# Primitive of Reciprocal of x cubed by a x + b cubed

## Theorem

$\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$

## Proof

A partial fraction expansion of the integrand gives:

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

From Linear Combination of Integrals it follows that the expression can be rendered as:

$\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {6 a^2} {b^5} \int \frac {\d x} x - \frac {3 a} {b^4} \int \frac {\d x} {x^2} + \frac 1 {b^3} \int \frac {\d x} {x^3} - \frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} } - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2} - \frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}$

This can be integrated termwise as follows (ignoring for now the arbitrary constant):

 $\displaystyle \frac {6 a^2} {b^5} \int \frac {\d x} x$ $=$ $\displaystyle \frac {6 a^2} {b^5} \ln \size x$ Primitive of Reciprocal $\displaystyle -\frac {3 a} {b^4} \int \frac {\d x} {x^2}$ $=$ $\displaystyle \frac {3 a} {b^4 x}$ Primitive of Power $\displaystyle \frac 1 {b^3} \int \frac {\d x} {x^3}$ $=$ $\displaystyle -\frac 1 {2 b^3 x^2}$ Primitive of Power $\displaystyle -\frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} }$ $=$ $\displaystyle -\frac {6 a^2} {b^5} \ln \size {a x + b}$ Primitive of $\dfrac 1 {a x + b}$ $\displaystyle - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2}$ $=$ $\displaystyle \frac {3 a^3} {b^4 a \paren {a x + b} }$ Primitive of $\dfrac 1 {\paren {a x + b}^2}$ $\displaystyle -\frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}$ $=$ $\displaystyle \frac {a^2} {2 b^3 \paren {a x + b}^2}$ Primitive of $\dfrac 1 {\paren {a x + b}^3}$

The terms with the logarithm can be combined thus:

 $\displaystyle \frac {6 a^2} {b^5} \ln \size x - \frac {6 a^2} {b^5} \ln \size {a x + b}$ $=$ $\displaystyle \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} }$ Difference of Logarithms

and so the solution to the integral can be expressed as:

 $\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3}$ $=$ $\displaystyle \frac {3 a} {b^4 x} - \frac 1 {2 b^3 x^2} + \frac {3 a^2} {b^4 \paren {a x + b} } + \frac {a^2} {2 b^3 \paren {a x + b}^2} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$

While this would usually be considered as an acceptable form to leave such an expression, there is some way to go to obtain the result requested.

So, placing the non-logarithmic terms over a common denominator and rearranging, with a view to our destination:

 $\displaystyle$  $\displaystyle \frac {3 a} {b^4 x} - \frac 1 {2 b^3 x^2} + \frac {3 a^2} {b^4 \paren {a x + b} } + \frac {a^2} {2 b^3 \paren {a x + b}^2}$ $\displaystyle$ $=$ $\displaystyle \frac {6 a x \paren {a x + b}^2} {2 b^4 x^2 \\paren {a x + b}^2} - \frac {b \paren {a x + b}^2} {2 b^4 x^2 \paren {a x + b}^2} + \frac {6 a^2 x^2 \paren {a x + b} } {2 b^4 x^2 \paren {a x + b}^2} + \frac {a^2 b x^2} {2 b^4 x^2 \paren {a x + b}^2}$ $\displaystyle$ $=$ $\displaystyle \frac {12 a^3 x^3 + 18 a^2 b x^2 + 4 a b^2 x - b^3} {2 b^4 \paren {a x + b}^2 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {a^4 x^4 + 4 a^3 b x^3 + 6 a^2 b^2 x^2 + 4 a b^3 x + b^4} {2 b^5 \paren {a x + b}^2 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^4} {2 b^5 \paren {a x + b}^2 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {8 a^3 b x^3 + 8 a^2 b^2 x^2} {2 b^5 \paren {a x + b}^2 x^2} + \frac {8 a^3 b x^3 + 16 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {8 a^2 b x^2 \paren {a x + b} } {2 b^5 \paren {a x + b}^2 x^2} + \frac {8 a b x \paren {a x + b}^2} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2} {b^4 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2 b} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2 \paren {a x + b} } {b^5 \paren {a x + b} } - \frac {4 a^3 x} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2} {b^5} - \frac {4 a^3 x} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}$

Thus the full solution to the integral can be assembled as follows, where $\dfrac {4 a^2} {b^5}$ is subsumed into the arbitrary constant:

 $\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3}$ $=$ $\displaystyle \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$

$\blacksquare$

## Sources

(in which a mistake apppears)