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

Notice that
$$
\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} = \frac{1}{4\sqrt{2}}\left(3(2x+1)^2  + 4(2x+1) -3\right)\sqrt{(2x+1)^2+1}
$$
under substitution $2x +1 = \tan(\alpha)$ simplifies your integral to
\begin{align*}
&\int\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} \, \mathrm{d}x \\
 &=\frac{1}{8\sqrt{2}} \int \left(3 \tan^2(\alpha) + 4\tan(\alpha) -3\right)\sec^3(\alpha) \, \mathrm{d}\alpha \\
& = \frac{3}{8\sqrt{2}} \int \left(\sec^2(\alpha)  -2\right)\sec^3(\alpha) \, \mathrm{d}\alpha  + \frac{1}{2\sqrt{2}} \int \left[\tan(\alpha)\sec(\alpha)\right]\sec^2(\alpha) \, \mathrm{d}\alpha\\
\end{align*}
Since the last integral is immediate under susbtitution $u = \sec(\alpha)$, you just need to solve $\int \sec^n(\alpha) \mathrm{d} \alpha$ for $n =3,5$. For this use the reduction formula for secant until you get to $
\int\sec(x) \mathrm{d}x = \ln|\sec(x) + \tan(x)| $, and afer returning the resulting expression to be written in terms of $x$ this concludes the problem.
