It is suspected that $$ {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) = \frac{8}{\pi} - \frac{4 \, \sqrt{2} \, \Gamma^2\left(\frac{3}{4}\right)}{\Gamma^3\left(\frac{1}{2}\right)} = 1.020959562466698996....$$ The ${}_{4}F_{3}$ is nicely set to use Clausen's formula, $${}_{4}F_{3}(a, b, c, d; e, f, g; 1) = \frac{(2 a)_{|d|} (a b)_{|d|} (a + b)_{|d|} }{(a)_{|d|} (b)_{|d|} (2a + 2b)_{|d|} }$$ where $\frac{1}{2} = a + b + c -d$, $e = a + b + \frac{1}{2}$, $a + f = d + 1 = b + g$. The condition that is not satisfied is that $d$ should be a negative integer. Another attempt made was to use $$ {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) = \frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma^2\left(\frac{3}{4}\right)} \, \int_{0}^{1} t^{-1/4} (1-t)^{-1/4} \, {}_{3}F_{2}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}; 1, \frac{3}{2}; t \right) \, dt$$ but switching around the ${}_{3}F_{2}$ seems to just cycle the result and not provide a reduction. Another method is to use $$ {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) = 2 \, {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{1}{2}, \frac{3}{2}; 1 \right) - {}_{4}F_{3}\left(\frac{5}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) $$ but doesn't seem to be able to fit a known reduction like Clausen's identity.
Might there be a better way to show this identity is true?
Maple says:
$$ \frac{8}{\pi}+\frac{8}{\pi}{\it EllipticK} \left( i \right) -\frac{8}{\pi}{\it EllipticE}(i) $$ This could also be written as
$$ \frac{8}{\pi} - \frac{8}{\pi} \int_0^1 \frac{t^2\; dt}{\sqrt{1-t^4}}$$
and Maple says that is indeed $$ \frac{8}{\pi} - \frac{2}{\pi} B(1/2, 3/4) = \frac{8}{\pi} - 4\,{\frac { \left( \Gamma \left( 3/4 \right) \right) ^{2}\sqrt {2}}{{ \pi}^{3/2}}} $$