Integral $\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$ I tried to integrate
$$\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$$
by multipling by $\sqrt{2x^2 + 2x + 1}$ in the numerator and the denominator to break it into $5$ fractions
The answer is possible but it's too long
Another better solution

A systematic approach after $t=2x+1$
\begin{align}
I=&\int (3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}\>dx\\
= &\frac1{8\sqrt2}\int (4t+3)(t^2+1)^{3/2}-6 (t^2+1)^{1/2}\>dt\tag1
\end{align}
Integrate below by parts
\begin{align}
&\int (t^2+1)^{3/2}dt=\frac14t(t^2+1)^{3/2}+\frac34\int(t^2+1)^{1/2}dt\\
 &\int (t^2+1)^{1/2}dt=\frac12 t(t^2+1)^{1/2}+\frac12\sinh^{-1}t
\end{align}
Plug into (1) to arrive at
$$I= \frac1{8\sqrt2}\left[\bigg(\frac34t + \frac43 \bigg)(t^2+1)^{3/2}-\frac{15}8 t(t^2+1)^{1/2}-\frac{15}8\sinh^{-1}t\right]
$$
