Integral $\int_{0}^{a_{j}} \int_{0}^{b_{j}} \frac{1}{\sqrt{(x-x_{i})^{2}+(y-y_{i})^{2}+z_{i}^{2}}} \, dy \, dx$

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For calculating electrostatic potential coefficients I need to solve the integral

$$I = \int_{S_j} \frac{1}{r_{ij}} \, dS',$$

where $S_{j}$ is a rectangular plane with $0 \le x \le a_{j}$, $0 \le y \le b_{j}$, $z = 0$ and $r_{ij}$ describes the distance between any point on $S_{j}$ and the point $P_{i}\, (x_{i}, \, y_{i}, \, z_{i})$:

$$r_{ij} = \sqrt{(x-x_{i})^{2}+(y-y_{i})^{2}+z_{i}^{2}}$$

After inserting $r_{ij}$ in $I$ and setting the limits, I've got

$$ I = \int_{0}^{a_{j}} \int_{0}^{b_{j}} \frac{1}{\sqrt{(x-x_{i})^{2}+(y-y_{i})^{2}+z_{i}^{2}}} \, dy \, dx$$

For the inner integral I've got (using an integral table)

$$ I_{\mathrm{inner}} = \int_{0}^{b_{j}} \frac{1}{\sqrt{(x-x_{i})^{2}+(y-y_{i})^{2}+z_{i}^{2}}} \, dy = \\ = \sinh^{-1}(\frac{b_{j}-y_{i}}{\sqrt{x^2-2x_{i}x+x_{i}^{2}+z_{i}^{2}}})-\sinh^{-1}(\frac{-y_{i}}{\sqrt{x^2-2x_{i}x+x_{i}^{2}+z_{i}^{2}}}).$$

Inserting into $I$ I've got for the outer integral:

$$ I = \int_{0}^{a_{j}} \sinh^{-1}(\frac{b_{j}-y_{i}}{\sqrt{x^2-2x_{i}x+x_{i}^{2}+z_{i}^{2}}})+\sinh^{-1}(\frac{-y_{i}}{\sqrt{x^2-2x_{i}x+x_{i}^{2}+z_{i}^{2}}}) \, dx $$

I tried to solve the outer integration with partial integration, but didn't get a solution. I've also been trying to solve the outer integral using MATLAB Symbolic Math Toolbox and Maxima, but I didn't get an answer with those tools.

My Questions are:

  1. Can anybody state, whether the solution of the inner integral is right?

  2. How can I solve the outer integral?

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Let $P= y^2+by+c$, $$\int \frac1{x^2+ax+y^2+by+c}dx = \ln(2\sqrt{x(x+a)+P}+2x+a)$$

Substituting in integration limits, we have to compute $$\int 1\cdot \ln(2\sqrt{y^2+by+A}+B)dy$$

Integrating by parts, $$=y\ln(...)-\int \frac{y(2y+b)}{(2\Delta +B)(\Delta)}dy$$ where $\Delta=\sqrt{y^2+by+A}$.

Performing partial fractions decomposition, the integral becomes $$\frac1{2B}(\int \frac{2y^2+by}{\Delta}dy - \int \frac{2y^2+by}{\Delta+B/2}dy)$$ where both integrals happen to have a closed form expression according to this integral calculator.

The second integral’s result is too long that Wolfram Alpha even refused to display the result!