Non-crossing X's and crossing non-X's

Question

You should write a program or function which given a text as input outputs or returns the sum of the following three integers:

• Number of lowercase x's which are not part of a cross-formation.
• Number of uppercase X's which are not part of a cross-formation.
• Number of cross-formations which are made of neither x nor X.

A cross-formation is 5 copy of the same character positioned to each other in the following manner:

  A A
   A
  A A

Cross-formations can overlap with each other.

Detailed example

Input:
x-x+x+X++
-x-x+x++asd
x-x+x+X+

Info:
There is 1 x and 2 X's not in cross-formation and 
there are 2 cross-formations of +'s as shown here:
   + +X+ 
    +x+    
   + +X+

Output:
5

Input

• A string containing only printable ascii characters (codepoint 32-126) and newlines.
• The string will contain at least one printable character.

Output

• A single positive integer, the sum of the number of occurrences for the three property.

Examples

After each output the values of the 3 properties (x's, X's and crosses) are provided in parens but note that those are not part of the output.

Input:
a
Output:
0 (0,0,0)

Input:
x xxx
 X x
x xxx
Output:
5 (4,1,0)

Input:
########
#####
########
Output:
4 (0,0,4)

Input:
a-a-a
0a0a
a-a-axXxX
0a0a
a-a-a
Output:
9 (2,2,5)

Input:
a c efghi
ab defgh
a c e ghij
abcd fgh
abc e g
Output:
2 (0,0,2)

Input:
/\/\/\/\/\/\/\
\/\/\/\/\/\/\/
#####x-x+x+X++
#####-x-x+x++###XX
#####x-x+x+X+
/\/\/\/\/\/\/\
\/\/\/\/\/\/\/
Output:
10 (1,4,5)

Input:
xxxxxxxxxx   
xxxxxxxXX
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxx
Output:
5 (3,2,0)

This is code-golf so the shortest entry wins.

Answer 1

JavaScript (ES6) 234 246

Longer than I expected ...

Run snippet to test (in Firefox)

// Golfed

F=t=>t.split('\n').map((r,j,t)=>(
  [...r].map((c,i)=>(
    f=c==r[i+2]&c==u[i+1]&c==v[i]&c==v[i+2],
    c!='x'&c!='X'?z+=f
    :(++z,f?k[j+[,i]]=k[2+j+[,i]]=k[1+j+[,i+1]]=k[j+[,i+2]]=k[2+j+[,i+2]]=1:0)
  )),v=u,u=r),
k={},u=v=z=0)&&z-Object.keys(k).length

// Ungolfed 

U=t=>{
  var k={}, u=0,v=0,z=0, t=t.split('\n')
  t.forEach((r,j)=> {
    [...r].forEach((c,i)=> {
      x = c=='x'|c=='X' // is any X
      f = c==r[i+2] & c==u[i+1] & c==v[i] & c==v[i+2] // is in a formation
      if (x) 
      {
        ++z
        if (f)
        { // remember the position of X involved in a formation, avoid counting duplicates more tha once
          k[j+[,i]]=k[2+j+[,i]]=k[1+j+[,i+1]]=k[j+[,i+2]]=k[2+j+[,i+2]]=1 
        }
      }
      else
      {
        z+=f 
      }                                                  
    })
    v=u,u=r
  })
  return z-Object.keys(k).length
}

// TEST

Out=s=>OUT.innerHTML=OUT.innerHTML + s;

['a','a a\n a\na a','x xxx\n X x\nx xxx', 'x-x+x+X++\n-x-x+x++asd\nx-x+x+X+'
,'########\n#####\n########','a-a-a\n0a0a\na-a-axXxX\n0a0a\na-a-a'
,'a c efghi\nab defgh\na c e ghij\nabcd fgh\nabc e g'
,'/\\/\\/\\/\\/\\/\\/\\\n\\/\\/\\/\\/\\/\\/\\/\n#####x-x+x+X++\n#####-x-x+x++###XX\n#####x-x+x+X+\n/\\/\\/\\/\\/\\/\\/\\\n\\/\\/\\/\\/\\/\\/\\/'
,'xxxxxxxxxx\nxxxxxxxXX\nxxxxxxxxxx\nxxxxxxxxxx\nxxxxxxxxxx\nxxxxxxxx']
.forEach(t=>Out(t+'\n<b>'+F(t)+'</b>\n\n'))

b { color: #00f }

<pre id=OUT></pre>

Answer 2

C (562 b)

---

Golfed Code:

int main(){wchar_t *q,*y,*w,*b,a[50][50];int l,c,t,i,j,u,v,h,s,x,r;l=c=i=0;do l=(t=wcslen(q=_getws(a[i++])))>l?t:l;while(t);for(j=0;j/l<i-1;j++)c+=(x=((t=*(b=a[u=j/l]+(v=j%l)))==120|t==88|t==344|t==376))&(s=(((t&(h=0xff))!=(*(b+2)&h))|((t&h)!=(*(w=a[u+1]+(v+1))&h))|((t&h)!=(*(y=a[u+2]+v)&h))|((t&h)!=(*(y+2)&h))))?(t==(t&h)):(x?0*(*b=*(b+2)=*w=*y=*(y+2)=(t+256)):(((t&h)==(*(b+2)&h))&((*w&h)==(*y&h))&((*(y+2)&h)==(*y&h))&((*y&h)==(t&h))?(r=(t==(t&h))+(*(b+2)==(t&h))+(*w==(t&h))+(*y==(t&h))+(*(y+2)==(t&h)))+0*(*b=*(b+2)=*w=*y=*(y+2)=t+256):0));printf("%d",c);}

unpacked Code:

#include <stdio.h>
#include <conio.h>
#include <string.h>
int main(){
wchar_t *q,*y,*w,*b, a[50][50];int l,c,t,i,j,u,v,h,s,x,r;l=c=i=0;
do
l=(t=wcslen(q=_getws(a[i++])))>l?t:l;
while (t);
for(j=0;j/l<i-1;j++) 
c+=(x=((t=*(b=a[u=j/l]+(v=j%l)))==120|t==88|t==344|t==376 ))&(s=(((t&(h=0xff))!=(*(b+2)&h))|((t&h)!=(*(w=a[u+1]+(v+1))&h))|((t&h)!=(*(y=a[u+2]+v)&h))|((t&h)!=(*(y+2)&h))))?(t==(t&h)):(x?0*(*b=*(b+2)=*w=*y=*(y+2)=(t+256)):(((t&h)==(*(b+2)&h))&((*w&h)==(*y&h))&((*(y+2)&h)==(*y&h))&((*y&h)==(t&h))?(r=(t==(t&h))+(*(b+2)==(t&h))+(*w==(t&h))+(*y==(t&h))+(*(y+2)==(t&h)))+0*(*b=*(b+2)=*w=*y=*(y+2)=t+256):0));
printf("%d",c);getch();
}