results are Fractional and long

i wrote this code:

clc

clear all

close all

%--------------------------------------------------------------------------

syms Alpha1 Betta2 Betta1 S1 S2 h q S

t=2;

R=27;

H=96;

h=13;

STRp=3.7;

STRy=360;

STRu=760;

Etta=0.75;

K=0.14;

n=1;

m=0.65;

Gamma0=2*n*h*Etta;

Alpha01=acos(m);

Betta02=acos(1-m);

STR0=sqrt((STRy*STRu)/(1+K));

Mp=(STR0*(t^2))/4;

Np=STR0*t;

%-----------------------------------------------

E_foam=(pi*((R-(t/2))^2)*STRp*2*h*Etta)

Eb=(4*pi*Mp)*(((R+(m*h))*acos(m))+(h*sqrt(1-m^2))+(R*asin(1-m))+((R-(h*(1-m)))*acos(1-m)+R*asin(m)))

Em=2*pi*Np*(h)^2

PART1=(STRp)*(2*pi)*(R);

Eint_AB1=int(int(-S1*PART1,S1,(m*h)/cos(Alpha1),h),Alpha1,acos(m),0)

PART2=(STRp)*(2*pi)*(R);

Eint_BC1=int(int((-S1*PART2),S1,0,h),Betta1,pi/2,acos(1-m))

PART3=(STRp)*(2*pi)*(R);

Eint_BC2=int(int((-S2*PART3),S2,0,((1-m)*h)/cos(Betta2)),Betta2,acos(1-m),0)

E_int=Eint_AB1+Eint_BC1+Eint_BC2

some of the results that matlab gives me are fractional and matlab will not solve them and show the result number, instead gives me some long answers

answers:

Eint_AB1 =

-(168831*pi*(95138542128201728*tan(7775121492917933/9007199254740992) - 194378037322948325))/2251799813685248000

Eint_BC1 =

(168831*pi*(pi/2 - 5463880662892063/4503599627370496))/10

Eint_BC2 =

(8272719*pi*tan(5463880662892063/4503599627370496))/4000

how can i simplify this results and find the final answer in a simpler form?

Answer 1

[Stephan]

double(Eint_AB1)

[John D'Errico]

Or if you want a solution in the form of a symbolic number, but still in a long precision result, then

vpa(Eint_AB1)