# How can $\int \frac{dx}{(x+a)^2(x+b)^2}$ be found?

Could you please suggest any hints or methods for solving $\int \frac{dx}{(x+a)^2(x+b)^2}$. I have used partial fractions to solve this integral but it is too long and complex solution. I'd like to know a simpler solution.
EDIT: $a\not= b$

Use $a-b=x+a-(x+b)$ and $$(a-b)^2=\{(x+a)-(x+b)\}^2=\cdots$$
$$\frac{1}{(x+a)^2 (x+b)^2 } = \frac{-2}{(a - b)^3 (b + x) } + \frac{1}{(a - b)^2 (b + x)^2} + \frac{2}{(a - b)^3 (a + x)} + \frac{1}{(a - b)^2 (a + x)^2}$$ by partial fraction decomposition.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int{\dd x \over \pars{x + a}^{2}\pars{x + b}^{2}} = {\partial^{2} \over \partial a\,\partial b} \int{\dd x \over \pars{x + a}\pars{x + b}} = {\partial^{2} \over \partial a\,\partial b} \int\pars{{1 \over x + a} - {1 \over x + b}}{\dd x \over b - a} \\[5mm] = &\ {\partial^{2} \over \partial a\,\partial b}\bracks{% \ln\pars{x + a \over x + b}\,{1 \over b - a}} \\[5mm] = &\ \bbx{\ds{\bracks{2\ln\pars{x + a \over x + b} - \pars{a - b}\,{2x + a + b \over \pars{x + a}\pars{x + b}}} {1 \over \pars{a - b}^{3}} + \pars{~\mbox{a constant}~}}} \end{align}