We'll give a closed-form to your quintic root. As Klein showed in his "Lectures on the Icosahedron", there is indeed such a thing though it goes beyond radicals.
Two special functions may be used: the Dedekind eta function $\eta(\tau)$ and the hypergeometric function $_2F_1$. The Bring quintic form will be avoided since that involves $2$nd-deg and $4$th-deg transformations, and to undo those will result in a mess.
Fortunately, there are other solutions. And the nice thing about the Klein-Brioschi result is one just uses nested square roots (unlike the Bring form which needs cube roots).
I. Principal quintic
First reduce the quintic to principal form, and yours is special in that its Tschirnhausen transformation needs only rational coefficients. Given,
$$x^5+3x^4+4x^3+x-1=0$$
Let $(x+1)^2-y=0$. Eliminating $x$ between the two using resultants in WA yields the principal quintic form,
$$y^5 - 8y^2 + 4y - 16=0$$
which has discriminant $d = 2^{23}\times47.$ The prime $47$ will appear in the second method.
II. First Method
We'll use Tyma Gaidash's method which is simpler than Klein's. Given,
$$y^5 - 8y^2 + 4y - 16=0$$
or equivalently,
$$\left(\tfrac{y}{2}\right)^5-\left(\tfrac{y}{2}-\tfrac{1+\sqrt{-31}}8\right)\left(\tfrac{y}{2}-\tfrac{1-\sqrt{-31}}8\right)=0$$
then the five roots are,
$$y_k=-2\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}5 \frac{(q)^\frac n5}{(p)^{\frac{4n}5-1}} \frac{\left(\frac n5\right)!}{n!\,\Gamma\big(2-\frac{4n}5\big)}\;{_2 F_1}\left(1-n,-\frac n5;2-\frac{4n}5;\frac pq\right)\qquad$$
for $k = 0,1,2,3,4,$ and where $p, q = \frac{1\pm\sqrt{-31}}8,$ with gamma function $\Gamma(n).$ Reversing the Tshirnhausen transformation, then the real root is,
$$x_2 = -1+\sqrt{y_2} \; \approx \; 0.45913372331$$
though one has to choose the correct sign of the square root.
III. Second Method
We can also use the Klein-Brioschi method. The real root is,
$$\quad x_2 = -1+\sqrt{\frac{pm+q}{rm^2-3}}\approx 0.45913372331$$
where,
$$m=-\frac{u^2-5}{w^3}\sqrt{\frac{u^2+2u+5}{-r\;}}$$
$$u =\frac{\eta(\tau/5)}{\eta(5\tau)},\quad w =\frac{\eta(\tau)}{\eta(5\tau)}$$
$$\tau = 2+\frac{_2F_1\big(\frac16,\frac56,1,1-\alpha\big)}{_2F_1\big(\frac16,\frac56,1,\alpha\big)}\sqrt{-1}$$
$$\alpha = \tfrac12\left(1-\sqrt{\frac{r}{r-1728}}\right)$$
and $(p,q,r)$ are the constants,
\begin{align}
\qquad p &= \frac{128\;}{259^3} (34323438 + 645563 \sqrt{470})\\[4pt]
q &= \frac{4}{259} (431 + 32 \sqrt{470})\\[4pt]
r &= \frac{320 (583125 + 30692 \sqrt{470})^3}{259^5 (2263 + 156 \sqrt{470})}+1728\\ \end{align}
where the formulas ($p = e,\, q = \lambda,\, r= 1/f$) in this Klein-Brioschi post.
P.S. In short, just choose which of the two methods you prefer.
