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Using the identity $\cosh \left( x\pm a \right)=\cosh x\cosh a\pm \sinh x\sinh a$ the integrand can be written as: $$ \begin{align} & ={{\left( \frac{1}{\cosh \left( x-a \right)}\frac{1}{\cosh \left( x+a \right)} \right)}^{2}} \\ & =\frac{1}{{{\left( {{\cosh }^{2}}x{{\cosh }^{2}}a-{{\sinh }^{2}}x{{\sinh }^{2}}a \right)}^{2}}} \\ & =\frac{{{\operatorname{sech}}^{4}}x\ {{\operatorname{sech}}^{4}}a}{{{\left( 1-{{\left( \tanh x\ \tanh a \right)}^{2}} \right)}^{2}}} \\ \end{align} $$ Enforcing $u=\tanh x\ \tanh a$ and the identity ${{\operatorname{sech}}^{2}}x=1-{{\tanh }^{2}}x$

$$ f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\int_{0}^{\tanh a}{\frac{{{\tanh }^{2}}a-{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} $$ the last integral equals

$$\begin{align} & ={{\tanh }^{2}}a\ \int{\frac{\ du}{\left( 1-{{u}^{2}} \right)}}+\left( {{\tanh }^{2}}a\ -1 \right)\int{\frac{{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} \\ & ={{\tanh }^{2}}a\ {{\tanh }^{-1}}u+\left( {{\tanh }^{2}}a\ -1 \right)\ \times -\frac{1}{2}\left( \frac{u}{{{u}^{2}}-1}+{{\tanh }^{-1}}u \right) \\ \end{align}$$ Finally $$f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\left( \frac{1}{2}\left( a\left( {{\tanh }^{2}}a+1 \right)-\tanh a \right) \right)$$

Using the identity $\cosh \left( x\pm a \right)=\cosh x\cosh a\pm \sinh x\sinh a$ the integrand can be written as: $$ \begin{align} & ={{\left( \frac{1}{\cosh \left( x-a \right)}\frac{1}{\cosh \left( x+a \right)} \right)}^{2}} \\ & =\frac{1}{{{\left( {{\cosh }^{2}}x{{\cosh }^{2}}a-{{\sinh }^{2}}x{{\sinh }^{2}}a \right)}^{2}}} \\ & =\frac{{{\operatorname{sech}}^{4}}x\ {{\operatorname{sech}}^{4}}a}{{{\left( 1-{{\left( \tanh x\ \tanh a \right)}^{2}} \right)}^{2}}} \\ \end{align} $$ Enforcing $u=\tanh x\ \tanh a$ where $a\ne 0$ and the identity ${{\operatorname{sech}}^{2}}x=1-{{\tanh }^{2}}x$

$$ f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\int_{0}^{\tanh a}{\frac{{{\tanh }^{2}}a-{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} $$ For the last integral(##)

$$\begin{align} & ={{\tanh }^{2}}a\ \int{\frac{\ du}{\left( 1-{{u}^{2}} \right)}}+\left( {{\tanh }^{2}}a\ -1 \right)\int{\frac{{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} \\ & ={{\tanh }^{2}}a\ {{\tanh }^{-1}}u+\left( {{\tanh }^{2}}a\ -1 \right)\ \times -\frac{1}{2}\left( \frac{u}{{{u}^{2}}-1}+{{\tanh }^{-1}}u \right) \\ \end{align}$$ Finally $$f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\left( \frac{1}{2}\left( a\left( {{\tanh }^{2}}a+1 \right)-\tanh a \right) \right)$$.

(##) I have used the following integration rules: $$\begin{align} & \int{\frac{\ du}{\left( 1-{{u}^{2}} \right)}}={{\tanh }^{-1}}u+C \\ & \int{\frac{{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du}=-\frac{1}{2}\left( \frac{u}{{{u}^{2}}-1}+{{\tanh }^{-1}}u \right)+C \\ \end{align}$$ Notice that the second one is easily verified using integration by parts.

Using the identity $\cosh \left( x\pm a \right)=\cosh x\cosh a\pm \sinh x\sinh a$ the integrand can be written as: $$ \begin{align} & ={{\left( \frac{1}{\cosh \left( x-a \right)}\frac{1}{\cosh \left( x+a \right)} \right)}^{2}} \\ & =\frac{1}{{{\left( {{\cosh }^{2}}x{{\cosh }^{2}}a-{{\sinh }^{2}}x{{\sinh }^{2}}a \right)}^{2}}} \\ & =\frac{{{\operatorname{sech}}^{4}}x\ {{\operatorname{sech}}^{4}}a}{{{\left( 1-{{\left( \tanh x\ \tanh a \right)}^{2}} \right)}^{2}}} \\ \end{align} $$ Enforcing $u=\tanh x\ \tanh a$ and the identity ${{\operatorname{sech}}^{2}}x=1-{{\tanh }^{2}}x$

$$ f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\int_{0}^{\tanh a}{\frac{{{\tanh }^{2}}a-{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} $$ the last integral equals

$$\begin{align} & ={{\tanh }^{2}}a\ \int{\frac{\ du}{\left( 1-{{u}^{2}} \right)}}+\left( {{\tanh }^{2}}a\ -1 \right)\int{\frac{{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} \\ & ={{\tanh }^{2}}a\ {{\tanh }^{-1}}u+\left( {{\tanh }^{2}}a\ -1 \right)\ \times -\frac{1}{2}\left( \frac{u}{{{u}^{2}}-1}+{{\tanh }^{-1}}u \right) \\ \end{align}$$

