How to compute the following integral $\int{\frac{x^2+d^2}{\sqrt{(x^4+b^2x^2+c^2)^3}}\mathrm dx}\,$? For some time, I have been struggling with the following integral:
$$\int{\frac{x^2+d^2}{\sqrt{\left(x^4+b^2x^2+c^2\right)^3}}\mathrm dx}\;,$$
where $b^4-4c^2>0$.
I did my best, but I am still far away from solving this problem. Does anyone have a clue how to compute it?
I would be very grateful for any hint.

Hint for first step: This just describes how the initial four parameter problem can simplified to a three parameter one. I find that in solving these types of multi-parameters problems, this should almost always be your first step. 
The integral in question may be written as a definite integral with variable lower limit of integration:
$$\mathcal{I}{\left(b,c,d,u\right)}:=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+b^2x^2+c^2)^{3/2}}\mathrm{d}x,$$
where $b,c,d,u\in\mathrm{R}$ with $b^4-4c^2>0$. For simplicity, we shall assume that the three parameters $b,c,d$ are positive.
Given $c>0$, we can rescale the other three parameters and the integration variable by $\sqrt{c}$, i.e.,
$$\begin{cases}
x=y\sqrt{c},\\
\frac{b}{\sqrt{c}}=:\sqrt{2}\,\beta,\\
\frac{d}{\sqrt{c}}=:\delta,\\
\frac{u}{\sqrt{c}}=:\eta,\\
\end{cases}$$
and in the process reduce the original four parameter integral to an integral with one fewer free parameters:
$$\begin{align}
\mathcal{I}{\left(b,c,d,u\right)}
&=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+b^2x^2+c^2)^{3/2}}\mathrm{d}x\\
&=\int_{u/\sqrt{c}}^{\infty}\frac{cy^2+d^2}{(c^2y^4+b^2cy^2+c^2)^{3/2}}\sqrt{c}\,\mathrm{d}y\\
&=c^{-3/2}\int_{\eta}^{\infty}\frac{y^2+\delta^2}{(y^4+2\beta^2y^2+1)^{3/2}}\,\mathrm{d}y\\
&=c^{-3/2}\mathcal{I}{\left(\sqrt{2}\,\beta,1,\delta,\eta\right)}.\\
\end{align}$$
That is, it is sufficient to solve,
$$\tilde{\mathcal{I}}{\left(b,d,u\right)}:=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+2b^2x^2+1)^{3/2}}\,\mathrm{d}x;~~\text{where }b>1.$$

More substantial hint:
Given two real parameters $u,b\in\mathbb{R}$ such that $0\le u$ and $1<b$, define the two-variable function $f{\left(u,b\right)}$ by the definite integral,
$$f{\left(u,b\right)}:=\int_{u}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^4+2b^2x^2+1}}.$$
We'll find it convenient to additionally define the auxiliary parameters,
$$\begin{cases}
\varphi:=\arcsin{\left(\frac{1-u^2}{u^2+1}\right)};\\
\kappa:=\sqrt{\frac{b^2-1}{1+b^2}}.\\
\end{cases}$$
Note in particular that, for $0\le u$, we have
$$-1<\frac{1-u^2}{u^2+1}\le 1,$$
and for $1<b$, we have
$$0<\frac{b^2-1}{1+b^2}<1;$$
which guarantee that our parameters $\varphi$ and $\kappa$ are well-defined real numbers.
The integral $f{\left(u,b\right)}$ can be reduced to elliptic integrals of the first kind with two substitutions, $x=\frac{\sqrt{1-y}}{\sqrt{y}}\implies y=\frac{1}{x^2+1}$ and $y=\frac{z+1}{2}\implies z=2y-1$:
$$\begin{align}
f{\left(u,b\right)}
&=\int_{u}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^4+2b^2x^2+1}}\\
&=\int_{\frac{1}{u^2+1}}^{0}\frac{\left(-\frac{y^{-3/2}}{2\sqrt{1-y}}\right)\,\mathrm{d}y}{\sqrt{\left(\frac{1-y}{y}\right)^2+2b^2\left(\frac{1-y}{y}\right)+1}}\\
&=\frac12\int_{0}^{\frac{1}{u^2+1}}\frac{\mathrm{d}y}{\sqrt{y(1-y)}\sqrt{1+2(b^2-1)y(1-y)}}\\
&=\frac{\sqrt{2}}{2\sqrt{1+b^2}}\int_{-1}^{\frac{1-u^2}{u^2+1}}\frac{\mathrm{d}z}{\sqrt{1-z^2}\sqrt{1-\left(\frac{b^2-1}{1+b^2}\right)z^2}}\\
&=\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{-1}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}\\
&=\small{\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{-1}^{0}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}+\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}}\\
&=\small{\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}+\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}}\\
&=\frac{K{\left(\kappa\right)}+F{\left(\varphi,\kappa\right)}}{\sqrt{2}\sqrt{1+b^2}}.\\
\end{align}$$
Now try to find a way to express $\mathcal{I}$ as a combination of derivatives of $f$.
