Integration of square root of quadratic

First we simplify the expression under the square root by completing the square:

$4{ x }^{ 2 }+8x-1=4({ x }^{ 2 }+2x)-1\\ \quad \quad \quad \quad \quad \quad \quad =4[{ (x+1) }^{ 2 }-1]-1\\ \qquad \qquad \quad =4{ (x+1) }^{ 2 }-5.$

Next, we make the substitution $x=\frac { \sqrt { 5 } }{ 2 } \sec { \theta -1 }$

$\Rightarrow dx=\frac { \sqrt { 5 } }{ 2 } \sec { \theta \tan { \theta d\theta } }$

With a change of variable, we must also change the limits of integration:

When $x=\frac { \sqrt { 15 } }{ 3 } -1, \theta =\frac { \pi }{ 6 }$ , and when $x=\sqrt { 5 } -1, \theta =\frac { \pi }{ 3 }.$

Hence the integral becomes:

$\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt { 4\cdot \frac { 5 }{ 4 } \sec ^{ 2 }{ \theta } -5 } }{ \frac { \sqrt { 5 } }{ 2 } \sec { \theta } } \cdot \frac { \sqrt { 5 } }{ 2 } \sec { \theta \tan { \theta d\theta } } } \\ =\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \sqrt { 5(\sec ^{ 2 }{ \theta -1) } } \cdot \tan { \theta d\theta } } \\ =\sqrt { 5 } \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \left| \tan { \theta } \right| } \cdot \tan { \theta d\theta }$

Since $\tan { \theta > 0 }$ for our limits, we take the positive root and drop the absolute value bars, giving $\\ \sqrt { 5 } \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \tan ^{ 2 }{ \theta d\theta } }$ .

Using the substitution $\tan ^{ 2 }{ \theta =\sec ^{ 2 }{ \theta -1 } }$ , the integral becomes

$\sqrt { 5 } { \left[ \tan { \theta -\theta } \right] }_{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }\\ =\frac { 4\sqrt { 15 } -\sqrt { 5 } \pi }{ 6 } .$

Hence, $a=4, b=15, c=5$ and $d=6,$ and $a+b+c+d=\boxed { 30 }$ .