With the interests of $$ I(a,b)=\int_{0}^{1} x^{a-1}(1-x)^{b-1}K\left ( \sqrt{x} \right ) \text{d}x, $$ where $K(k)$ represents the complete elliptic integral with modulus $k$ and $K^\prime(k)$ its complementary. Many $I(a,b)$ at rational points are known now. They are drawn in my latest edit of my note below.
The questions come here:
We summarize our results here. We'll firstly put forth several means to evaluate the integral and then make a list of all the known results.
For $\Re(a,b)>0$, we have the following hypergeometric representations, $$ \begin{aligned} I(a,b) &=\frac{\pi}{2} \frac{\Gamma\left ( a \right ) \Gamma\left ( b \right ) }{\Gamma\left (a+b \right )} \,_3F_2\left ( \frac{1}2,\frac12,a;a+b,1;1 \right ),\\&= \frac{\pi}{2} \frac{\Gamma\left ( b \right )^2}{\Gamma\left (b+\frac12 \right )^2} \,_3F_2\left ( b,b,1-a;b+\frac12,b+\frac12;1 \right ). \end{aligned} $$ The first is by definition $K(\sqrt{x} ) =\frac{\pi}{2} \,_2F_1\left ( \frac12,\frac12;1;x \right )$ and for the second we use the evaluation of $I(1,b)$. Now that the $\,_3F_2$ doesn't have a general formula, however, we observe two special cases where $\,_3F_2$ degenerates into a $\,_2F_1$, which allows us to apply Gauss's well-known $\,_2F_1$ formula. They are $a=1$ and $a+b=\frac12$, where we always assume $a,b$ satisfy $\Re(a,b)>0$ without specific notifications. To go further, we note that $I(a,b)$ at special points cater to classical hypergeometric identities like Dixon's $\,_3F_2$ and Watson's $\,_3F_2$. And by repeated applications of the hypergeometric transformations displayed here, we compute the closed-forms for $a=b,a+2b=\frac12,\frac32,2a+b=\frac32,\frac52.$ Also we have an obvious induction $I(a,b+1)=I(a,b)-I(a+1,b)$, thence to the general cases where $a+2b,2a+b\in\mathbb{Z}+\frac12,a-b\in\mathbb{Z}$. Notably these evaluations don't exhaust all the closed-froms. It's also possible to prove that$$\small \begin{aligned} &{\left(\frac2\pi\right)^2\int_{0}^{1} \left ( \frac{K^\prime}{K} \right )^{s-1} \frac{(1-2k^2)K(k)}{\sqrt{k}(1-k^2)^{3/4}}\text{d}k =2^{2s}\pi^{-s}\Gamma(s)[L_8(s)L_{-8}(s-2)+L_{-8}(s)L_{8}(s-2)]},\\ &{\left(\frac2\pi\right)^2\int_{0}^{1} \left ( \frac{K^\prime}{K} \right )^{s-1} \frac{\sqrt{k}K(k)}{(1-k^2)^{1/4}}\text{d}k =2^{2s-1}\pi^{-s}\Gamma(s)[L_8(s)L_{-8}(s-2)-L_{-8}(s)L_{8}(s-2)]}. \end{aligned}$$ This gives $I(\frac34,\frac34),I(\frac54,\frac14),I(\frac14,\frac54).$ The three cases are also do-able by discovering the relation $$ I\left ( s,\frac32-s \right ) =\frac{\sin\left ( \pi s \right ) }{\sqrt{\pi} } \Gamma\left ( s \right ) \Gamma\left ( \frac{3}{2}-s \right ) \int_{0}^{1} x^{s-\frac32}(1-x)^{-s}\arcsin\left ( \sqrt{x} \right ) \mathrm{d}x. $$ And however, two special cases $I(\frac78,\frac58),I(\frac58,\frac78)$ are excluded here. They have evaluations at here, also by standard hypergeometric transformations. At the same time, we go on another direction regarding these integrals as critical $L$-values of modular forms and empolying the $q$-series expansions to arrive at some double sums, which evaluates to be some gamma factors. For example, setting $q=\exp(-\pi K^\prime(k)/K(k))$ we have $$ f(q)=\sum_{m,n\in\mathbb{Z}} (-1)^m\left [ 3\left ( m+\frac16 \right )^2-n^2 \right ] q^{3\left ( m+\frac16 \right )^2+n^2 }=\frac{2^{2/3}k^{1/6}(1-k^2)^{1/12}(1-2k^2)}{ 3\pi^3}K(k)^3.$$ And using the modular parametrization $x=K^\prime(k)/K(k)$, $$\int_{0}^{\infty}xf(q)\text{d}x =\frac{1}{\pi^2} \sum_{m,n\in\mathbb{Z}} \frac{(-1)^m}{\left ( \sqrt{3}\left ( m+\frac16 \right )+ni \right )^2 } \Leftrightarrow\int_{0}^{1} \frac{(2k^2-1)K(k)}{k^{5/6}(1-k^2)^{11/12}}\text{d}k =\frac{3^{7/4}\left ( 1+\sqrt{3} \right )\Gamma\left ( \frac13 \right )^6 }{ 16\pi^2\cdot2^{1/3} }.$$ This is already known in our hypergeometric cases. The cases $I(\frac18,1),I(\frac38,1)$ obtained from here are unique, though. Gathering all these we make the following list of known evaluations therefore:
Cases where $a=1$ and $a+b=\frac12$. $\displaystyle I(1,b)=\frac{\pi}{2} \frac{\Gamma\left ( b \right )^2 }{ \Gamma\left ( b+\frac12 \right )^2 }$, $\displaystyle I\left ( a,\frac{1}{2}-a \right ) =\frac{\sin\left ( \pi a \right ) }{2\pi} \Gamma\left ( a \right )^2\Gamma\left ( \frac12-a \right )^2$.
