How about this one... $f'(x) = ((x-5)/(x+1))^2$. So (according to Maple) $$ f(x) = -\frac{-37x - x^{2} + 12 \ln (x + 1) + 12 \ln (x + 1) x}{x + 1} $$ Now $f$ is not concave, since $f'$ is increasing above $5$.
EDIT Next attempt ... $$ g(x) = \frac{x \bigl(53140 x^{3} + 212550 x^{2} + 314500 x + 240625 + x^{4}\bigr)}{(x + 1)^{4}} $$ so that $$ g'(x) = \frac{(x + 25) \bigl(x^{2} - 10 x + 385\bigr) (x - 5)^{2}}{(x + 1)^{5}} $$ and $$ g''(x) = \frac{-25920(x - 5) (x - 10)}{(x + 1)^{6}} $$
