A mixed bonanza!

The substitution $u = x^8$ makes the integral $I \; = \; \tfrac{1}{64} \int_0^\infty e^{-u} \,\cos u \,u^{-\frac12} \, \ln u\,du \; = \; \tfrac{1}{64}A'(\tfrac12)$ where $A$ is the function $A(a) \; = \; \int_0^\infty e^{-u} \,\cos u\, u^{a-1}\,du \; = \; 2^{-\frac12a} \,\cos\tfrac{\pi a}{4}\, \Gamma(a) \hspace{1cm} a > 0 \;.$ Differentiating and simplifying, we obtain $\begin{array}{rcl} A'(\tfrac12) & = & \displaystyle -\frac{\sqrt{\pi}}{2^{\frac54}} \cos \tfrac{\pi}{8}\, \ln 2 + \frac{\sqrt{\pi}}{2^{\frac14}} \cos \tfrac{\pi}{8}\, \psi(\tfrac12) -\frac{\pi\sqrt{\pi}}{2^{\frac{17}{4}}}\sin\tfrac{\pi}{8} \\ I & = & \displaystyle-\frac{\sqrt{\pi}}{128}\left[ \sqrt{\sqrt{2}+1}\left(\gamma + \tfrac52\ln2\right) + \tfrac14\pi\sqrt{\sqrt{2}-1}\right] \end{array}$ making the answer $2 + 128 + 1 + 5 + 4 = \boxed{140}$ .

The answer is: $\dfrac{-\pi^{\frac{1}{2}}}{128}\left[ \sqrt{\sqrt{2}+1}\left(\gamma+\dfrac{5}{2}\ln(2)\right)+\dfrac{\pi}{4}\sqrt{\sqrt{2}-1}\right]$ Will post the solution later