Distribution of $x_4(x_1-x_3)+x_5(x_2-x_1)$ with iid $x_i \sim N(0,1)$ Problem
Given are 5 independent standard normal variables $x_1,x_2,x_3,x_4,x_5$.
What is the pdf of $x_4(x_1-x_3)+x_5(x_2-x_1)$ ?
What I know
$$x_1-x_3\sim \mathcal{N}\left(0,\sqrt{2}\right)\tag{1}$$
$$x_2-x_1\sim \mathcal{N}\left(0,\sqrt{2}\right)\tag{2}$$
$$x_4(x_1-x_3)\sim \frac{1}{\pi \sqrt{2}}K_0\left(\frac{\left|z\right|}{\sqrt{2}}\right)\tag{3}$$
$$x_5(x_2-x_1)\sim \frac{1}{\pi \sqrt{2}}K_0\left(\frac{\left|z\right|}{\sqrt{2}}\right)\tag{4}$$
where $K_0$ is the Bessel function and eq.(3,4) is a normal product distribution.
Remains to add the 2 summands $x_4(x_1-x_3)$ and $x_5(x_2-x_1)$ but they are dependent and cannot be convolved.
Related problem from literature
If two variables are transformed like $x_1(a x_1+b x_2)$ with $a\in \mathbb{R},b\in\mathbb{R}^+$ then the pdf is known $[1]$ but this is not applicable here.
Simulation
Simulation shows that the distribution of $x_4(x_1-x_3)+x_5(x_2-x_1)$ can be approximated with a Laplace distribution with parameter $(0,1.34)$. But the сorrect answer is not a Laplace distribution.

$[1]$ R. Gaunt: A note on the distribution of the product of zero-mean correlated normal random variables, Statistica Neerlandica, 2018
 A: Linear algebra shows
$$2(x_4(x_1-x_3)+x_5(x_2-x_1)) =(x_2-x_3+x_4+x_5)^2/4-(-x_2+x_3+x_4+x_5)^2/4+\sqrt{3}(-\sqrt{1/3}x_1+\sqrt{1/12}(x_2+x_3)+(1/2)(-x_4+x_5))^2-\sqrt{3}(-\sqrt{1/3}x_1+\sqrt{1/12}(x_2+x_3)+(1/2)(x_4-x_5))^2.$$
Each squared term is a linear combination of independent standard Normal variables scaled to have a variance of $1,$ whence each of those squares has a $\chi^2(1)$ distribution.  The four linear combinations are also orthogonal (as a quick check confirms), whence uncorrelated; and because they are uncorrelated joint random variables, they are independent.

Thus, the distribution is that of (a) half the difference of two iid $\chi^2(1)$ variables plus (b) $\sqrt{3}$ times half the difference of independent iid $\chi^2(1)$ variables.

(Differences of iid $\chi^2(1)$ variables have Laplace distributions, so this equivalently is the sum of two independent Laplace distributions of different variances.)
Because the characteristic function of a $\chi^2(1)$ variable is
$$\psi(t) = \frac{1}{\sqrt{1-2it}},$$
the characteristic function of this distribution is
$$\psi(t/2) \psi(-t/2) \psi(t\sqrt{3}/2) \psi(-t\sqrt{3}/2) = \left[(1+t^2)(1+3t^2)\right]^{-1/2}.$$
This is not the characteristic function of any Laplace variable -- nor is it recognizable as the c.f. of any standard statistical distribution.  I have been unable to find a closed form for its inverse Fourier transform, which would be proportional to the pdf.
Here is a plot of the formula (in red) superimposed on an estimate of $\psi$ based on a sample of 10,000 values (real part in black, imaginary part in gray dots):

The agreement is excellent.

Edit
There remain questions of what the PDF $f$ looks like.  It can be computed by numerically inverting the Fourier Transform by computing
$$f(x) = \frac{1}{2\pi}\int_{\mathbb R} e^{-i x t} \psi(t)\,\mathrm{d}t = \frac{1}{2\pi}\int_{\mathbb R} \frac{e^{-i x t}}{\sqrt{(1+t^2)(1+3t^2)}}\,\mathrm{d}t.$$
This expression, by the way, fully answers the original question.  The aim of the rest of this section is to show it is a practical answer.
Numerical integration will become problematic once $|x|$ exceeds $10$ or $15,$ but with a little patience can be accurately computed.
In light of the analysis of differences of Gamma variables at https://stats.stackexchange.com/a/72486/919, it is tempting to approximate the result by a mixture of the two Laplace distributions.  The best approximation near the middle of the distribution is approximately $0.4$ times Laplace$(1)$ plus $0.6$ times Laplace$(\sqrt{3}).$  However, the tails of this approximation are a little too heavy.

