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Let $K(x) = \int_0^{\pi/2}\frac{1}{\sqrt{1-x^2 \sin^2 \theta}}d\theta$ be the complete elliptic integral of first kind. It could be shown that $$ \int_{0}^{1} \frac{K\left ( x \right ) }{\sqrt{3-x} } \text{d}x=\frac{\Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right)\Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24}\right ) }{96\pi\sqrt{3}} $$ thanks to the known results here.
With its simple appearance, I wonder whether a brief approach exists so that we can understand them better. Appreciate your creative efforts.

Applying the well-known quadratic transformation formula $$ K\left ( \frac{1-x}{1+x} \right ) =\frac{1+x}{2} K^\prime(x),K^\prime(x):=K\left ( \sqrt{1-x^2} \right ) $$ we get $$ \int_{0}^{1} \frac{K(x)}{\sqrt{3-x} }\text{d}x =\frac{1}{\sqrt{2} } \int_{0}^{1} \frac{K^\prime(x)}{\sqrt{\left ( 1+x\right )\left ( 1+2x \right ) } } \text{d} x. $$ If using the expansion $$ \frac{1}{\sqrt{\left ( 1+x\right )\left ( 1+2x \right ) } } =\sum_{n\ge0} a_n x^n $$ and $$ \int_{0}^{1}x^n K^\prime(x)\text{d}x =\frac\pi4\frac{\Gamma\left ( \frac{n+1}{2} \right )^2 }{ \Gamma\left ( \frac{n+2}{2} \right )^2},$$ we could obtain a series representation for the integral. This works more on the case mentioned by @MiracleInvoker, because the series could split into two hypergeometric series, both of which are normally evaluated.

FD_bfa
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Setness Ramesory
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    How about the simpler case, $$\int_0^1\frac{K}{\sqrt{1-k}}dk=\frac{\Gamma^2(1/8)\Gamma^2(3/8)}{2^{9/2}\pi}$$ – Miracle Invoker Feb 20 '24 at 17:05
  • I'm unsure how helpful this is for what you are asking, but maybe someone else can make some headway with it... $$\int_{0}^{1}\frac{K(x)}{\sqrt{b-x}}dx=\sqrt{\frac{\pi}{4b}}\sum_{n=0}^{\infty}\bigg(\frac{\Gamma\big(\frac{1}{2}+n\big)}{n!}\bigg)^2\sum_{m=0}^{\infty}\frac{(-1/b)^m}{m!(2n+m+1)\Gamma\big(\frac{1}{2}-m\big)}$$ – Volk Feb 25 '24 at 15:46
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    It's very interesting , I want to know how to transform about $$ \boxed{\int_0^1\frac{\mathbf K'(\frac{1-x^2}{2})}{\sqrt{1+x^2}},\mathrm dx\overset{?}{=} \frac12\int_0^1\frac{\mathbf K(x)}{\sqrt{3-x}},\mathrm dx=\frac1{2\sqrt2}\int_0^1\frac{\mathbf K'(x)}{\sqrt{(1+x)(1+2x)}},\mathrm dx\ \ } $$ Thanks very much ~ – Jay Hung Feb 04 '25 at 07:35

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