Rescaling the integrand to $\sqrt{1+x^4}$ is a simple matter so let's start there. First, some trickery with integration by parts can reduce the desired integral to one closer to the form of an elliptic integral of the first kind:
$$\begin{align}
\int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x
&=\int_{0}^{a}\frac{1+x^4}{\sqrt{1+x^4}}\,\mathrm{d}x\\
&=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\int_{0}^{a}\frac{x^4}{\sqrt{1+x^4}}\,\mathrm{d}x\\
&=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\int_{0}^{a}x\cdot\frac{x^3}{\sqrt{1+x^4}}\,\mathrm{d}x\\
&=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\left[\frac12x\sqrt{1+x^4}\right]_{0}^{a}-\frac12\int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x\\
&=\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}+\frac12a\sqrt{1+a^4}-\frac12\int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x\\
\implies \frac32\int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x
&=\frac{a}{2}\sqrt{1+a^4}+\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}\\
\implies \int_{0}^{a}\sqrt{1+x^4}\,\mathrm{d}x
&=\frac{a}{3}\sqrt{1+a^4}+\frac23\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}.\\
\end{align}$$
Focusing now on the integral $\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}$, applying a Landen transformation of the form $x=\frac{1-y}{1+y}$ yields,
$$\begin{align}
\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}
&=\int_{1}^{\frac{1-a}{1+a}}\frac{(1+y)^2}{\sqrt{2(1+6y^2+y^4)}}\cdot\frac{(-2)\,\mathrm{d}y}{(1+y)^2}\\
&=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{1+6y^2+y^4}}\\
&=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{\left(3+2\sqrt{2}+y^2\right)\left(3-2\sqrt{2}+y^2\right)}}.\\
\end{align}$$
Note that the constant terms in the last line above have the useful properties
$$(3+2\sqrt{2})^{-1}=3-2\sqrt{2};\\
\sqrt{3+2\sqrt{2}}=1+\sqrt{2}.$$
Scaling the integral by substituting $(\sqrt{2}+1)y=t$,
$$\begin{align}
\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}
&=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{\left[(1+\sqrt{2})^2+y^2\right]\left[(1-\sqrt{2})^2+y^2\right]}}
\\
&=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{(\sqrt{2}+1)(\sqrt{2}-1)\sqrt{\left[1+(\sqrt{2}-1)^2y^2\right]\left[1+(\sqrt{2}+1)^2y^2\right]}}
\\
&=\sqrt{2}\int_{\frac{1-a}{1+a}}^{1}\frac{\mathrm{d}y}{\sqrt{\left[1+(\sqrt{2}-1)^2y^2\right]\left[1+(\sqrt{2}+1)^2y^2\right]}}
\\
&=(2-\sqrt{2})\int_{(\sqrt{2}+1)\frac{1-a}{1+a}}^{\sqrt{2}+1}\frac{\mathrm{d}t}{\sqrt{\left[1+(\sqrt{2}-1)^4t^2\right]\left(1+t^2\right)}}
\\
&=(2-\sqrt{2})\int_{(\sqrt{2}+1)\frac{1-a}{1+a}}^{\sqrt{2}+1}\frac{\mathrm{d}t}{\sqrt{1+t^2}\sqrt{1+(\sqrt{2}-1)^4t^2}}.
\\
\end{align}$$
Now it's time for trigonometric substitution. For compactness of notation, write $(\sqrt{2}-1)^2=b$. Using $t=\tan{\theta}$,
$$\begin{align}
\int_{0}^{a}\frac{\mathrm{d}x}{\sqrt{1+x^4}}
&=(2-\sqrt{2})\int_{(\sqrt{2}+1)\frac{1-a}{1+a}}^{\sqrt{2}+1}\frac{\mathrm{d}t}{\sqrt{1+t^2}\sqrt{1+b^2t^2}}
\\
&=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\sec^2{\theta}\,\mathrm{d}\theta}{\sqrt{\sec^2{\theta}}\sqrt{1+b^2\tan^2}{\theta}}
\\
&=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\sec^2{\theta}\,\mathrm{d}\theta}{\sqrt{\sec^4{\theta}}\sqrt{\cos^2{\theta}+b^2\sin^2}{\theta}}
\\
&=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\mathrm{d}\theta}{\sqrt{\cos^2{\theta}+b^2\sin^2}{\theta}}
\\
&=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\mathrm{d}\theta}{\sqrt{1-(1-b^2)\sin^2}{\theta}}
\\
&=(2-\sqrt{2})\int_{\tan^{-1}{\left[(\sqrt{2}+1)\frac{1-a}{1+a}\right]}}^{\frac{3\pi}{8}}\frac{\mathrm{d}\theta}{\sqrt{1-b^{\prime\,2}\sin^2}{\theta}}.
\\
\end{align}$$
And presto chango, an elliptic integral of kind numero uno! I presume I can safely leave the remaining details to you, but let me know if should make anything clearer.
I completely agree with Paras Khosla's comment. Further, quoting "Calculus" volume 1, 2nd Edition, 1966, page 266 (by Tom Apostol), integrals of the form $\;\int\sqrt{(cx+d)^2 - a^2}\,dx\;$ should be attacked via the substitution $\;cx + d = a \sec t.$
I am (superficially redundantly) answering because on the one hand you showed a good effort but on the other hand (apparently through no fault of your own), you have made a serious workflow mistake. This is not the type of problem whose solution a Calculus student should be attempting to derive from scratch. If you are in a Calculus class, then your class materials (e.g. textbook) should have explicitly provided the information in this answer's first paragraph.
Do not try to attack problems like this on your own. Instead, buy a moderately priced Calculus book. To determine which book to buy, ask your teacher (if available) or heavily research user comments (e.g. Amazon.com's customer reviews).
Buying the right math book can be tricky; it needs to be customized to your experience, goals, and budget. Generically, try to buy one with a lot of exercises, don't be in a hurry, don't skip any exercises, and (for the exercises where you are having trouble) post a query on a math forum like this one (showing heavy preliminary effort, just as you did with this query).
I am upvoting because of the good preliminary (though misguided) effort that you made. Note that the whole issue of when to go for it, as you did is tricky. Math students need to look for a balance between making a reasonable preliminary effort and never trying to re-invent the wheel.
Best Answer
What you should end with is
$\sec\theta = \dfrac x3\quad$ and $\quad\tan \theta = \dfrac{\sqrt{ x^2 - 9}}{3}$.
Then you have $$\begin{align} \log \Big| \frac 13\left(x + \sqrt{x^2 - 9}\right)\Big| + C & = \log|x +\sqrt{x^2 - 9}| -\log 3 + C \\ \\ & = \log|x + \sqrt{x^2 - 9}| + C'\end{align}$$