= Solution
Take $\mathbf r=\mathbf x_1-\mathbf x_2$, $r=|\mathbf r|$ and $M=M_1+M_2$. <Newton's law of universal gravitation> gives $\ddot{\mathbf x}_1=-GM_2\mathbf r/r^3$ and $\ddot{\mathbf x}_2=GM_1\mathbf r/r^3$. Subtracting yields the \b[relative two-body equation]
$$
\boxed{\ddot{\mathbf r}=-\frac{GM}{r^3}\mathbf r.}
$$
The <reduced mass> is $\mu=M_1M_2/M$. The orbital energy and <angular momentum> in the <center of mass> frame are
$$
E=\frac\mu2|\dot{\mathbf r}|^2-\frac{GM\mu}{r},\qquad
\mathbf J=\mu\mathbf r\times\dot{\mathbf r}.
$$
Differentiation gives
$$
\dot E=\mu\dot{\mathbf r}\cdot\ddot{\mathbf r}+\frac{GM\mu}{r^3}\mathbf r\cdot\dot{\mathbf r}=0,
\qquad
\dot{\mathbf J}=\mu\mathbf r\times\ddot{\mathbf r}=0.
$$
Thus both are conserved. The <eccentricity vector> is
$$
\mathbf e=\frac{\dot{\mathbf r}\times(\mathbf r\times\dot{\mathbf r})}{GM}-\frac{\mathbf r}{r}.
$$
For $\mathbf h=\mathbf J/\mu$, one has $\ddot{\mathbf r}\times\mathbf h/(GM)=\dot{\mathbf r}/r-\mathbf r(\mathbf r\cdot\dot{\mathbf r})/r^3=d(\mathbf r/r)/dt$, so $\dot{\mathbf e}=0$. In particular the \b[<orbital eccentricity> is constant], with
$$
\boxed{e^2=1+\frac{2EJ^2}{G^2M^2\mu^3}.}
$$
For a bound <Kepler orbit>, $E=-GM\mu/(2a)$ and $J^2=GM\mu^2a(1-e^2)$.
For the far-field <multipole expansion>, set the <center of mass> to zero. Then $\mathbf x_1=(M_2/M)\mathbf r$ and $\mathbf x_2=-(M_1/M)\mathbf r$. At $x=|\mathbf x|\gg r$, expand each point-mass <Newtonian gravitational potential>:
$$
\frac1{|\mathbf x-\mathbf x_A|}=\frac1x+\frac{x_i x_{Ai}}{x^3}
+\frac12x_{Ai}x_{Aj}\frac{3x_ix_j-x^2\delta_{ij}}{x^5}+O(r^3/x^4).
$$
Repeated indices are summed. The dipole vanishes since $\sum_A M_A\mathbf x_A=0$, and $\sum_A M_Ax_{Ai}x_{Aj}=\mu r_ir_j$. The second-order <gravitational quadrupole potential of a point-mass binary> is therefore
$$
\phi(\mathbf x)=-\frac{GM}{x}-\frac{G\mu}{2x^5}\left[3(\mathbf x\cdot\mathbf r)^2-x^2r^2\right]+O(GMr^3/x^4).
$$
Both tensors specified in the question are trace-free. Direct contraction gives
$$
q_{ij}l_{ij}=\frac{\mu}{2x^5}\left[3(\mathbf x\cdot\mathbf r)^2-x^2r^2\right],
$$
so \b[to the requested order]
$$
\boxed{\phi(\mathbf x)=-\frac{GM}{x}-Gq_{ij}l_{ij}(\mathbf x).}
$$
For equal masses the cubic multipole also vanishes, but it need not vanish for unequal masses.
To evaluate the <quadrupole formula>, use $k=GM$, $\mathbf v=\dot{\mathbf r}$, $\mathbf n=\mathbf r/r$, $u=\dot r=\mathbf n\cdot\mathbf v$, and $\mathbf w=\mathbf v-u\mathbf n$. The acceleration and its derivative are
$$
\mathbf b=-\frac{k\mathbf n}{r^2},\qquad
\dot{\mathbf b}=-\frac{k\mathbf v}{r^3}+\frac{3ku\mathbf n}{r^3}.
$$
The third derivative of $r_ir_j$ is $\dot b_i r_j+r_i\dot b_j+3b_i v_j+3v_i b_j$. Its trace is $d^3(r^2)/dt^3=-2ku/r^2$. Substituting in $q_{ij}=\mu(3r_ir_j-r^2\delta_{ij})/2$ gives
$$
\frac{d^3q_{ij}}{dt^3}=\frac{\mu k}{r^2}\left[u(\delta_{ij}-3n_in_j)-6(n_iw_j+w_in_j)\right].
$$
Because $\mathbf n\cdot\mathbf w=0$, the cross <tensor contraction> is zero. The squared norms of the two bracketed pieces are $6u^2$ and $72w^2$, respectively. Thus
$$
\dddot q_{ij}\dddot q_{ij}=\frac{72\mu^2k^2}{r^4}\left(v^2-\frac{11}{12}u^2\right),
$$
and the \b[instantaneous energy loss] is
$$
\boxed{\dot E=-\frac{32G^3M^2\mu^2}{5c^5r^4}\left(\dot{\mathbf r}\cdot\dot{\mathbf r}-\frac{11}{12}\dot r^2\right).}
$$
The parenthesis is $w^2+u^2/12\geq0$, as required for positive radiated power. This is the <instantaneous quadrupole luminosity of a Kepler binary> with loss sign for the orbital energy.
A slowly evolving initially circular binary is expected to follow a quasi-circular inspiral at leading radiation order. One needs angular-momentum loss as well as energy loss for this conclusion: the rotating quadrupole has frequency $2\Omega$ and azimuthal harmonic number two, so its fluxes satisfy the <gravitational-wave energy and angular-momentum balance> $\dot E=\Omega\dot J$. Along the circular family,
$$
E=-\frac{G^2M^2\mu^3}{2J^2},\qquad\frac{dE}{dJ}=\Omega.
$$
The flux relation is precisely the tangent condition for remaining on that family; equivalently it keeps $e^2=1+2EJ^2/(G^2M^2\mu^3)$ equal to zero at leading secular order. The shrinking orbit is not an exact fixed-radius circle: its small radial drift is neglected at this adiabatic order.
For a circular orbit of instantaneous separation $a$, $u=0$ and $v^2=GM/a$. Therefore
$$
\dot E=-\frac{32G^4M^3\mu^2}{5c^5a^5}.
$$
Differentiating $E=-GM\mu/(2a)$ at fixed masses now gives
$$
\boxed{\frac{\dot a}{a}=-\frac{64G^3M^2\mu}{5c^5a^4}.}
$$
Writing $C=64G^3M^2\mu/(5c^5)$, the <circular gravitational-wave inspiral> obeys $a^4=a_0^4-4Ct$. The formal point-mass coalescence time is $a_0^4/(4C)$, and $\dot P/P=(3/2)\dot a/a$ gives a decreasing <orbital period> and increasing <gravitational-wave frequency>.
\b[Gravitational radiation tightens a detached close binary and becomes more effective at small separation.] It can bring a stellar donor into <Roche-lobe overflow> or drive compact objects toward merger. Finite radii, tides and mass transfer can then alter the evolution; the fixed-mass inspiral law is not the entire evolution law once these matter. Near merger the weak-field, slow-motion <quadrupole formula> also ceases to be sufficient.
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