Bounding an elliptic-type integral

In general one has the following bound on your integral $$ \frac{2}{3}L^{3/2}+ \frac{L^{1/2}}{\sqrt{2}}\ln \left(\frac{1+2K}{2+2(K-L)}\right)\leq \mathrm{integral}\leq \frac{2}{3}L^{3/2}+ L^{1/2} \ln\left(\frac{3+4K}{1+4(K-L)}\right) $$

Let us make change of variables $x=Ly$. Then we want to estimate from above $$ L^{3/2}\int_{0}^{1}\sqrt{y \left(1+ \frac{1}{K-Ly}\right)} - \sqrt{y}\; dy $$ We have $$ \int_{0}^{1}\sqrt{y}\left(\frac{\sqrt{K-Ly+1}-\sqrt{K-Ly}}{\sqrt{K-Ly}} \right)\; dy = \frac{1}{K}\times \int_{0}^{1}\sqrt{y}\left(\frac{1}{\sqrt{1-(L/K)y}(\sqrt{1-(L/K)y+(1/K)}+\sqrt{1-(L/K)y})} \right)\; dy \leq \frac{1}{K} \times \int_{0}^{1}\left(\frac{1}{\sqrt{1-(L/K)y}\sqrt{1-(L/K)y+(1/K)}} \right)\; dy = \frac{1}{L}\ln\left( \frac{1+2K+2\sqrt{K}\sqrt{K+1}}{1+2(K-L)+2\sqrt{K-L}\sqrt{K-L+1}}\right)\leq \frac{1}{L}\ln\left(\frac{3+4K}{1+4(K-L)}\right) $$

Update (lower bound): $$ \int_{0}^{1}\sqrt{y}\left(\frac{1}{\sqrt{1-(L/K)y}(\sqrt{1-(L/K)y+(1/K)}+\sqrt{1-(L/K)y})} \right)\; dy \geq \frac{1}{\sqrt{2}}\int_{1/2}^{1}\left(\frac{1}{\sqrt{1-(L/K)y}\sqrt{1-(L/K)y+(1/K)}} \right)\; dy = \frac{1}{\sqrt{2}}\frac{K}{L}\int_{1-\frac{L}{K}}^{1-\frac{L}{2K}} \frac{1}{\sqrt{z}\sqrt{z+\frac{1}{K}}}dz $$ Now let us use the fact that $$ \int \frac{1}{\sqrt{z^{2}+\frac{z}{K}}} = \ln \left(1+ 2Kz +2\sqrt{Kz}\sqrt{Kz+1}\right) - \ln K, \quad z>0 $$ which can be checked by direct differentiation (but I will skip it), then we can continuo our chain of inequalities $$ =\frac{1}{\sqrt{2}}\frac{K}{L}\,\ln \frac{1+2(K-\frac{L}{2}) +2\sqrt{K-\frac{L}{2}}\sqrt{K-\frac{L}{2}+1}}{1+2(K-L) +2\sqrt{K-L}\sqrt{K-L+1}}\geq \frac{1}{\sqrt{2}}\frac{K}{L}\,\ln \frac{1+4(K-\frac{L}{2})}{2+2(K-L)}\geq \frac{1}{\sqrt{2}}\frac{K}{L}\,\ln \frac{1+2K}{2+2(K-L)} $$

which is pretty much the same.