Let the three consecutive binomial coefficients in $(1+x)^{n+5}$ be $\binom{n+5}{r-1},\ \binom{n+5}{r},\ \binom{n+5}{r+1}$ for some integer $r$.

They are given to be in the ratio $5:10:14.$

So, $\binom{n+5}{r-1}:\binom{n+5}{r}:\binom{n+5}{r+1}=5:10:14.$

Use ratios of consecutive binomial coefficients: $\frac{\binom{N}{r}}{\binom{N}{r-1}}=\frac{N-r+1}{r}, \qquad \frac{\binom{N}{r+1}}{\binom{N}{r}}=\frac{N-r}{r+1}$ where $N=n+5$.

From the given ratio, $\frac{\binom{N}{r}}{\binom{N}{r-1}}=\frac{10}{5}=2$ so $\frac{N-r+1}{r}=2$ $N-r+1=2r$ $N+1=3r$ $r=\frac{N+1}{3}.$

Also, $\frac{\binom{N}{r+1}}{\binom{N}{r}}=\frac{14}{10}=\frac{7}{5}$ so $\frac{N-r}{r+1}=\frac{7}{5}$ $5(N-r)=7(r+1)$ $5N-5r=7r+7$ $5N-7=12r$ $r=\frac{5N-7}{12}.$

Equate the two expressions for $r$: $\frac{N+1}{3}=\frac{5N-7}{12}$ $4(N+1)=5N-7$ $4N+4=5N-7$ $N=11.$

Since $N=n+5$, $n+5=11$ $n=6.$

Therefore, the required integer is $\boxed{6}.$