Primitive of Reciprocal of Cube of Root of a x squared plus b x plus c

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Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\ds \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + C$


Proof

For $a > 0$:

\(\ds \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) \(=\) \(\ds \int \frac {\d x} {\paren {\sqrt {\frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a} } }^3}\) Completing the Square
\(\ds \) \(=\) \(\ds \int \frac {8 a \sqrt a \rd x} {\paren {\sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} }^3}\) simplifying


Then:

\(\ds z\) \(=\) \(\ds 2 a x + b\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a\) Derivative of Power
\(\ds \leadsto \ \ \) \(\ds \int \frac {8 a \sqrt a \rd x} {\paren {\sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} }^3}\) \(=\) \(\ds \int \frac {8 a \sqrt a \rd z} {2 a \paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) Integration by Substitution
\(\ds \) \(=\) \(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) Primitive of Constant Multiple of Function


Let $4 a c - b^2 > 0$.

Then:

\(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) \(=\) \(\ds 4 \sqrt a \paren {\frac z {\paren {4 a c - b^2} \sqrt {z^2 + 4 a c - b^2} } } + C\) Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$


Let $4 a c - b^2 < 0$.

Then:

\(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) \(=\) \(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 - \paren {b^2 - 4 a c} } }^3}\)
\(\ds \) \(=\) \(\ds 4 \sqrt a \paren {\frac {-z} {\paren {b^2 - 4 a c} \sqrt {z^2 - \paren {b^2 - 4 a c} } } } + C\) Primitive of $\dfrac 1 {\paren {\sqrt {x^2 - a^2} }^3}$
\(\ds \) \(=\) \(\ds 4 \sqrt a \paren {\frac z {\paren {4 a c - b^2} \sqrt {z^2 + 4 a c - b^2} } } + C\) simplifying


Thus in both cases the same result applies, and so:

\(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) \(=\) \(\ds \frac {4 \sqrt a \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} } + C\) substituting for $z$ and simplifying
\(\ds \) \(=\) \(\ds \frac {4 \sqrt a \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {4 a \paren {a x^2 + b x + c} } } + C\) Completing the Square
\(\ds \) \(=\) \(\ds \frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + C\) simplifying

$\Box$


This needs considerable tedious hard slog to complete it.
In particular: Similarly for $a < 0$, unless there's a quick way to cover both cases
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$\blacksquare$


Sources

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