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

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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)