The following shows how to transform the quartic
\begin{equation}
y^2=ax^4+bx^3+cx^2+dx+e ,
\end{equation}
with $a,b,c,d,e \in \mathbb{Q}$, into an equivalent elliptic curve.
Case 1:
We first consider the case when $a=1$, which is very common. If $a=\alpha^2$ for some rational $\alpha$, we
substitute $y=Y/\alpha$ and $x=X/\alpha$, giving
\begin{equation*}
Y^2=X^4+\frac{b}{\alpha}X^3+cX^2+\alpha d X+\alpha^2 e
\end{equation*} Thus, suppose
\begin{equation}
y^2=x^4+bx^3+cx^2+dx+e
\end{equation}
We describe the method given by Mordell on page $77$ of his book Diophantine Equations, with some minor modifications.
We first get rid of the cubic term by making the standard substitution $x=z-b/4$ giving
\begin{equation*}
y^2=z^4+fz^2+gz+h
\end{equation*}
where
\begin{equation*}
f=\frac{8c-3b^2}{8} \hspace{1cm} g=\frac{b^3-4bc+8d}{8} \hspace{1cm}
h=\frac{-(3b^4-16b^2c+64bd-256e)}{256}
\end{equation*}
We now get rid off the quartic term by defining $y=z^2+u+k$, where $u$ is a new variable and $k$ is a constant to be
determined. This gives the quadratic in $z$
\begin{equation*}
(f-2(k+u))z^2+gz+h-k^2-u(2k+u)=0
\end{equation*}
For $x$ to be rational, then $z$ should be rational, so the discriminant of this quadratic would have to be a rational square.
The discriminant is a cubic in $u$. We do not get a term in $u^2$ if we make $k=f/6$, giving
\begin{equation*}
D^2=-8u^3+2\frac{f^2+12h}{3}u+\frac{2f^3-72fh+27g^2}{27}
\end{equation*}
and, if we substitute the formulae for $f,g,h$, and clear denominators we have
\begin{equation}
G^2=H^3+27(3bd-c^2-12e)H+27(27b^2e-9bcd+2c^3-72ce+27d^2)
\end{equation}
with
\begin{equation}
x=\frac{2G-3bH+9(bc-6d)}{12H-9(3b^2-8c)}
\end{equation}
and
\begin{equation}
y= \pm \frac{18x^2+9bx+3c-H}{18}
\end{equation}
Case 2:
If $a \ne \alpha^2$, we need a rational point $(p,q)$ lying on the quartic.
Let $z=1/(x-p)$, so that $x=p+1/z$ giving
\begin{equation*}
y^2z^4=(ap^4+bp^3+cp^2+dp+e)z^4+(4ap^3+3bp^2+2cp+d)z^3+
\end{equation*}
\begin{equation*}
(6ap^2+3bp+c)z^2+(4ap+b)z+a
\end{equation*}
where the coefficient of $z^4$ is, in fact, $q^2$. Define, $y=w/z^2$, and then
$z=u/q^2, \, w=v/q^3$
giving
\begin{equation}
v^2=u^4+(4ap^3+3bp^2+2cp+d)u^3+q^2(6ap^2+3bp+c)u^2+
\end{equation}
\begin{equation*}
q^4(4ap+b)u+aq^6 \equiv u^4+fu^3+gu^2+hu+k
\end{equation*}
We now, essentially, complete the square. We can write
\begin{equation*}
y^2=u^4+fu^3+gu^2+hu+k=(u^2+mu+n)^2+(su+t)
\end{equation*}
if we set
\begin{equation*}
m=\frac{f}{2} \hspace{1cm} n=\frac{4g-f^2}{8} \hspace{1cm} s=\frac{f^3-4fg+8h}{8}
\end{equation*}
and
\begin{equation*}
t=\frac{-(f^4-8f^2g+16(g^2-4k))}{64}
\end{equation*}
This gives
\begin{equation*}
(y+u^2+mu+n)(y-u^2-mu-n)=su+t
\end{equation*}
and if we define $y+u^2+mu+n=Z$ we have
\begin{equation}
2(u^2+mu+n)=Z-\frac{su+t}{Z}
\end{equation}
Multiply both sides by $Z^2$, giving
\begin{equation*}
2u^2Z^2+2muZ^2+suZ=Z^3-2nZ^2-tZ
\end{equation*}
which, on defining $W=uZ$, gives
\begin{equation*}
2W^2+2mWZ+sW=Z^3-2nZ^2-tZ
\end{equation*}
Define, $Z=X/2$ and $W=Y/4$ giving
\begin{equation}
Y^2+2mXY+2sY=X^3-4nX^2-4tX
\end{equation}
which is an elliptic curve. If we define $Y=G-s-mX$, we transform to the form
\begin{equation}
G^2=X^3+(m^2-4n)X^2+(2ms-4t)X+s^2
\end{equation}
All of the above are easy to program in any symbolic package. Hopefully, I have all the formulae correct.
