Bash count number of numbers

Could you please try following(tested and written with shown samples), this should include negative numbers too, since OP has mentioned about integers so I have not considered counting floats. It gives output as 3 3 3 for mentioned 3 lines of samples.

awk '
{
  for(i=1;i<=NF;i++){
    if(length(int($i))==length($i)){ count++ }
  }
  print count
  count=""
}' Input_file

Another awk:

$ awk '{for(i=1;i<=NF;i++)if($i+0||$i==0)c++;print c;c=0}' file

Output:

3
3
3

Edit: After my morning coffee I noticed pure integer strings so beware, this one counts decimal numbers also.

Bash count number of numbers

echo $string | grep -Po '(^|^-| | -)[0-9]+((?= )|$)' | wc -w

wc -w will count the number of space separated sub-strings in a given input
We can use grep -Po (-P for Perl regex, -o output ONLY matching) to match pure number sub-strings

We can use Regex to define a pattern of sub-strings we want to keep:

1. (^|^-| | -) Starts with one of the four possible starting conditions
2. [0-9]+ Contains only one or more integer characters
3. ((?= )|$) Ends with a space or end of line character,(?= ) looks ahead for a space character but doesn't match, allowing overlap