During some calculations for a program I came upon this Integral which I am not able to solve. I already tried Matlab but it didn't help me. Here is the Integral: $$\int\left(\sqrt{\sum_{0}^{5} 9 a_i^2 t^4 +12 a_i b_i t^3 + 6 a_i c_i t^2 + 4 b_i^2 t^2 + 4 b_i c_i t +c_i^2}\right) dt$$
I already found this post but couldn't solve my problem.
Thanks in advance
Edit: Rearranging the formula $$\int\left(\sqrt{\sum_{0}^{5} (3 a_i t^2 + c_i)^2 + 4 b_i t^2 (3 a_i t + b_i) + 4 b_i c_i t}\right) dt$$
I ended up with this formula $$\sum\left(u\left(15 c_i^2 \sqrt{\frac{\sqrt{- a_i^2 b_i^2 (c_i^2 -1)}-6a_i^2 b_i^2 u^{3/2} - a_i b_i}{\sqrt{-a_i^2 b_i^2(c_i^2-1)}-a_i b_i}} \sqrt{\frac{\sqrt{-a_i^2 b_i^2(c_i^2-1)}+6 a_i^2 b_i^2 u^{3/2}+a_i b_i}{\sqrt{-a_i^2 b_i^2(c_i^2 -1)}+a_i b_i}}\\ F_1\left(\frac{2}{3};\frac{1}{2};\frac{1}{2};\frac{5}{3};-\frac{6 a_i^2 b_i^2 u^{3/2}}{a_i b_i + \sqrt{- a_i^2 b_i^2(c_i^2 -1)}},-\frac{6 a_i^2 b_i^2 u^{3/2}}{a_i b_i - \sqrt{-a_i^2 b_i^2 - (c_i^2 -1)}}\right)+36 a_i b_i u^{3/2}\sqrt{\frac{\sqrt{- a_i^2 b_i^2(c_i^2 -1)}-6 a_i^2 b_i^2 u^{3/2}- a_i b_i}{\sqrt{-a_i^2 b_i^2(c_i^2 -1)}-a_i b_i}}\sqrt{\frac{\sqrt{- a_i^2 b_i^2(c_i^2 -1)}+6 a_i^2 b_i^2 u^{3/2}+a_i b_i}{\sqrt{-a_i^2 b_i^2(c_i^2 -1)}-a_i b_i}}\\ F_1\left(\frac{5}{3};\frac{1}{2};\frac{1}{2};\frac{8}{3};-\frac{6 a_i^2 b_i^2 u^{3/2}}{a_i b_i+\sqrt{- a_i^2 b_i^2(c_i^2-1)}},-\frac{6 a_i^2 b_i^2 u^{3/2}}{a_i b_i -\sqrt{- a_i^2 b_i^2(c_i^2-1)}}\right)\\ +10(12 a_i b_i u^{3/2}(3 a_i b_i u^{3/2}+1)+c_i^2)\right)\right) /\left(25 \sqrt{12 a_i b_i u^{3/2}(3 a_i b_i u^{3/2}+1)+c_i^2}\right) (+ constant)$$
Were $F_1$ is the Appell hypergeometric function and $u=t^2$.
Could that be right or are there any errors (probably...)?