The left hand plot in this figure is a histogram of 100,000 realizations of $x_4(x_1-x_3) + x_5(x_2-x_1).$  On it are superimposed (in black) the numerical calculation of $f$ and then, in red, its mixture approximation.  The approximation is so good it coincides with $f.$  However, it's not perfect, as the related plot at right shows.  This plots $f$ and its approximation on a logarithmic scale.  The decreasing accuracy of the approximation in the tails is clear.
Here is an R function for computing values of a PDF that is specified by its characteristic function.  It will work for any numerically well-behaved CF (especially one that decays rapidly).
cf <- Vectorize(function(x, psi, lower=-Inf, upper=Inf, ...) {
  g <- function(y) Re(psi(y) * exp(-1i * x * y)) / (2 * pi)
  integrate(g, lower, upper, ...)$value
}, "x")

As an example of its use, here is how the black graphs in the figure were computed.
f <- function(t) ((1 + t^2) * (1 + 3*t^2)) ^ (-1/2)
x <- seq(0, 15), length.out=101)
y <- cf(x, f, rel.tol=1e-12, abs.tol=1e-14, stop.on.error=FALSE, subdivisions=2e3)

The graph is constructed by connecting all these $(x,y)$ values.
This calculation for $101$ values of $|x|$ between $0$ and $15$ takes about one second.  It is massively parallelizable.
For more accuracy, increase the subdivisions argument--but expect the computation time to increase proportionally.  (The figure used subdivisions=1e4.)
A: An almost exact expression could be derived using link between $\chi^{2}$, $\Gamma$ and symmetric $VG$ variance gamma distributions. Given very useful results above by @whuber, we can proceed first by the link $\chi^{2}(1)\sim\Gamma(\frac{1}{2},2)$. Therefore, from result that the product in question is distributed as $\frac{1}{2}[X_{1}-X_{2}]+\frac{\sqrt{3}}{2}[Y_{1}-Y_{2}]$ and knowing that $X_{1,2}\sim\chi^{2}(1)$, $Y_{1,2}\sim\chi^{2}(1)$ and $\chi^{2}(1)\sim\Gamma(\frac{1}{2},2)$. We will have,
$$\begin{equation}
\frac{1}{2}[X_{1}-X_{2}]+\frac{\sqrt{3}}{2}[Y_{1}-Y_{2}]\sim\left[\Gamma(\frac{1}{2},1)-\Gamma(\frac{1}{2},1)+\Gamma(\frac{1}{2},\sqrt{3})-\Gamma(\frac{1}{2},\sqrt{3})\right]
\end{equation}$$
which is the sum of difference of $\Gamma(\alpha,\nu)$ rvs. Now, symmmetric $VG$, which was introduced by Madan and Seneta (1990) is a distribution with $\sigma W\sqrt{\Gamma(\alpha_{G},\nu_{G})}$ where $W\sim N(0,1)$ and could be written as a difference of two $\Gamma$ distributed rvs. This has a characteristic function (CF)
$$\begin{equation}
\phi(u,\sigma,\nu)=\left(\frac{1}{1+\frac{\nu u^{2}\sigma^{2}}{2}}\right)^{\frac{1}{\nu}}
\end{equation}$$
Difference of $\Gamma(\alpha,\nu)$ rvs will have CF
$$\begin{equation}
\phi(u,\alpha,\nu_{G})=\left(\frac{1}{1+\nu_{G}^{2}}\right)^{\alpha}
\end{equation}$$
Equating these two will give us
$$\begin{equation}
\sigma=\nu_{G}\sqrt{2\alpha}\\
\nu=\frac{1}{\alpha}
\end{equation}$$
Therefore, $\frac{1}{2}[X_{1}-X_{2}]+\frac{\sqrt{3}}{2}[Y_{1}-Y_{2}] \sim VG(\sigma=1,\nu=2)+VG(\sigma=\sqrt{3},\nu=2)$. Here we have some result, but still we need the convolution of these two $VG$ random variables. There is following convolution rule for $VG$ distribution, $VG(\sigma_{1},\nu_{1})+VG(\sigma_{2},\nu_{2})\sim VG\left(\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}},\frac{\nu_{1}+\nu_{2}}{\nu_{1}\nu_{2}}\right)$, which has additional constraint $\sigma_{1}^{2}\nu_{1}=\sigma_{2}^{2}\nu_{2}$. Here, we slightly diverge from this additional condition and use the first one, which results an approx. exact density. Therefore, final result is  $\frac{1}{2}[X_{1}-X_{2}]+\frac{\sqrt{3}}{2}[Y_{1}-Y_{2}] \sim VG(\sigma=2, \nu=1)$.
$$\begin{equation}
f_{X}\left(x\right)=\frac{x}{4\sqrt{\pi}}K_{\frac{1}{2}}\left(4\sqrt{2}|x|\right),
\end{equation}$$
After we check this result with MC simulation, we see that the result is fairly successful to have a closed form density and MC simulation is caught almost perfectly by both pdf and cdf.  
Hope this will be useful.
