Evaluate $\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$ 
Evaluate :
$$\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$$

This is my attempt in below ... but what I want is simplify more to answer....thanks.
My own attempt:

 A: I really do not see how you could simplify the monster. I really wonder if a closed form expression could exist. I would try numerical integration.
We can get an approximate solution using the following series expansions at $x=0$.
$$\log (\Gamma (x))=-\log (x)-\gamma  x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
   x^4}{360}+O\left(x^5\right)$$
$$\log (\Gamma (1-x))=\gamma  x+\frac{\pi ^2 x^2}{12}-\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
   x^4}{360}+O\left(x^5\right)$$
$$\log (\Gamma (1+x))=-\gamma  x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
   x^4}{360}+O\left(x^5\right)$$ which would make the integrand to be
$$\log ^4(x)+4 \gamma  x \log ^3(x)+4 \gamma ^2 x^2 \log ^2(x)-\frac{2}{3} x^3
   \left(\psi ^{(2)}(1) \log ^3(x)\right)-\frac{4}{3} x^4 \left(\gamma  \psi
   ^{(2)}(1) \log ^2(x)\right)+O\left(x^5\right)$$ and the antiderivative
$$x \left(\log ^4(x)-4 \log ^3(x)+12 \log ^2(x)-24 \log (x)+24\right)+\frac{1}{2}
   \gamma  x^2 \left(4 \log ^3(x)-6 \log ^2(x)+6 \log (x)-3\right)+\frac{4}{27}
   \gamma ^2 x^3 \left(9 \log ^2(x)-6 \log (x)+2\right)+\frac{1}{192} x^4 \psi
   ^{(2)}(1) \left(-32 \log ^3(x)+24 \log ^2(x)-12 \log (x)+3\right)-\frac{4}{375}
   x^5 \left(\gamma  \psi ^{(2)}(1) \left(25 \log ^2(x)-10 \log
   (x)+2\right)\right)+O\left(x^6\right)$$ Then, for the definite integral
$$\gamma  \left(\frac{16 \zeta (3)}{375}-\frac{3}{2}\right)-\frac{\zeta
   (3)}{32}+24+\frac{8 \gamma ^2}{27}\approx 23.2249$$ while the numerical integration leads to $23.2372$ that is to say in error by $0.05$%.