Using the identity $\cosh \left( x\pm a \right)=\cosh x\cosh a\pm \sinh x\sinh a$ the integrand can be written as: $$ \begin{align} & ={{\left( \frac{1}{\cosh \left( x-a \right)}\frac{1}{\cosh \left( x+a \right)} \right)}^{2}} \\ & =\frac{1}{{{\left( {{\cosh }^{2}}x{{\cosh }^{2}}a-{{\sinh }^{2}}x{{\sinh }^{2}}a \right)}^{2}}} \\ & =\frac{{{\operatorname{sech}}^{4}}x\ {{\operatorname{sech}}^{4}}a}{{{\left( 1-{{\left( \tanh x\ \tanh a \right)}^{2}} \right)}^{2}}} \\ \end{align} $$ Enforcing $u=\tanh x\ \tanh a$ and the identity ${{\operatorname{sech}}^{2}}x=1-{{\tanh }^{2}}x$

$$ f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\int_{0}^{\tanh a}{\frac{{{\tanh }^{2}}a-{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} $$ the last integral equals

$$\begin{align} & ={{\tanh }^{2}}a\ \int{\frac{\ du}{\left( 1-{{u}^{2}} \right)}}+\left( {{\tanh }^{2}}a\ -1 \right)\int{\frac{{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} \\ & ={{\tanh }^{2}}a\ {{\tanh }^{-1}}u+\left( {{\tanh }^{2}}a\ -1 \right)\ \times -\frac{1}{2}\left( \frac{u}{{{u}^{2}}-1}+{{\tanh }^{-1}}u \right) \\ \end{align}$$ Finally $$f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\left( \frac{1}{2}\left( a\left( {{\tanh }^{2}}a+1 \right)-\tanh a \right) \right)$$

Using the identity $\cosh \left( x\pm a \right)=\cosh x\cosh a\pm \sinh x\sinh a$ the integrand can be written as: $$ \begin{align} & ={{\left( \frac{1}{\cosh \left( x-a \right)}\frac{1}{\cosh \left( x+a \right)} \right)}^{2}} \\ & =\frac{1}{{{\left( {{\cosh }^{2}}x{{\cosh }^{2}}a-{{\sinh }^{2}}x{{\sinh }^{2}}a \right)}^{2}}} \\ & =\frac{{{\operatorname{sech}}^{4}}x\ {{\operatorname{sech}}^{4}}a}{{{\left( 1-{{\left( \tanh x\ \tanh a \right)}^{2}} \right)}^{2}}} \\ \end{align} $$ Enforcing $u=\tanh x\ \tanh a$ and the identity ${{\operatorname{sech}}^{2}}x=1-{{\tanh }^{2}}x$

$$ f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\int_{0}^{\tanh a}{\frac{{{\tanh }^{2}}a-{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} $$ the last integral equals

$$\begin{align} & ={{\tanh }^{2}}a\ \int{\frac{\ du}{\left( 1-{{u}^{2}} \right)}}+\left( {{\tanh }^{2}}a\ -1 \right)\int{\frac{{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} \\ & ={{\tanh }^{2}}a\ {{\tanh }^{-1}}u+\left( {{\tanh }^{2}}a\ -1 \right)\ \left( -u+{{\tanh }^{-1}}u \right) \\ \end{align}$$ Finally, $$f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\left( a{{\tanh }^{2}}a\ +\left( {{\tanh }^{2}}a\ -1 \right)\ \left( a-\tanh a \right) \right)$$

Using the identity $\cosh \left( x\pm a \right)=\cosh x\cosh a\pm \sinh x\sinh a$ the integrand can be written as: $$ \begin{align} & ={{\left( \frac{1}{\cosh \left( x-a \right)}\frac{1}{\cosh \left( x+a \right)} \right)}^{2}} \\ & =\frac{1}{{{\left( {{\cosh }^{2}}x{{\cosh }^{2}}a-{{\sinh }^{2}}x{{\sinh }^{2}}a \right)}^{2}}} \\ & =\frac{{{\operatorname{sech}}^{4}}x\ {{\operatorname{sech}}^{4}}a}{{{\left( 1-{{\left( \tanh x\ \tanh a \right)}^{2}} \right)}^{2}}} \\ \end{align} $$ Enforcing $u=\tanh x\ \tanh a$ and the identity ${{\operatorname{sech}}^{2}}x=1-{{\tanh }^{2}}x$

$$ f\left( a \right)=2\frac{{{\operatorname{sech}}^{4}}a}{{{\tanh }^{3}}a}\int_{0}^{\tanh a}{\frac{{{\tanh }^{2}}a-{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} $$ the last integral equals

$$\begin{align} & ={{\tanh }^{2}}a\ \int{\frac{\ du}{\left( 1-{{u}^{2}} \right)}}+\left( {{\tanh }^{2}}a\ -1 \right)\int{\frac{{{u}^{2}}\ }{{{\left( 1-{{u}^{2}} \right)}^{2}}}du} \\ & ={{\tanh }^{2}}a\ {{\tanh }^{-1}}u+\left( {{\tanh }^{2}}a\ -1 \right)\ \times -\frac{1}{2}\left( \frac{u}{{{u}^{2}}-1}+{{\tanh }^{-1}}u \right) \\ \end{align}$$