I was wondering if there is a possibility of using some special functions in order to get the closed-form solution to the integral equations given below. I have tried expanding the bracket using binomial expansion, but to no avail. Any help very much appreciated! $$\int\left(1+f(x)^2\right)f(x)\ \text{d}x$$ where $$f(x)=\sqrt{A^2+Bx+Cx^2}$$ $A, B, C$ are constants

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    $\begingroup$ It's not an equation if there is no equality sign. $\endgroup$ – Ennar Nov 7 '17 at 23:47
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    $\begingroup$ This isn't actually an "integral equation", it's just computing the antiderivatives of a class of functions. $\endgroup$ – Ian Nov 7 '17 at 23:48

For illustration, consider $A=1,C=-1$. Then you're looking at $\int (2-x^2)\sqrt{1-x^2} dx$. That can be done by trig substitution: if you have a right triangle with legs $x$ and $\sqrt{1-x^2}$ then the hypotenuse is $1$ and so you can choose $\theta$ so $\sqrt{1-x^2}=\cos(\theta)$. In this case $x=\sin(\theta)$ so $dx=\cos(\theta) d \theta$, so you have

$$\int (1+\cos(\theta)^2)\cos(\theta)^2 d \theta$$

which is straightforward, albeit a bit tedious, to evaluate. Then back-substitute by setting $\theta=\sin^{-1}(x)$.

Integrands of your general form should be tractable but you'll need to take cases depending on the sign of $b^2-4ac$ (taking the more common form $ax^2+bx+c$ instead of your form). In my example this was positive, which gives trigonometric substitutions. When it is negative, you have hyperbolic substitutions. When it is zero, the problem is trivial.


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