= Solution
A <stellar polytrope> has a spatially constant coefficient $K$ in the relation $P=K\rho^{1+1/n}$; $n$ is its <polytropic index>. This relates <pressure> to <mass density> and need not describe an adiabatic gas. Combining <hydrostatic equilibrium> with <mass conservation> eliminates the enclosed <mass>:
$$
\frac{1}{r^2}\frac{d}{dr}\left(\frac{r^2}{\rho}\frac{dP}{dr}\right)=-4\pi G\rho.
$$
For $\rho=\rho_c\theta^n$, the <polytropic equation of state> gives $(1/\rho)dP/dr=(n+1)K\rho_c^{1/n}d\theta/dr$. Choose the <Lane-Emden variables for a stellar polytrope> with
$$
\alpha^2=\frac{(n+1)K}{4\pi G}\rho_c^{1/n-1},\qquad r=\alpha\xi.
$$
Substitution then gives the dimensionless <Lane-Emden equation>:
$$
\boxed{\frac{1}{\xi^2}(\xi^2\theta')'=-\theta^n.}
$$
A regular stellar centre has $\theta(0)=1$, $\theta'(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta=1-\xi^2/6+O(\xi^4)$. For an isolated finite-radius <stellar polytrope> with negligible external <pressure>, the surface is the first zero $\xi_1$ of $\theta$, and $R=\alpha\xi_1$. A core embedded in an envelope can instead end at a positive matching <pressure>, before this zero.
The integrated <Lane-Emden equation> is $\xi^2\theta'=-\int_0^\xi s^2\theta(s)^n ds$. Thus <mass conservation> gives the <Lane-Emden mass formula> directly:
$$
\boxed{m(\xi)=-4\pi\rho_c\alpha^3\xi^2\theta'(\xi).}
$$
For the proposed functional form, differentiation gives
$$
\theta''+\frac{2}{\xi}\theta'=2\beta\gamma(1+\beta\xi^2)^{\gamma-2}\bigl[3+(2\gamma+1)\beta\xi^2\bigr].
$$
Choosing $\gamma=-1/2$ removes the extra polynomial factor; equality to $-\theta^5$ then requires $3\beta=1$. The regular <polytrope of index five> is therefore
$$
\boxed{\beta=\frac13,\quad\gamma=-\frac12,\quad\theta=(1+\xi^2/3)^{-1/2}.}
$$
Its enclosed <mass> and enclosed mean <mass density> are
$$
m(\xi)=\frac{4\pi\rho_c\alpha^3\xi^3}{3(1+\xi^2/3)^{3/2}},\qquad
\overline\rho(\xi)=\frac{3m}{4\pi r^3}=\rho_c(1+\xi^2/3)^{-3/2}.
$$
There is no finite zero: its isolated radius is infinite. Nevertheless its total <mass> converges, since the outer <mass density> falls as $r^{-5}$:
$$
\boxed{M=4\pi\sqrt3\rho_c\alpha^3,\qquad R=\infty,\qquad\lim_{r\to\infty}\overline\rho(r)=0.}
$$
The zero mean <mass density> is a limiting volume average, not a statement that the material has zero <mass density> at finite radius.
To turn this density model into a <CNO cycle> burning model, use gas-dominated <pressure> and uniform <mean molecular weight>. The <ideal gas law> and the <polytropic equation of state> then imply $T/T_c=\theta$. Without this thermal assumption a <polytropic equation of state> for total <pressure> does not alone fix the <temperature> profile. Since the <stellar energy-generation rate> is power per unit <mass>, its contribution to <luminosity> is $\epsilon\,dm$, giving the <luminosity integral of an index-five polytrope>:
$$
L=4\pi\epsilon_0\alpha^3\rho_c^2T_c^{11}\int_0^\infty\xi^2(1+\xi^2/3)^{-21/2}\,d\xi.
$$
Put $u=\xi/\sqrt{3+\xi^2}$. Then $\xi=\sqrt3u/\sqrt{1-u^2}$ and the dimensionless integral becomes $3\sqrt3\int_0^1u^2(1-u^2)^8du$. Evaluation gives
$$
A=12\pi\sqrt3\frac{8!\,9!\,2^{17}}{19!}=1.029415\ldots,\qquad
\boxed{L=A\epsilon_0\alpha^3\rho_c^2T_c^{11},\quad A\sim1.}
$$
The PDF's power of $\alpha$ in the proposed <luminosity> scaling is a genuine dimensional error: it must be three, rather than two. The volume element supplies $\alpha^3$; replacing it by $\alpha^2$ would give power divided by length, and could not be repaired by a dimensionless numerical constant.
The <polytrope of index five> is unlikely to model a real burning <stellar core> well. An embedded core has a finite boundary, and a complete isolated star cannot have this infinite radius. More importantly, strongly <temperature>-sensitive <CNO cycle> heating concentrates the <luminosity> centrally and tends to produce a <convection zone> there. A well-mixed gas-pressure-dominated convective core is closer to an <adiabatic stellar polytrope> of index $3/2$ than to index five. A finite truncation would also change the integral defining $A$; its value above pertains to the full mathematical model.
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