Integral $\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$

Question

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

Answer 1

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.

Answer 2

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 $$I= \frac1{8\sqrt2}\left[\bigg(\frac34t + \frac43 \bigg)(t^2+1)^{3/2}-\frac{15}8 t(t^2+1)^{1/2}-\frac{15}8\sinh^{-1}t\right] $$