We have \begin{align} I & = \int_0^{\infty} \dfrac{dx}{\cos(x) + \cosh(x)}\\ & = \int_0^{\infty} \dfrac{\text{sech}(x)dx}{1 + \text{sech}(x)\cos(x)}\\ & = \sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx \end{align} where \begin{align} \int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx & = 2^{k+1}\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx \end{align} Now we have $$\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx = \int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx$$ Further, $$\dfrac1{(1+e^{-2x})^{k+1}} = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l e^{-2lx}$$ Hence, $$\int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx$$ Hence, \begin{align} \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx & = \int_0^{\infty} e^{-(2l+k+1)x}\left(\dfrac{e^{ix}+e^{-ix}}2 \right)^k dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{imx}e^{-i(k-m)x}dx\\ & = \dfrac{e^{-ikx}}{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{2imx}dx\\ & = \dfrac{e^{-ikx}}{2^k} \sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \end{align}\begin{align} \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx & = \int_0^{\infty} e^{-(2l+k+1)x}\left(\dfrac{e^{ix}+e^{-ix}}2 \right)^k dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{imx}e^{-i(k-m)x}dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{2imx-ikx}dx\\ & = \dfrac1{2^k} \sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \end{align} Hence, $$I = 2 \sum_{k=0}^{\infty}(-e^{ix})^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \right)$$$$I = 2 \sum_{k=0}^{\infty}(-1)^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \right)$$ The integral you are after is twice $I$ and hence $$\color{blue}{4 \sum_{k=0}^{\infty}(-e^{ix})^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \right)}$$$$\color{blue}{4 \sum_{k=0}^{\infty}(-1)^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \right)}$$ You can truncate the two infinite sums (since both are alternating sums) to get arbitrary accuracy.
We have \begin{align} I & = \int_0^{\infty} \dfrac{dx}{\cos(x) + \cosh(x)}\\ & = \int_0^{\infty} \dfrac{\text{sech}(x)dx}{1 + \text{sech}(x)\cos(x)}\\ & = \sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx \end{align} where \begin{align} \int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx & = 2^{k+1}\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx \end{align} Now we have $$\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx = \int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx$$ Further, $$\dfrac1{(1+e^{-2x})^{k+1}} = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l e^{-2lx}$$ Hence, $$\int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx$$ Hence, \begin{align} \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx & = \int_0^{\infty} e^{-(2l+k+1)x}\left(\dfrac{e^{ix}+e^{-ix}}2 \right)^k dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{imx}e^{-i(k-m)x}dx\\ & = \dfrac{e^{-ikx}}{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{2imx}dx\\ & = \dfrac{e^{-ikx}}{2^k} \sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \end{align} Hence, $$I = 2 \sum_{k=0}^{\infty}(-e^{ix})^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \right)$$ The integral you are after is twice $I$ and hence $$\color{blue}{4 \sum_{k=0}^{\infty}(-e^{ix})^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \right)}$$ You can truncate the two infinite sums (since both are alternating sums) to get arbitrary accuracy.
We have \begin{align} I & = \int_0^{\infty} \dfrac{dx}{\cos(x) + \cosh(x)}\\ & = \int_0^{\infty} \dfrac{\text{sech}(x)dx}{1 + \text{sech}(x)\cos(x)}\\ & = \sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx \end{align} where \begin{align} \int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx & = 2^{k+1}\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx \end{align} Now we have $$\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx = \int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx$$ Further, $$\dfrac1{(1+e^{-2x})^{k+1}} = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l e^{-2lx}$$ Hence, $$\int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx$$ Hence, \begin{align} \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx & = \int_0^{\infty} e^{-(2l+k+1)x}\left(\dfrac{e^{ix}+e^{-ix}}2 \right)^k dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{imx}e^{-i(k-m)x}dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{2imx-ikx}dx\\ & = \dfrac1{2^k} \sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \end{align} Hence, $$I = 2 \sum_{k=0}^{\infty}(-1)^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \right)$$ The integral you are after is twice $I$ and hence $$\color{blue}{4 \sum_{k=0}^{\infty}(-1)^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+ik+1} \right)}$$ You can truncate the two infinite sums (since both are alternating sums) to get arbitrary accuracy.
We have \begin{align} I & = \int_0^{\infty} \dfrac{dx}{\cos(x) + \cosh(x)}\\ & = \int_0^{\infty} \dfrac{\text{sech}(x)dx}{1 + \text{sech}(x)\cos(x)}\\ & = \sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx \end{align} where \begin{align} \int_0^{\infty} \text{sech}^{k+1}(x) \cos^k(x) dx & = 2^{k+1}\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx \end{align} Now we have $$\int_0^{\infty} \dfrac{\cos^k(x)}{\left(e^x + e^{-x}\right)^{k+1}} dx = \int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx$$ Further, $$\dfrac1{(1+e^{-2x})^{k+1}} = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l e^{-2lx}$$ Hence, $$\int_0^{\infty} \dfrac{e^{-(k+1)x} \cos^k(x)}{(1+e^{-2x})^{k+1}} dx = \sum_{l=0}^{\infty} \dbinom{k+l}{l}(-1)^l \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx$$ Hence, \begin{align} \int_0^{\infty} e^{-(2l+k+1)x} \cos^k(x) dx & = \int_0^{\infty} e^{-(2l+k+1)x}\left(\dfrac{e^{ix}+e^{-ix}}2 \right)^k dx\\ & = \dfrac1{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{imx}e^{-i(k-m)x}dx\\ & = \dfrac{e^{-ikx}}{2^k} \int_0^{\infty} \sum_{m=0}^k \dbinom{k}m e^{-(2l+k+1)x}e^{2imx}dx\\ & = \dfrac{e^{-ikx}}{2^k} \sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \end{align} Hence, $$I = 2 \sum_{k=0}^{\infty}(-e^{ix})^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \right)$$ The integral you are after is twice $I$ and hence $$\color{blue}{4 \sum_{k=0}^{\infty}(-e^{ix})^k \sum_{l=0}^{\infty}(-1)^l \dbinom{k+l}l \left(\sum_{m=0}^k \dfrac{\dbinom{k}m}{k+2l-2im+1} \right)}$$ You can truncate the two infinite sums (since both are alternating sums) to get arbitrary accuracy.