Primitive of Reciprocal of x cubed by a x + b squared/Partial Fraction Expansion

Lemma for Primitive of $\dfrac 1 {x^3 \paren {a x + b}^2}$

 * $\dfrac 1 {x^3 \paren {a x + b}^2} \equiv \dfrac {3 a^2} {b^4 x} - \dfrac {2 a} {b^3 x^2} + \dfrac 1 {b^2 x^3} - \dfrac {3 a^3} {b^4 \paren {a x + b} } - \dfrac {a^3} {b^3 \paren {a x + b}^2}$

Proof
Setting $a x + b = 0$ in $(1)$:

Equating constants in $(1)$:

Equating $1$st powers of $x$ in $(1)$:

Equating $2$nd powers of $x$ in $(1)$:

Equating $4$th powers of $x$:

Summarising:

Hence the result.