 5 & -4.53928 \\
 6 & -4.53948 \\
 7 & -4.53955 \\
 8 & -4.53958 \\
 9 & -4.53959 \\
 10 & -4.53960 \\
\end{array}
\right)$$
A: In his thesis, Anil evaluated tons of integrals like these (without $\log$-term), even those having fourth degree polynomials in the square-root, by $x=x_0+\tan^2\theta$ substitutions.
$$I=\int_0^1\frac{\log x}{\sqrt{x(1-x)(1-cx)}}\,dx\overset{x\rightarrow 1/x}{=}-\int_1^{\infty}\frac{\log x}{\sqrt{x(x-1)(x-c)}}\,dx\overset{x\rightarrow 1+\tan^2\theta}=4\int_0^{\pi/2}\frac{\ln\cos\theta\,d\theta}{\sqrt{1-c\cos^2\theta}}$$
and now with $\ln\cos\theta=\frac12\ln(1-\sin^2\theta)=-\frac12\sum_{n=1}^{\infty}\frac1n\sin^{2n}\theta$ we get
$$I=-2\sum_{n=1}^{\infty}\frac1n\color{blue}{\int_0^{\pi/2}\frac{\sin^{2n}\theta\,d\theta}{\sqrt{1-c\cos^2\theta}}}=-2\sum_{n=1}^{\infty}\frac1n \color{blue}{K_{2n}(\sqrt c)}$$
where $K_{2n}(\sqrt c)$ can be calculated recursively in terms of $K(\sqrt c)$ and $E(\sqrt c)$ not just $K(\sqrt c)$ alone, as OP requested. Also, the convergence is slow. My numerical experience so far is that: the sum is approaching to $-4$ at $c=\frac12$.
A: Firstly,
$$I(c)=\int\limits_0^1 \dfrac{\ln x\,\text dx}{\sqrt{x(1-x)(1-cx)\large\mathstrut}}
=4\int\limits_0^1 \dfrac{\ln\sqrt x\,\text d\sqrt x}{\sqrt{(1-x)(1-cx)\large\mathstrut}}
=4\int\limits_0^1 \dfrac{\ln y\,\text dy}{\sqrt{(1-y^2)(1-cy^2)\large\mathstrut}}.$$
If $\mathbf{c=0},$ then
$$I(0) = 4\int\limits_0^1\dfrac{\ln y}{1-y^2}\,\text dy =-\dfrac{\pi^2}2
\approx-4.9348.$$
If $\mathbf{c=1},$ then
$$I(1) = 4\int\limits_0^1\dfrac{\ln y}{\sqrt{1-y^2}}\,\text dy =-2\pi\ln2\approx-4.35517.$$
Assume $\mathbf{c\in(0,1).}$
Using elliptic antiderivative
$$\int \dfrac{\text dy}{\sqrt{(1-y^2)(1-c y^2)}} = \operatorname{F}\big(\arcsin y\ |\,c\big)\tag1$$
and integration by parts, one can get:
$$I(c)=4\int\limits_0^1 \ln y\,\text d\operatorname{F}\big(\arcsin(y)\,|\,c\big)
=4\ln y\operatorname{F}\big(\arcsin(y)\,|\,c\big)\bigg|_0^1
-4\int\limits_0^1 \operatorname{F}\big(\arcsin(y)\,|\,c\big)\,\dfrac{\text dy}{y},$$
$$\color{brown}{\mathbf{I(c)=-4\int\limits_0^1 \operatorname{F}\big(\arcsin(y)\,|\,c\big)\,\dfrac{\text dy}{y}.}}\tag1$$
Applying Maclaurin series in the form of
$$\operatorname{F}\big(\arcsin(y)\,|\,c\big)=
\dfrac{c^k \left(1/2\right)_{(k)}}{(1 + 2 k) k!}   
\operatorname{_2F_1}\left(\dfrac12, -k, \dfrac12 - k, \dfrac1c\right)
y^{2k+1}, \qquad (|y|<1)$$
easily tu get
$$I(c)=-4\int_0^1\dfrac{c^k \left(1/2\right)_{(k)}}{(1 + 2 k)^2 k!}   
\operatorname{_2F_1}\left(\dfrac12, -k, \dfrac12 - k, \dfrac1c\right)
y^{2k+1}, \qquad (|y|<1)\tag3.$$
Sadly, obtained series converges badly (5 digs by 200 terms).
Alternative way is based on the orthogonal polynomials technic.
Let $c=p^2,\; z=py^2,\; t=\dfrac{1+p^2}{2p},$ then
$$(1-y^2)(1-p^2y^2) = 1-2 t z+z^2,$$
and from the generating function representation it follows
$$\dfrac1{\sqrt{1-2tz+z^2}} = \sum\limits_{k=0}^\infty \operatorname P_k(t) z^k =\sum\limits_{k=0}^\infty c^{k}\operatorname P_k\left(\dfrac{1+c}{2\sqrt c}\right) y^{2k},\tag4$$
where
$$\operatorname P_k(t)= \sum_{m=0}^n \dbinom nm \dbinom{-n-1}m \left(\dfrac{1-x}2\right)^m\tag5$$
are the Legendre orthogonal polynomials.
Taking in account definite integration of
$$\int\limits_0^1 x^k\ln x dx = -\dfrac1{(k+1)^2},\qquad  (k>-1),\tag6$$
finally we can obtain
$$\color{brown}{\mathrm{I(c) = -4\,\sum_{k=0}^\infty \binom{k-\frac12}k \dfrac1{(k+1)^2}  \operatorname{_1F_2}\left(\dfrac12, -k, -\dfrac12-k, \dfrac1c\right)}}.\tag7$$
However, this series converges similarly bad, as the previous one.
