Prove this closed-form of sum of ${_4F_3}$ hypergeometric functions I think the following identity is true. How could we prove it?
$${_4F_3}\left(\begin{array}c 1,1,1,1 \\\tfrac54,2,2\end{array}\middle|\,1\right) + 3\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right) = \frac{\pi^3}{8} + \frac{
3\pi^2}{4}\ln 2 - \frac 72 \zeta (3).$$
Here ${_pF_q}$ is the generalized hypergeometric function and $\zeta(3)$ is Apéry's constant. 
A numerical approximation of the identity:
$$4.7994017718717349316710651835631714860755433336848275119882157\dots$$
 A: $\def\Li{{\mathrm{Li}}}$These can be evaluated in closed form with a CAS (I used Mathematica).
The main tool is the (Euler) integral
representation
$$ F\left({a,\cdots\atop b,\cdots}\middle|\, z\right) =
\int_0^1\frac{t^{a-1}(1-t)^{b-a-1}}{B(a,b-a)}
F\left({\cdots\atop\cdots}\middle|\,zt\right)\,dt. $$
Call the hypergeometric values $F_1$ and $F_2$. Then
$$ 4F_1 = \int_0^1 \frac{\Li_2(t)\,dt}{t(1-t)^{3/4}} = \int_0^1
\frac{\Li_2(1-t)\,dt}{(1-t)t^{3/4}} = \int_0^1
\frac{dt}{(1-t)t^{3/4}}\int_0^t \frac{du}{u}(-1)\log(1-u). $$
Then substitute $t=s^4$, interchange the order of summation, integrate
over $u^{1/4} < s < 1$, and substitute $u=v^4$, giving
$$ 4F_1 = -\int_0^1 \frac{16v^3\,dv}{1-v^4} \log
v\left(\log\frac{1+v}{1-v} + i \log\frac{1-i v}{1+i v}\right). $$
After expanding the logs, and expanding the rational the rational function
$\frac{v^3}{1-v^4}$ in partial fractions, each individual integrand
has an elementary antiderivative (which a good CAS should be able to
find). I omit the explicit antiderivatives, because they are very
long. The final value is
$$ 4F_1 = \tfrac{7}{8} \pi ^3-48
\Im\Li_3(\tfrac12+\tfrac i2)+\tfrac{3}{2} \pi (\log 2)^2+3\pi ^2
\log2-14\zeta(3). $$
The second hypergeometric is similar. Choosing $a$ and $b$
appropriately, and using a CAS to simplify the hypergeometric term in
the integrand, then substituting $t=s^2$, we have
$$ F_2 = \int_0^1 \frac{ds}{2\sqrt{1-s^2}}\big(\Li_2(s) -
\Li_2(-s)\big). $$
Now, the same expression with a "$+$" instead of "$-$" can be
evaluated by a CAS easily:
$$ \int_0^1 \frac{dt}{2\sqrt{1-t^2}}\big(\Li_2(t)+\Li_2(-t)\big) =
\tfrac1{48}\pi^3 - \tfrac14\pi(\log2)^2. $$
So it's only necessary to find the integral
$$ \int_0^1 \frac{\Li_2(t)\,dt}{2\sqrt{1-t^2}}. $$
Using, like before, the integral representation
$$ \Li_2(t) = -\int_0^t \frac{du}{u}\log(1-u), $$
we find that the integral is equal to
$$ -\int_0^1 \frac{dt}{2\sqrt{1-t^2}} \int_0^t \frac{du}{u}\log(1-u) =
\int_0^1 \frac{du}{2u}\log(1-u)\big(\arcsin u - \tfrac12\pi\big), $$
where we've done the integral over $t$ first.
Expanding the arcsine in logarithms to get
$$ -\frac\pi4 \frac{\log(1-u)}{u} - \frac{i}{2} \frac{\log(1-u)
  \log(iu+\sqrt{1-u^2})}{u}, $$
my CAS was able to find a very long elementary (with polylogarithms)
antiderivative for this expression. Then taking limits $u=0,1$ and
subtracting gives the expression
$$ \int_0^1 \frac{\Li_2(t)\,dt}{2\sqrt{1-t^2}} = \tfrac5{48}\pi^3 +
2\Im \Li_3(1-i). $$
Using the identity
$$ 0 = - \Li_3(z) + \Li_3(1/z) -
\tfrac16\log^3(-z)-\tfrac16\pi^2\log(-z) $$
with $z=1-i$, $1/z = \frac{1+i}{2}$, and putting things together gives
$$ F_2 = -\tfrac1{32}\pi^3 + 4\Im
\Li_3(\tfrac12+\tfrac{i}2)-\tfrac18\pi(\log2)^2. $$
Finally we reach
$$ F_1 + 3F_2 = \tfrac18\pi^3 + \tfrac34\pi^2\log2 -
\tfrac72\zeta(3). $$
