The authors provide a closed-form expression for the integral \[ \begin{aligned} N_{0,4}(a;m) & := \int^\infty_0 {dz\over (z^4+ 2az^2+ 1)^{m+1}},\quad \text{where }m\in \mathbb{N}\text{ and }a\in (-1,\infty):\\N_{0,4}(a;m) &= {\pi\over 2^{m+{3\over 2}}(a+ 1)^{m+{1\over 2}}} P^{(m+{1\over 2},-m-{1\over 2})}(a)\\ & = {\pi\over 2^{3m+{3\over 2}}(a+ 1)^{m+{1\over 2}}} \sum^m_{k= 0} 2^k {2m-2k\choose m-k}{m+k\choose m}(a+ 1)^k.\end{aligned} \] They have given the CPU table of \(N_{0,4}(4,m)\) by comparing CPU times of the direct calculation with the CPU time of their formula. There is a significance difference.