A consideration of Aryabhata's answer to the linked question shows that there is a map from the elliptic curve $y^2 = P(x)$ to the conic $v^2 = u^2 + 2$ given by
$$(x,y) \mapsto \left(x - \dfrac{1}{x}, \dfrac{y}{x}\right),$$
and the differential
$$\dfrac{1+x^2}{(1-x^2)\sqrt{1 + x^4}} \,\mathrm dx$$
on the elliptic curve is the pull-back of the differential
$$\dfrac{1}{u v}\,\mathrm du$$
on the conic.
Since a conic has genus zero (i.e. it can be parameterized by a single variable,
using a classical "$t$-substitution"), the integral of a differential on a conic can always be expressed via elementary functions. Thus the same is true
of the integral of the original differential on the elliptic curve.
The answer to the general question is the same: if the differential in question can be pulled back from a map to a rational curve (i.e. a genus zero curve),
then the "elliptic integral" in question can be in fact integrated via elementary functions.
For example, any elliptic curve $y^2 = P(x)$ has a map to the the $x$-line given
by $(x,y) \mapsto x$. So if the integral only involves rational functions of $x$ (which will be the case when $y$ appears to even powers, since we can always
substitute $P(x)$ for $y^2$) then it can be computed in elementary terms. Also,
if $P(x)$ has repeated roots, then the curve $y^2 = P(x)$ itself is actually rational (it can be parameterized by a variation of the classical $t$-substitution for conics), and so any "elliptic integral" is actually elementarily integrable.
P.S. I have used some geometric terminology here (pull-back, differential, elliptic curve, rational curve) because the modern point of view on this material is via algebraic geometry. If some of this is unfamiliar, leave a comment and I (or someone else) can elaborate.
Added: If we have a curve $C$ (which could be our elliptic curve $y^2 = P(x)$,
or our rational curve $v^2 = u^2 + 2$, or just the $x$-line, or ...) and if $\omega$ is a differential on $C$, then finding the indefinite integral of $\omega$ means finding some function $f$ on $C$ such that $\omega = df$.
Now if $\varphi: C' \to C$ is a map of curves, then $\varphi^* \omega
= \varphi^* d f = d (f\circ \varphi).$ So $f\circ \varphi$ is an indefinite
integral of the pulled back differential $\varphi^*\omega$.
In particular, if $f$ is an elementary function of the coordinates on $C$,
and $\varphi$ is given by expressions which are elementary functions of the
coordinates, than the composite $f\circ \varphi$ will again be given by
elementary functions of the coordinates.
This is what is happening in your example.
Explicitly, on our curve $v^2 = u^2 + u,$ we had for example the differential
$$\dfrac{1}{u v} \,\mathrm du = \frac{1}{2 u^2 v}\,\mathrm d (u^2 + 2) = \frac{1}{2(v^2-2)v}\,\mathrm d(v^2) = \dfrac{1}{v^2-2}\,\mathrm dv = \dfrac{1}{2\sqrt{2}}\mathrm d\log\bigl( \frac{v-\sqrt{2}}{v+\sqrt{2}}\bigr).$$
Now pulling back this differential via our map $\varphi:(x,y)\mapsto \left(x-\dfrac{1}{x}, \dfrac{y}{x}\right)$ we obtain
$$\dfrac{1 + x^2}{(1-x^2)y}\,\mathrm dx = \dfrac{1}{2\sqrt{2}}\mathrm d\log\bigl(\frac{y-\sqrt{2}x}{y+\sqrt{2}x} \bigr).$$
As this example shows, pulling back is just the theoretical terminology
for making a substitution, just as a map of curves is just theoretical terminology for a change of variables.
If memory serves, Miles Reid's wonderful book Undergraduate algebraic geometry discusses some of this, and in particular gives some of the history of how the analytic theory of elliptic integrals turned into the
algebro-geometric theory of elliptic curves. (If you don't know this book, don't be fooled by the title --- it is a great introduction to the subject for someone at any level, not just undergraduates!) A much more detailed history can be found in Dieudonne's book on the history of algebraic geometry, but that book is probably not very readable unless you already have some feeling for algebraic geometry as a subject.
(might as well...)
As a quick review: anytime you see an integral that involves the square root of a cubic or a quartic polynomial, it is quite likely that an elliptic integral will be needed (except in some very special cases, termed "pseudoelliptic integrals").