By classic hypergeometric identities. $$ I(s,s)=\frac{\Gamma\left ( \frac14 \right )^2 }{2^{2s+1}} \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac14 \right )^2}, \qquad I\left ( s,\frac32-2s \right ) =\frac{\sin(\pi s)\Gamma\left ( \frac14 \right )^2}{2^{2s+2}\sin\left ( \pi \left ( s+\frac14 \right ) \right )^2 } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac14 \right )^2}.$$
Cases where $a-b\in\mathbb{Z}$(by induction $I(a,b+1)=I(a,b)-I(a+1,b)$). $$\begin{aligned} &I(s+1,s)= \frac{\Gamma\left ( \frac14 \right )^2 }{2^{2s+2}} \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac14 \right )^2} +\frac{\Gamma\left ( \frac34 \right )^2 }{2^{2s+2}} \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac34 \right )^2},\\ &I(s,s+1)= \frac{\Gamma\left ( \frac14 \right )^2 }{2^{2s+2}} \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac14 \right )^2} -\frac{\Gamma\left ( \frac34 \right )^2 }{2^{2s+2}} \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac34 \right )^2}. \end{aligned} $$
Cases where $2a+b,a+2b\in\mathbb{Z}+\frac12$(by induction $I(a,b+1)=I(a,b)-I(a+1,b)$). $$\begin{aligned} &I\left ( s,\frac12-2s \right ) =\frac{\sin(\pi s)\Gamma\left ( \frac14 \right )^2}{2^{2s+3}\sin\left ( \pi \left ( s+\frac14 \right ) \right )^2 } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac14 \right )^2} +\frac{\sin(\pi s)\Gamma\left ( \frac34 \right )^2}{2^{2s+3}\sin\left ( \pi \left ( s+\frac34 \right ) \right )^2 } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac34 \right )^2},\\ &I\left ( s+1,\frac12-2s \right ) =-\frac{\sin(\pi s)\Gamma\left ( \frac14 \right )^2}{2^{2s+3}\sin\left ( \pi \left ( s+\frac14 \right ) \right )^2 } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac14 \right )^2} +\frac{\sin(\pi s)\Gamma\left ( \frac34 \right )^2}{2^{2s+3}\sin\left ( \pi \left ( s+\frac34 \right ) \right )^2 } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac34 \right )^2},\\ &I\left ( \frac12-2s,s+1 \right ) =-\frac{\Gamma\left ( \frac14 \right )^2}{2^{2s+5/2}\sin\left ( \pi \left ( s+\frac14 \right ) \right ) } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac14 \right )^2} +\frac{\Gamma\left ( \frac34 \right )^2}{2^{2s+5/2}\sin\left ( \pi \left ( s+\frac34 \right ) \right ) } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s+\frac34 \right )^2},\\ &I\left ( \frac32-2s,s \right ) =\frac{\Gamma\left ( \frac14 \right )^2}{2^{2s+3/2}\sin\left ( \pi \left ( \frac34 -s\right ) \right ) } \frac{\Gamma\left ( s \right )^2 }{\Gamma\left (s-\frac34 \right )^2}. \end{aligned}$$
Special cases regarding modular forms. $$ \begin{aligned}\small &I\left ( \frac54,\frac14 \right )= \frac{\pi^2}{2\sqrt{2} }+\frac{\Gamma\left ( \frac14 \right )^4 }{4\pi\sqrt{2} }, \qquad I\left ( \frac14,\frac54 \right )= -\frac{\pi^2}{2\sqrt{2} }+\frac{\Gamma\left ( \frac14 \right )^4 }{4\pi\sqrt{2} },\\ &I\left (\frac18,1 \right ) =\frac{\left ( 3+\sqrt{2} \right ) \Gamma\left ( \frac18 \right )^2\Gamma\left ( \frac38 \right )^2 }{ 24\pi\sqrt{2} },\qquad I\left ( \frac38,1 \right )= \frac{\left ( 3-\sqrt{2} \right ) \Gamma\left ( \frac18 \right )^2\Gamma\left ( \frac38 \right )^2 }{ 24\pi\sqrt{2} },\\ &4I\left ( \frac16,\frac7{24} \right ) +I\left ( \frac76,\frac7{24} \right ) =\frac{\sqrt{\left ( \sqrt{2}-1 \right )^3 \left ( \sqrt{3} +\sqrt{2} \right )^3}\cdot3^{1/4} \left ( 1+\left ( 2-\sqrt{3} \right ) \left ( \sqrt{3} -\sqrt{2} \right ) \right )^2 }{2^{7/3}\pi} \Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right ) \Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24} \right ). \end{aligned} $$
