Symbolic integration of c/(b*z-a)

Dear all,

I want to integrate c/(b*tau-a) from tau=0 to tau=z with the assumptions a>0, b>0 c>0 and z<a/b.

In Maple it works:

eq1 := c/(b*tau-a); eq2 := int(eq1,tau=0..z, AllSolutions) assuming a::positive, b::positive, c::positive, z<a/b;

with the result:

c*(ln(-b*z+a)-ln(a))/b

How can I do this in MATLAB?

Thanks a lot!

[John D'Errico]

Please stop adding answers every time you wish to make a comment. Use the comment button instead.

Answer 1

[Star Strider]

Use the simplify function:

syms a b c tau z real
assume(a>0)
assume(b>0)
assume(c>0)
assume(z<a/b)
out1 = int( c/(b*tau-a),tau,0,z);
out2 = simplify(out1, 'Steps',20)

out2 =
(c*log(a - b*z))/b - (c*log(a))/b

[Sean de Wolski]

In a live script with pretty output:

Answer 2

[Torsten]

Use "int".

https://de.mathworks.com/help/symbolic/int.html

Best wishes

Torsten.

Answer 3

[Ludwig]

Thanks for the answer, I did this already but I am not sattisfied with the result:

 clear 
syms a b c tau z
assume(a>0)
assume(b>0)
assume(c>0)
assume(z<a/b)
int( c/(b*tau-a),tau,0,z)

Answer 4

[Ludwig]

Thanks!

[Star Strider]

My pleasure!