Integral (close form?) I'm struggling to evaluate the following integral:
$\displaystyle \int_{-1}^{1}\frac{1+2x^2+3x^4+4x^6+5x^8+6x^{10}+7x^{12}}{\sqrt{\left ( 1+x^2 \right )\left ( 1+x^4 \right )\left ( 1+x^6 \right )}}\,dx$.
I see that the integrand is even, thus the integral can be re-written as: $\displaystyle \int_{-1}^{1}=2\int_{0}^{1}$, which does not help at all. Applying the sub $u=1-x$ is useless because it would result in a giant mess.
Do you see a pattern or an alternate approach?
 A: I'll play and see if
I can find anything interesting.
$I
=\int_{0}^{1}\dfrac{1+2x^2+3x^4+4x^6+5x^8+6x^{10}+7x^{12}}{\sqrt{( 1+x^2 ) ( 1+x^4  ) ( 1+x^6  )}}\,dx$
Let $y=x^2$.
Then
$dy = 2x\, dx
$
or
$dx
=\frac{dy}{2x}
=\frac{dy}{2\sqrt{y}}
$.
Then
$\begin{array}\\
I
&=\frac12\int_{0}^{1}\dfrac{1+2y+3y^2+4y^3+5y^4+6y^5+7y^6}{\sqrt{y( 1+y ) ( 1+y^2  ) ( 1+y^3  )}}
dy\\
&=\frac12\int_{0}^{1}\dfrac{g(y)dy}{\sqrt{f(y)}}
\\
\end{array}
$
where
$g(y)
=1+2y+3y^2+4y^3+5y^4+6y^5+7y^6
$
and
$\begin{array}\\
f(y)
&=y( 1+y ) ( 1+y^2  ) ( 1+y^3  )\\
&=y\left( ( 1+y^2  ) ( 1+y^3  )+y ( 1+y^2  ) ( 1+y^3  )\right)\\
&=y\left( 1+y^2+y^3+y^5+y(1+y^2+y^3+y^5)\right)\\
&=y\left( 1+y^2+y^3+y^5+y+y^3+y^4+y^6\right)\\
&=y\left(1+y+y^2+2y^3+y^4+y^5+y^6\right)\\
&=y+y^2+y^3+2y^4+y^5+y^6+y^7\\
\end{array}
$
Noting that
$\begin{array}\\
f'(y)
&=1+2y+3y^2+8y^3+5y^4+6y^5+7y^6\\
&=(1+2y+3y^2+4y^3+5y^4+6y^5+7y^6)+4y^3\\
&=g(y)+4y^3\\
\end{array}
$
$\begin{array}\\
I
&=\frac12\int_0^1 \dfrac{g(y)dy}{\sqrt{f(y)}}\\
&=\frac12\int_0^1 \dfrac{(f'(y)-4y^3)dy}{\sqrt{f(y)}}\\
&=\frac12\int_0^1 \dfrac{f'(y)dy}{\sqrt{f(y)}}
-2\int_0^1 \dfrac{y^3\,dy}{\sqrt{f(y)}}\\
\end{array}
$
I'll stop here;
this seems like a good start.
Any errors here
should be reported
to the appropriate committee
for immediate remediation.