Cases $a+b=\frac32.$ $$ \begin{aligned} &I\left ( \frac12,1 \right )=4G,\qquad I\left ( \frac34,\frac34 \right ) = \frac{\pi^2}{2\sqrt{2} } ,\\ &I\left ( \frac78,\frac58 \right )=\frac{\pi^2}{6} \sqrt{5+\frac{1}{\sqrt{2} } } ,\qquad I\left ( \frac58,\frac78 \right )=\frac{\pi^2}{6} \sqrt{5-\frac{1}{\sqrt{2} } }. \end{aligned} $$
Some particular examples. Note that $$(n+1)^2\int_{0}^{1}k^{n+1}K(k)\text{d}k -n^2\int_{0}^{1}k^{n-1}K(k) \text{d}k=1,$$ $$I(s,1)+I\left ( \frac12-s ,1\right ) =\frac{\pi}{2} \tan(\pi s) \frac{\Gamma\left ( s \right )^2 }{ \Gamma\left ( s+\frac12 \right )^2}.$$ Substituting $n=-\frac12,s=\frac34$ gives two equations over $I(\frac34,1),\operatorname{reg}I(-\frac14,0)$, and we have $I\left ( \frac34,1 \right ) = 4-\frac{2\,\Gamma\left ( \frac34 \right )^4}{\pi}.$ Interestingly, differentiating with respect to $s$ where $s=\frac14$ gives a series of closed-forms of the hypergeometric type $L$-values $\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^2 }{(1)_n^2} \frac{1}{\left ( 4n+1 \right )^{2k+1} },k\in\mathbb{N}$. For instance, $$ \sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^2 }{(1)_n^2} \frac{1}{\left ( 4n+1 \right )^{5} } =\frac{\Gamma\left ( \frac14 \right )^4 }{32\pi^2} \left ( G^2+\beta(4) \right ) . $$
Connections between $b=1$ and $a+b=\frac32$. $$ \begin{aligned} I(s,1) & = \frac{\sin(\pi s)}{\pi} \frac{\Gamma\left ( s \right )^2 }{ \Gamma\left ( s+\frac12 \right )^2 } I\left(1-s,s+\frac12\right) ,\\ &=\frac{8\sin(\pi s)\Gamma(s)}{\sqrt{\pi}\,\Gamma\left ( s+\frac12 \right ) } \int_{0}^{1} \left ( \frac{1-x^2}{2x} \right )^{2s} \frac{\arctan(x)}{1-x^2}\mathrm{d}x. \end{aligned} $$
Two transformations. $$ I(a,b) =\cos\left ( \pi a \right ) I\left ( a,\frac32-a-b \right ) +\sin\left ( \pi b \right ) I\left ( b,\frac32-a-b \right ) +\cos\left ( \pi b \right ) I\left ( \frac32-a-b ,b\right ), $$ $$ \sin\left ( \pi b \right ) I\left ( \frac32-a-b ,b\right ) =\sin\left ( \pi a \right ) I\left ( a,\frac32-a-b \right ) +\cos\left ( \pi b \right ) I\left ( b,\frac32-a-b \right ). $$
In addition to classical Dixon's and Watson's ${}_3F_2$ identities, Karlsson's one, cited in paper by Miller and Paris eq. (1.2) is \begin{equation} { }_3 F_2\left(u, v, z+m ;w,z; 1\right)=\frac{\Gamma(w) \Gamma(w-u-v)}{\Gamma(w-u) \Gamma(w-v)} \sum_{k=0}^m\binom{m}{k} \frac{(-1)^k(u)_k(v)_k}{(z)_k(1+u+v-w)_k} \end{equation} where $m$ is a non-negative integer and which is valid when the hypergeometric series converges and the summation formula for it makes sense. With the representation \begin{equation} I(a,b)=\frac{\pi}{2} \frac{\Gamma\left ( a \right ) \Gamma\left ( b \right ) }{\Gamma\left (a+b \right )} \,_3F_2\left ( \frac{1}2,\frac12,a;a+b,1;1 \right ) \end{equation} for $a=m+1$ and by choosing $u=v=1/2,w=b+m+1,z=1$, one deduces \begin{equation} I(m+1,b)=\frac\pi2 \frac{m!\Gamma(b)\Gamma(b+m)}{\Gamma(b+m+1/2)^2}\sum_{k=0}^m\binom{m}{k} \frac{(-1)^k\left[(1/2)_k\right]^2}{k!(1-m-b)_k} \end{equation} which gives closed forms for the integral when $a$ is a positive integer.
