Primitive of x squared by Root of x squared plus a squared/Also presented as

Primitive of $x^2 \sqrt {x^2 + a^2}$: Also presented as

This result is also seen presented in the form:

$\ds \int x^2 \sqrt {a^2 + x^2} \rd x = \frac {x \paren {a^2 + 2 x^2} \sqrt {a^2 + x^2} } 8 - \frac {a^4} 8 \arsinh \frac x a + C$

$\ds \int x^2 \sqrt {x^2 + a^2} \rd x$

$=$

$\ds \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 - \frac {a^2 x \sqrt {x^2 + a^2} } 8 - \frac {a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C$

$\ds $

$=$

$\ds \frac {2 x \paren {\sqrt {x^2 + a^2} }^3 - a^2 x \sqrt {x^2 + a^2} } 8 + \frac {a^4} 8 \arsinh \frac x a + C$

$\ds $

$=$

$\ds \frac {2 x \paren {x^2 + a^2} \sqrt {x^2 + a^2} - a^2 x \sqrt {x^2 + a^2} } 8 + \frac {a^4} 8 \arsinh \frac x a + C$

$\ds $

$=$

$\ds \frac {x \paren {\paren {2 x^2 + 2 a^2} - a^2} \sqrt {x^2 + a^2} } 8 + \frac {a^4} 8 \arsinh \frac x a + C$

$\ds $

$=$

$\ds \frac {x \paren {a^2 + 2 x^2} \sqrt {a^2 + x^2} } 8 + \frac {a^4} 8 \arsinh \frac x a + C$

$\blacksquare$