Four variable expression

$\begin{aligned} X & = \frac {1+a+ab}{1+a+ab+abc} + \frac {1+b+bc}{1+b+bc+bcd} + \frac {1+c+cd}{1+c+cd+cda} + \frac {1+d+da}{1+d+da+dab} \\ & = \frac {1+a+ab}{1+a+ab+abc} + \frac {1+b+bc}{1+b+bc+bcd} \blue{\times \frac aa} + \frac {1+c+cd}{1+c+cd+cda} \blue{\times \frac {ab}{ab}} + \frac {1+d+da}{1+d+da+dab} \blue{\times \frac {abc}{abc}} \\ & = \frac {1+a+ab}{1+a+ab+abc} + \frac {a+ab+abc}{a+ab+abc+1} + \frac {ab+abc+1}{ab+abc+1+a} + \frac {abc+1+a}{abc+1+a+ab} \\ & = \frac {3+3a+3ab+3abc}{1+a+ab+abc} \\ & = \boxed 3 \end{aligned}$

Since $abcd=1$ we see that $\begin{aligned} \frac{1+b+bc}{1+b+bc+bcd} & = \; \frac{a(1+b+bc)}{1 + a + ab + abc} \; = \; \frac{a + ab + abc}{1 + a + ab + abc} \\ \frac{1 + c + cd}{1 + c + cd + cda} & = \; \frac{ab(1 + c + cd)}{1 + a + ab + abc} \; =\; \frac{1 + ab + abc}{1 + a + ab + abc} \\ \frac{1 + d + da}{1 + d + da + dab} & = \; \frac{abc(1 + d + da)}{1 + a + ab +abc} \; = \; \frac{1 + a + abc}{1 + a + ab + abc} \end{aligned}$ and hence $\begin{aligned} \frac{1+a+ab}{1 + a + ab + abc} + \frac{1+b+bc}{1+b+bc+bcd} + \frac{1 + c + cd}{1 + c + cd + cda} + \frac{1 + d + da}{1 + d + da + dab} & = \; \frac{(1+a+ab)+(a + ab + abc)+(1 + ab+abc) + (1 + a +abc)}{1 + a + ab + abc} \\ & =\; \boxed{3} \end{aligned}$

The simple way: set $a=b=c=d=1$ and the expression evaluates to 3. So if any answer is right, it must be 3.

In proving that this holds for arbitrary positive $a,b,c,d$ , I can't improve other solutions.