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

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

 * $\dfrac 1 {x^3 \paren {a x + b}^3} \equiv \dfrac {6 a^2} {b^5 x} - \dfrac {3 a} {b^4 x^2} + \dfrac 1 {b^3 x^3} - \dfrac {6 a^3} {b^5 \paren {a x + b} } -\dfrac {3 a^3} {b^4 \paren {a x + b}^2} - \dfrac {a^3} {b^3 \paren {a x + b}^3}$

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 $5$th powers of $x$ in $(1)$:

Equating $4$th powers of $x$ in $(1)$:

Summarising:

Hence the result.