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): {{begin-eqn}} {{eqn | l = \frac {6 a^2} {b^5} \int \frac {\d x} x        | r = \frac {6 a^2} {b^5} \ln \size x       | c = Primitive of Reciprocal }} {{eqn | l = -\frac {3 a} {b^4} \int \frac {\d x} {x^2}       | r = \frac {3 a} {b^4 x}       | c = Primitive of Power }} {{eqn | l = \frac 1 {b^3} \int \frac {\d x} {x^3}       | r = -\frac 1 {2 b^3 x^2}       | c = Primitive of Power }} {{eqn | l = -\frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} }       | r = -\frac {6 a^2} {b^5} \ln \size {a x + b}       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | l = - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2}       | r = \frac {3 a^3} {b^4 a \paren {a x + b} }       | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$ }} {{eqn | l = -\frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}       | r = \frac {a^2} {2 b^3 \paren {a x + b}^2}       | c = Primitive of $\dfrac 1 {\paren {a x + b}^3}$ }} {{end-eqn}} The terms with the logarithm can be combined thus: {{begin-eqn}} {{eqn | l = \frac {6 a^2} {b^5} \ln \size x - \frac {6 a^2} {b^5} \ln \size {a x + b}       | r = \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} }       | c = Difference of Logarithms }} {{end-eqn}} and so the solution to the integral can be expressed as: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b}^3}       | r = \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 }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | r = \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}       | o =  }} {{eqn | r = \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} }} {{eqn | r = \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} }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = \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}       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b}^3}       | r = \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 }} {{end-eqn}} {{qed}}