Primitive of Reciprocal of x cubed by a x + b cubed/Partial Fraction Expansion
== Lemma for Primitive of $\dfrac 1 {x^3 \left({a x + b}\right)^3}$ == :$\dfrac 1 {x^3 \left({a x + b}\right)^3} \equiv \dfrac {6 a^2} {b^5 x} - \dfrac {3 a} {b^4 x^2} + \dfrac 1 {b^3 x^3} - \dfrac {6 a^3} {b^5 \left({a x + b}\right)} -\dfrac {3 a^3} {b^4 \left({a x + b}\right)^2} - \dfrac {a^3} {b^3 \left({a x + b}\right)^3}$ == Proof == {{begin-eqn}} {{eqn | l = \dfrac 1 {x^3 \left({a x + b}\right)^3}       | o = \equiv       | r = \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b} + \dfrac E {\left({a x + b}\right)^2} + \dfrac F {\left({a x + b}\right)^3}       | c =  }} {{eqn | n = 1       | ll= \implies       | l = 1       | o = \equiv       | r = A x^2 \left({a x + b}\right)^3 + B x \left({a x + b}\right)^3 + C \left({a x + b}\right)^3 + D x^3 \left({a x + b}\right)^2 + E x^2 \left({a x + b}\right) + F x^2       | c = multiplying through by $x^2 \left({a x + b}\right)^3$ }} {{eqn | o = \equiv       | r = A a^3 x^5 + 3 A a^2 b x^4 + 3 A a b^2 x^3 + A b^3 x^2       | c = multiplying everything out }} {{eqn | o =        | ro= +       | r = B a^3 x^4 + 3 B a^2 b x^3 + 3 B a b^2 x^2 + B b^3 x       | c =  }} {{eqn | o =        | ro= +       | r = C a^3 x^3 + 3 C a^2 b x^2 + 3 C a b^2 x + C b^3       | c =  }} {{eqn | o =        | ro= +       | r = D a^2 x^5 + 2 D a b x^4 + D b^2 x^3 + E a x^4 + E b x^3 + F x^3       | c =  }} {{end-eqn}} Setting $a x + b = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = a x + b       | r = 0       | c =  }} {{eqn | ll= \implies       | l = x       | r = -\frac b a       | c =  }} {{eqn | ll= \implies       | l = F \left({-\frac b a}\right)^3       | r = 1       | c = substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ }} {{eqn | ll= \implies       | l = F       | r = -\frac {a^3} {b^3}       | c =  }} {{end-eqn}} Equating constants in $(1)$: {{begin-eqn}} {{eqn | l = 1       | r = C b^3       | c =  }} {{eqn | n = 2       | ll= \implies       | l = C       | r = \frac 1 {b^3}       | c =  }} {{end-eqn}} Equating $1$st powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = B b^3 + 3 C a b^2       | c =  }} {{eqn | ll= \implies       | l = B b^3       | r = -3 \frac 1 {b^3} a b^2       | c = substituting for $C$ from $(2)$ }} {{eqn | n = 3       | ll= \implies       | l = B       | r = \frac {-3 a} {b^4}       | c = simplifying }} {{end-eqn}} Equating $2$nd powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A b^3 + 3 B a b^2 + 3 C a^2 b       | c =  }} {{eqn | ll= \implies       | l = A b^3       | r = -3 \frac {-3 a} {b^4} a b^2 - 3 \frac 1 {b^3} a^2 b       | c = substituting for $C$ from $(2)$ and $B$ from $(3)$ }} {{eqn | ll= \implies       | l = A       | r = \frac {9 a^2} {b^5} - 3 \frac {a^2} {b^5}       | c = simplifying }} {{eqn | n = 4       | r = \frac {6 a^2} {b^5}       | c = simplifying }} {{end-eqn}} Equating $5$th powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = A a^3 + D a^2       | c =  }} {{eqn | ll= \implies       | l = D a^2       | r = -\frac {6 a^2} {b^5} a^3       | c = substituting for $A$ from $(4)$ }} {{eqn | n = 5       | ll= \implies       | l = D       | r = -\frac {6 a^3} {b^5}       | c = simplifying }} {{end-eqn}} Equating $4$th powers of $x$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = 3 A a^2 b + B a^3 + 2 D a b + E a       | c =  }} {{eqn | ll= \implies       | l = E       | r = -3 A a b - B a^2 - 2 D b       | c = rearranging }} {{eqn | r = -3 a b \frac {6 a^2} {b^5} - \frac {-3 a} {b^4} a^2 - 2 \frac {-6 a^3} {b^5} b       | c = substituting for $A$ from $(4)$, $B$ from $(3)$ and $D$ from $(5)$ }} {{eqn | r = -\frac {18 a^3} {b^4} + \frac {3 a^3} {b^4} + \frac {12 a^3} {b^4}       | c = simplifying }} {{eqn | r = -\frac {3 a^3} {b^4}       | c = simplifying }} {{end-eqn}} Summarising: {{begin-eqn}} {{eqn | l = A       | r = \frac {6 a^2} {b^5} }} {{eqn | l = B       | r = -\frac {3 a} {b^4} }} {{eqn | l = C       | r = \frac 1 {b^3} }} {{eqn | l = D       | r = -\frac {6 a^3} {b^5} }} {{eqn | l = E       | r = -\frac {3 a^3} {b^4} }} {{eqn | l = F       | r = -\frac {a^3} {b^3} }} {{end-eqn}} Hence the result.