Closed form of improper integral $\int_{-\infty}^{\infty}{\frac{1}{\sqrt{(x^2+1)^3\cdot [(x - x_d)^2+(1-z_d)^2]^3}}}dx$ I would like to evaluate an integrale which depend on 2 parameters. 
The goal is to obtain an expression of the integrale depending on theses 2 parameters ($x_d$ and $z_d$) such as $f(x_d,z_d)= \ldots$.
$$ \int_{-\infty}^{\infty}{\frac{1}{\sqrt{(x^2+1)^3\cdot [(x - x_d)^2+(1-z_d)^2]^3}}}dx$$
The range of $z_d$ is [-6,0.8] and the range of $x_d$ is [-10,10]. The idea is to have the value of this integral for any combination of $(x_d,z_d).$
I tried integration by parts, partial fraction decomposition, integration by susbstitution..
Do you have any idea ? Do you think that such an expression $f(x_d,z_d)$ can be obtained ?
 A: It's possible to express this integral in terms of elliptic integrals. First, we will rename the parameters and introduce a new parameter, getting the integral into the following form:
$$J(a,b,c)=\int_{-\infty}^{\infty}{\frac{dx}{\sqrt{(x^2+a^2)^3\cdot [(x - b)^2+c^2]^3}}}$$
Here we introduced a new parameter $a$ (which is equal to $1$ in the OP) and set for convenience $x_d=b$, and $1-z_d=c$.
Now we introduce a more simple integral:
$$I(a,b,c)=\int_0^{\infty}{\frac{dx}{\sqrt{(x^2+a^2)\cdot [(x - b)^2+c^2]}}}$$
We notice that:
$$J(a,b,c)=\frac{1}{ac} \left( \frac{\partial^2 I(a,b,c)}{\partial a \partial c}+\frac{\partial^2 I(a,-b,c)}{\partial a \partial c} \right)$$
So it's enough to find $I(a,b,c)$. Now we factorize the polynomial under the root:

$$I(a,b,c)=\int_0^{\infty}{\frac{dx}{\sqrt{(x+ia)(x-ia) [x-(b+ic)][x-(b-ic)]}}}$$

We get an integral in the general form:
$$R(A,B,C,D)=\int_0^{\infty}{\frac{dx}{\sqrt{(x+A)(x+B) (x+C)(x+D)}}}$$
I've just recently asked a question about this integral and got a very thorough answer, which links this integral to the incomplete elliptic integral:
$$R(A,B,C,D)=\frac{2}{\sqrt{\left(D-B\right)\left(C-A\right)}}\int_{\alpha}^{\beta}\frac{\mathrm{d}u}{\sqrt{\left(1-u^2\right)\left(1-m u^2\right)}}$$
$$\alpha=\sqrt{\frac{A\left(D-B\right)}{B\left(D-A\right)}}$$
$$\beta=\sqrt{\frac{D-B}{D-A}}$$
$$m=\frac{\left(D-A\right)\left(C-B\right)}{\left(D-B\right)\left(C-A\right)}$$

Mind, it was assumed in the linked question that all the parameters are real. Here we have complex parameters. But the general form should still work.

I also advise the OP to look into this paper by Carlson which introduces some algorithms to efficiently compute elliptic integrals, even with complex parameters.
