Just for fun!

$f(a,x)=\displaystyle\int (a^{x}+x^{a}+a^{a}) dx$

$f(a,x)=\dfrac{a^{x}}{ln\ a}+\dfrac{x^{a+1}}{a+1}+a^{a}x+c$

Substituting $a=e,\ x=0\ \Rightarrow c=0$

Substituting $a=4,\ x=3$ ,

$f(4,3)=\dfrac{4083}{5}+\dfrac{32}{ln\ 2}$

$\Rightarrow a-\frac{25}{2}bcd=4083-\frac{25}{2}×5×32×2=\boxed{83}$

$\begin{aligned} f(k,x) & = \int \left(e^{x\ln k} + e^{k\ln x} + e^{k\ln k}\right) dx \\ & = \int \left(k^x + x^k + k^k \right) dx \\ & = \frac {k^x}{\ln k} + \frac {x^{k+1}}{k+1} + k^k x + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ f(e, 0) & = \frac 11 + 0 + 0 + C = 1 & \small \color{#3D99F6} \implies C = 0 \\ \implies f(k,x) & = \frac {k^x}{\ln k} + \frac {x^{k+1}}{k+1} + k^k x \\ f(4,3) & = \frac {4^3}{\ln 4} + \frac {3^{4+1}}{4+1} + 4^4 \times 3 \\ & = \frac {64}{2\ln 2} + \frac {243}5 + 768 \\ & = \frac {4083}5 + \frac {32}{\ln 2} \end{aligned}$

Therefore, $a-25bc = 4083 - 25(5)(32) = \boxed{83}$ .