How to find closed-form of $\int_{0}^{+\infty} \operatorname{sech}^2 (x^2)\,dx$ 
How to find this integral closed form:
$$I=\int_{0}^{+\infty}\operatorname{sech}^2{(x^2)}\,dx$$

where $\operatorname{sech}{(x)}$  is defined as secant of hyperbolic function.
This problem form is very simple and it's interesting problem, but I use computer to help me to find its closed-form and W|A turns its numerical result $$I=\int_{0}^{+\infty}\operatorname{sech}^2{(x^2)}\,dx\approx 0.952781\ldots$$
Thank you for your help.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}\sech^{2}\pars{x^{2}}\,\dd x}
=\int_{x\ =\ 0}^{x\ \to\ \infty}{\dd\tanh\pars{x^{2}} \over 2x}
=\half\int_{0}^{\infty}{\tanh\pars{x^{2}} \over x^{2}}\,\dd x
\\[5mm]&=4\sum_{n\ =\ 0}^{\infty}\ \overbrace{%
\int_{0}^{\infty}{\dd x \over 4x^{4} + \bracks{\pars{2n + 1}\pi}^{2}}}
^{\ds{\color{#c00000}{1 \over 4\root{\pi}\pars{2n + 1}^{3/2}}}}\ =\
{1 \over \root{\pi}}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{2n + 1}^{3/2}}
\\[5mm]&={1 \over \root{\pi}}\bracks{%
\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}-
\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{3/2}}}
={1 \over \root{\pi}}\pars{%
\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}-
2^{-3/2}\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}}
\\[5mm]&={1 - 2^{-3/2} \over \root{\pi}}\sum_{n\ =\ 1}^{\infty}{1 \over n^{3/2}}
=\color{#66f}{\large{1 - 2^{-3/2} \over \root{\pi}}\,\zeta\pars{3 \over 2}}
\approx {\tt 0.9528}
\end{align}

We used the well known identity:
  $$
{\tanh\pars{z} \over z}
=8\sum_{n\ =\ 0}^{\infty}{1 \over 4z^{2} + \bracks{\pars{2n + 1}\pi}^{2}}
$$
  and the $\ds{\color{#c00000}{\mbox{red result}}}$ was found with
  one of my  previous answers .

A: We have
\begin{equation}
\int_{0}^{\infty} \frac{x^{a}}{\cosh^{2} x} \ dx=\frac{2 \Gamma(a+1) \eta(a)}{2^{a}}\qquad,\qquad\mbox{for}\,\,a>-1
\end{equation}
 where $\eta(a)$ is the Dirichlet eta function. Proof can be seen here.
Making substitution $x^2\mapsto x$ and using ${\rm{sech}}^2 x=\dfrac{1}{{\rm{cosh}}^2 x}$, then the considered integral can be rewritten as
\begin{equation}
\frac{1}{2}\int_{0}^{\infty} \frac{1}{\cosh^{2} x} \, \frac{dx}{\sqrt{x}}
\end{equation}
which is evaluated to

\begin{equation}
\sqrt{2}\, \Gamma\left(\frac{1}{2}\right) \eta\left(-\frac{1}{2}\right)=\frac{\left(2\sqrt{2}-1\right)}{2\sqrt{2\pi}}\zeta\left(\frac{3}{2}\right)
\end{equation}

where we use Hardy's formula to evaluate the negative argument of Dirichlet eta function
\begin{equation}
\eta(-s) = \frac{s}{\pi^{1+s}} \frac{2^{1+s}-1}{2^{s}-1}  \sin\left({\pi s \over 2}\right) \Gamma(s)\eta(s+1)
\end{equation}
and the relation of the Dirichlet eta function and the Riemann zeta function
\begin{equation}
\eta(s) = \left(1-2^{1-s}\right)\zeta(s)
\end{equation}
