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$$ \int_{-\infty}^{\infty} [c_1 + c_2 (x-c_3)^2 + (x - c_4)]^{-c_5} \, dx $$

with $c_1, c_2, c_3, c_4, c_5$ known real constant.

Can you help me to solve this integral?

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  • $\begingroup$ This probably involves nasty hypergeometric functions. Any conditions about the $c_i$'s ? $\endgroup$ Jul 27 '15 at 10:26
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If we let $$ z = \frac{x+\frac{1-2c_2c_3}{2c_2}}{\sqrt{\frac{c_1+c_2c_3^2}{c_2}-\left(\frac{1-2c_2c_3}{2c_2}\right)^2}} $$ (this was for brevity) we then obtain $$ \left[c_2\left(\frac{c_1+c_2c_3^2}{c_2}-\left(\frac{1-2c_2c_3}{2c_2}\right)^2\right)^{1-\frac{1}{2c_5}}\right]^{-c_5}\int_{-\infty}^{\infty}\left(z^2+1\right)^{-c_5}dz $$ so now you have to integrate $$ \int_{-\infty}^{\infty}\left(z^2+1\right)^{-c_5}dz $$ letting $z=\tan \theta$ we get $$ \int_{-\infty}^{\infty}\left(z^2+1\right)^{-c_5}dz = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^{-2c_5}\theta \sec^2\theta \,d\theta $$ if we have $c_5>1$ $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^{-2c_5}\theta \sec^2\theta \,d\theta = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos^{2\left(c_5-1\right)}\theta \,d\theta $$ or if we have $c_5<1$ $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^{-2c_5}\theta \sec^2\theta \,d\theta = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^{2\left(c_5-1\right)}\theta \,d\theta $$ let me know if you can proceed..or if indeed I made a mistake!

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