Difference in getting symbolic integral result

Hi. I'm new to symbolic toolbox.

Can anyone please explain why there is a different result obtained from what is virtually two identical expressions?

How is Matlab interpreting f1, that it produces an "unexpected" result for the integral_f1?

syms a b t
f1 = a*(1-t/b);
integral_f1 = int(f1, t)

integral_f1 = 

f2 = a - a*t/b;
integral_f2 = int(f2, t)

integral_f2 = 

Answer 1

[Torsten]

If you had specified a definite integral, the results would have been the same.

Indefinite integrals are only unique up to an arbitrary constant. And the constant in the first call to int is -ab/2, in the second call 0. So everything in order.

The question "why" MATLAB decided to integrate the way it did can only be explained by the software developers, I guess. You could ask Technical Support if you are really interested in the technical details.

[Nicolae Preda]

Thank you for the answer Torsten.

Well, the issue is that I was expecting MATLAB to default to a zero constant, and i was puzzled by the first call.

I don't understand the underlying rule it uses for picking the random constant i.e. how is the form (arrangement) of the expresion to be integrated influencing the outcome.

[Dyuman Joshi]

"I don't understand the underlying rule it uses"

No one other than people who wrote it knows the workings of SymEngine or the set of instructions it follows to give the output for computations involving symbolic variables and expressions.

As Torsten suggested, you can ask TMW technical support regarding this.

[Torsten]

Well, the issue is that I was expecting MATLAB to default to a zero constant, and i was puzzled by the first call.

If you always want an integration constant of 0, use

syms a b t
f1 = a*(1-t/b);
integral_f1 = int(f1, t, 0, t)

integral_f1 = 

f2 = a - a*t/b;
integral_f2 = int(f2, t, 0, t)

integral_f2 =