Numerical evaluation conjecture for a Berndt-type integral with denominator power six

Let $\Gamma=\Gamma(1/4)$. The Berndt-type integral

$$\displaystyle\!\int_0^\infty \displaystyle\!\frac{x^9\,dx}{(\cos x+\cosh x)^6}$$

should equal

$$-\displaystyle\!\frac{63}{5\cdot 2^{10}}+\displaystyle\!\frac{1071\Gamma^{8}}{5^2\cdot 2^{13}\pi^{2}}-\displaystyle\!\frac{21\Gamma^{8}}{2^{12}\pi^{3}}+\displaystyle\!\frac{63\Gamma^{8}}{2^{16}\pi^{4}}-\displaystyle\!\frac{21\Gamma^{16}}{5^3\cdot 2^{13}\pi^{4}}+\displaystyle\!\frac{3\Gamma^{16}}{5\cdot 2^{13}\pi^{5}}-\displaystyle\!\frac{161\Gamma^{16}}{5\cdot 2^{19}\pi^{6}}+\displaystyle\!\frac{21\Gamma^{16}}{2^{21}\pi^{8}}+\displaystyle\!\frac{\Gamma^{24}}{5^2\cdot 2^{19}\pi^{8}}-\displaystyle\!\frac{\Gamma^{24}}{3\cdot 2^{20}\pi^{9}}+\displaystyle\!\frac{69\Gamma^{24}}{5\cdot 2^{25}\pi^{10}}-\displaystyle\!\frac{21\Gamma^{24}}{5\cdot 2^{24}\pi^{11}}+\displaystyle\!\frac{63\Gamma^{24}}{5\cdot 2^{27}\pi^{12}}-\displaystyle\!\frac{17\Gamma^{32}}{3\cdot 5^2\cdot 2^{31}\pi^{14}}+\displaystyle\!\frac{13\Gamma^{32}}{5\cdot 2^{34}\pi^{16}}+\displaystyle\!\frac{3\Gamma^{40}}{5^2\cdot 2^{40}\pi^{20}}.$$

This is a numerically verified special case of the paper's broader arithmetic-pattern results, but no proof of this exact evaluation is supplied here.