How can I solve the following integral?
$$\int_0 ^L \frac{1}{||\mathbf{x}-(\mathbf{a}+\mathbf{T} s)||}ds$$
The vector in the denominator represents the distance between a point in space ($\mathbf{x}$) and a straight line, over which we are integrating. The line is parameterized by its initial point ($\mathbf{a}$), its tangent vector $\mathbf{T}$ and the coordinate s, for $s \in (0,L)$
$$\int_0^L \frac{1}{\sqrt{(\mathbf{x}-\mathbf{a} - \mathbf{T}s)^2}}\text{d}s = \int_0^L \frac{1}{\sqrt{(\mathbf{x}-\mathbf{a})^2 - 2s\mathbf{T}\cdot(\mathbf{x}-\mathbf{a})+ T^2s^2}}\text{d}s.$$
Now we do the substitution $s:= t + \frac{\mathbf{T}\cdot(\mathbf{x}-\mathbf{a})}{T^2}$, d$s$=d$t$ (write $h:=\frac{\mathbf{T}\cdot(\mathbf{x}-\mathbf{a})}{T^2}$) and find
$$=\int_{h}^{L+h}\frac{1}{\sqrt{T^2t^2 - T^2h^2 + (\mathbf{x}-\mathbf{a})^2}}\text{d}t.$$
Due to Cauchy-Schwarz, we know that $T^2h^2\le (\mathbf{x}-\mathbf{a})^2$, hence
$$=\frac{1}{\sqrt{(\mathbf{x}-\mathbf{a})^2 - T^2h^2}}\int_h^{L+h} \frac{1}{\sqrt{1 + t^2\frac{T^2}{(\mathbf{x}-\mathbf{a})^2-T^2h^2}}}\text{d}t.$$
Now, we substitute $u:=t\sqrt{\frac{T^2}{(\mathbf{x}-\mathbf{a})^2-T^2h^2}}$ and obtain
$$=\frac{1}{T}\int_{\sqrt{\frac{T^2h^2}{(x-a)^2-T^2h^2}}}^{\sqrt{\frac{T^2(L+h)^2}{(x-a)^2-T^2h^2}}}\frac{1}{\sqrt{1+u^2}}\text{d}u = \frac{1}{T} (arsinh(..)-arsinh(..)).$$ but I'm sure there is a nice way to write it out in terms of logarithms or similar.