Evaluate $\int_{0}^{1}\frac{K(x)\ln\left(1-x^2\right)}{\sqrt{1-x^2}}\text{d}x,\int_{0}^{1}\frac{xK(x)^2\ln\left(1-x^2\right)}{\sqrt{1-x^2}}\text{d}x$ I am recently interested in integrals containing an elliptic integral $K(x)$, which is defined by $\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2}  }\text{d}t$ for $|x|<1$ and $x$ is the elliptic modulus.
Then I come across the following two integrals
$$
\begin{aligned}
&\int_{0}^{1} \frac{K(x)\ln\left ( 1-x^2 \right ) }{
\sqrt{1-x^2} }\text{d}x=-\frac{\Gamma\left(\frac14\right)^4}{24},\\
&\int_{0}^{1} \frac{xK(x)^2\ln\left ( 1-x^2 \right ) }{
\sqrt{1-x^2} }\text{d}x=-\frac{\pi^4}{4}\,_4F_3\left ( \frac12,\frac12,\frac12,\frac12;1,1,1;1 \right ).
\end{aligned}
$$
by estimation. Where generalized hypergeometric function($\,_pF_q$) is used. For the first one, I write
$$
\begin{aligned}
\int_{0}^{1} \frac{K(x)\ln\left ( 1-x^2 \right ) }{
\sqrt{1-x^2} }\text{d}x
&=\frac{\pi}{2} \frac{\mathrm{d}}{\mathrm{d}n} \frac{\sqrt{\pi}\,\Gamma\left ( \frac{n+2}{2}  \right ) }{
\Gamma\left ( \frac{n+3}{2}  \right ) }\,_3F_2
\left ( \frac12,\frac12,\frac12;1,\frac{3+n}{2} ;1 \right )\Bigg|_{n=-1}\\
&=\frac{\pi}{2}\left [ -\pi\ln(2)\cdot\frac{\pi}{\Gamma\left ( \frac34 \right )^2 } 
+\pi \frac{\mathrm{d}}{\mathrm{d} n} \,_3F_2
\left ( \frac12,\frac12,\frac12;1,\frac{3+n}{2} ;1 \right )\Bigg|_{n=-1}\right ] .
\end{aligned}
$$

Edit: We just need to prove
$$
\,_3F_2
\left ( \frac12,\frac12,\frac12;1,\frac{3+n}{2} ;1 \right )
\Bigg|_{n=-1}
=-\frac{(\pi-3\ln(2))\Gamma\left ( \frac14 \right )^4 }{12\pi^3}.
$$
Or we equivalently write($H_n$ denotes harmonic numbers)
$$
\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^3 }{
(n!)^3}H_n=\frac{(\pi-3\ln(2))\Gamma\left ( \frac14 \right )^4 }{6\pi^3}.
$$
Which has been evaluated here.

Question: Can we verify these closed-forms? Thanks for reaching my hand.
 A: For the first integral we can start by substituting $1-x^2\to x^2$ followed by expanding the denominator into power series.
$$I=\int_0^1 \frac{K(x)\ln (1-x^2)}{\sqrt{1-x^2}}dx=2\int_0^1 \frac{K'(x)\ln x}{\sqrt{1-x^2}}dx$$
$$=2\sum_{k=0}^\infty \frac{\binom{2k}{k}}{2^{2k}}\int_0^1 x^{2k}K'(x)\ln x\,dx$$
We have:
$$\int_0^1 x^{2k}K'(x)dx=\frac{\pi}{4}\frac{\Gamma\left(k+\frac12\right)^2}{\Gamma\left(k+1\right)^2}$$
$$\Rightarrow \int_0^1 x^{2k}K'(x)\ln x\,dx=\frac{\pi}{4}\frac{\Gamma\left(k+\frac12\right)^2\left(\psi\left(k+\frac12\right)-\psi\left(k+1\right)\right)}{\Gamma\left(k+1\right)^2}$$
Therefore we desire to find:
$$I=\frac{\pi^2}{2}\sum_{k=0}^\infty \frac{\binom{2k}{k}^3}{\left(2^{2k}\right)^3}\left(\psi\left(k+\frac12\right)-\psi\left(k+1\right)\right)$$
And using $(*)$ from here, gives us:
$$I=-\frac{\pi^3}{6} {_2F_1}{\left( {\frac{1}{2},\frac{1}{2};1;\frac{1}{2}} \right)^2} = -\frac{\Gamma\left(\frac14\right)^4}{24}$$

Note that:
$$\binom{2k}{k}=\frac{2^{2k}}{\sqrt \pi}\frac{\Gamma\left(k+\frac12\right)}{\Gamma\left(k+1\right)}$$