Now, for the integral at hand (treating the indefinite case for now, and worrying about the limits later): your integrand is a quartic that consists of only even powers of the dummy variable, so the first thing to do is to perform a (modified) Weierstrass substitution, $t=2\tan\dfrac{u}{2}$:
$$\begin{align*}
\int \sqrt{t^4+4t^2+16} \,\mathrm dt&=8\int \frac{\sqrt{3+\cos^2 u}}{(1+\cos\,u)^2} \mathrm du=8\int \frac{\sqrt{4-\sin^2 u}}{(1+\cos\,u)^2} \mathrm du\\
&=16\int \frac{\sqrt{1-\frac14 \sin^2 u}}{(1+\cos\,u)^2} \mathrm du
\end{align*}$$
(the additional factor of $2$ in the substitution used comes from the fourth root of the polynomial's constant term.)
Why did I make those last few transformations, you ask? This is because the standard elliptic integrals are made to pop out easily when the part within the square root takes the form $\sqrt{1-m\,\sin^2 u}$.
In any event, we now need to perform a Jacobian substitution. We let $u=\mathrm{am}(v\mid m)$, where $\mathrm{am}$ is the Jacobian amplitude function and $m$ is a constant (in elliptic integral parlance, the parameter) to be determined. The usual Jacobian elliptic functions come from this basic function:
$$\begin{align*}
\mathrm{sn}(v\mid m)&=\sin(\mathrm{am}(v\mid m))\\
\mathrm{cn}(v\mid m)&=\cos(\mathrm{am}(v\mid m))\\
\mathrm{dn}(v\mid m)&=\frac{\mathrm d}{\mathrm dv}\mathrm{am}(v\mid m)=\sqrt{1-m\,\sin^2(\mathrm{am}(v\mid m))}
\end{align*}$$
Consideration of the last relation involving $\mathrm{dn}$ indicates that we can choose $m=\frac14$ for convenience; thus,
$$16\int \frac{\sqrt{1-\frac14 \sin^2 u}}{(1+\cos\,u)^2} \mathrm du=16\int \frac{\mathrm{dn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv=4\int \frac{3+\mathrm{cn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv$$
Unfortunately, things get rather messy at this juncture, so I had to ask for some assistance from Mathematica. A little bit of massaging of the results from Mathematica yielded
$$4\int \frac{3+\mathrm{cn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv=\frac23\left(6v-4\,\varepsilon\left(v \mid \frac14\right)+\frac{4\left(3+\mathrm{cn}\left(v\mid \frac14\right)\right)\mathrm{dn}\left(v\mid \frac14\right)\mathrm{sn}\left(v\mid \frac14\right)}{\left(1+\mathrm{cn}\left(v\mid \frac14\right)\right)^2}\right)$$
where $\varepsilon(v\mid m)$ is the Jacobi epsilon function.
One could of course use the appropriate formulae from Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Scientists if one insists on a wholly manual evaluation, but the evaluation involves a rather messy set of recurrence relations. (If I find more time, I'll eventually edit this answer to include the full solution, bloody guts and all...)
In any event, to now consider the definite integral, we undo the two substitutions in turn and apply those to the limits of the original integral. In particular, the inverse of $\phi=\mathrm{am}(v\mid m)$ is the incomplete elliptic integral of the first kind, $v=F(\phi\mid m)$, while the Jacobi epsilon function satisfies the relation
$\varepsilon(F(\phi\mid m)\mid m)=E(\phi\mid m)$
where $E(\phi\mid m)$ is the incomplete elliptic integral of the second kind. The transformed limits are then $v=0$ and $v=F\left(2\arctan\frac32\mid\frac14\right)$
We finally end up with
$$\int_0^3 \sqrt{t^4+4t^2+16} \,\mathrm dt=\frac{17}{13}\sqrt{133}+4\,F\left(2\arctan\frac32\mid\frac14\right)-\frac83 \,E\left(2\arctan\frac32\mid\frac14\right)$$
which is a fair bit simpler than what you get if you directly input the integral into Mathematica.
If it wasn't already apparent in the paragraphs above: elliptic integrals and elliptic functions are very different things; one is (very roughly) the inverse of the other. Too many people tend to conflate these two families of functions, though they are definitely related...
In addition to Byrd/Friedman, you'll also want to look into Greenhill's The Applications of Elliptic Functions, which, contrary to the title, also deals a bit with the elliptic integrals as well.
Best Answer
Using the substitution $x=\sin(u)$, you get your problem in the form $\int\!dx\, \frac{\sqrt{(1-mx^2)(1-x^2)}}{x^2 (1-x^2)}$.
Legendre studied the problem of solving integrals which involve rational functions of $x$ and the square root of a polynomial of degree less or equal four in $x$. You will find a recipe on the page http://everything2.com/title/elliptic+integral+standard+forms. Following the argument, noting that in your case $A=0, B=1, C=x^2-x^4, D=1, R=(1-mx^2)(1-x^2)$, you will obtain the result as a function of the three elliptic integrals.