partfrac() function returning same expression that was given as input

I have the following expression:

(4*l^2*m5 - 2*dl^2*m5*cos(2*THETA1(t) - 4*THETA2(t) + 4*THETA3(t) - 4*THETA4(t) + 2*THETA5(t)))/(12*dl^2)

I want the denominator (i.e. 12*dl^2) to be divided across all the numerator terms so that the resulting expression is the sum of a bunch of terms.

partfrac((4*l^2*m5 - 2*dl^2*m5*cos(2*THETA1(t) - 4*THETA2(t) + 4*THETA3(t) - 4*THETA4(t) + 2*THETA5(t)))/(12*dl^2))

ans =
 
(- m5*cos(2*THETA1(t) - 4*THETA2(t) + 4*THETA3(t) - 4*THETA4(t) + 2*THETA5(t))*dl^2 + 2*m5*l^2)/(6*dl^2)

When I use the partfrac() function in MATLAB to do this, it gives me almost the same expression again.

Please help.

Answer 1

[Star Strider]

The partfrac function needs more information, specifically the variable (not expression) that you want it to use.

This (note that I provided the variable ‘dl’ and a useful name-value pair):

syms dl l m5 t THETA1(t) THETA2(t) THETA3(t) THETA4(t) THETA5(t) 
Eq = (4*l^2*m5 - 2*dl^2*m5*cos(2*THETA1(t) - 4*THETA2(t) + 4*THETA3(t) - 4*THETA4(t) + 2*THETA5(t)))/(12*dl^2);
Eq_pf = partfrac(Eq, dl, 'FactorMode','full')

produces:

Eq_pf =
(l^2*m5)/(3*dl^2) - (m5*cos(2*THETA1(t) - 4*THETA2(t) + 4*THETA3(t) - 4*THETA4(t) + 2*THETA5(t)))/6

That is as much as it can do.

.

[Thejus Jose]

Thank you. This works.

[Star Strider]

As always, my pleasure!