I'm not sure about the Mathematica expressions, but I'm pretty sure you can solve this using partial fractions (it won't be pretty though)...
Hi Guys,
I need to solve the following integral using mathematica, but i do not know how to identify the constants and do symbolic integration, so i get a very nasty expression .
dw/dt=(t-r)/{(t-b)*(t-p)*sqrt((t-a)*(t-c))}
a,b ,p,c,r are constants.
any help with this!
Thanks
mopen80
$$\frac{dw}{dt}=\frac{t-r}{(t-b)(t-p)\sqrt{(t-a)(t-c)}}$$
The output from Mathematica is
$$\int \frac{t-r}{(t-b) (t-p) \sqrt{(t-a) (t-c)}} \, dt=$$
$$\frac{\sqrt{t-a} \sqrt{t-c} \left(i \sqrt{a-p} (b-r) \sqrt{c-p} \log \left(\frac{i (b-p) \left(a (b-2 c+t)+2 i \sqrt{a-b} \sqrt{t-a} \sqrt{b-c} \sqrt{t-c}+b (c-2 t)+c t\right)}{\sqrt{a-b} \sqrt{b-c} (b-r) (b-t)}\right)+\sqrt{a-b} \sqrt{b-c} (p-r) \log (p-t)+\sqrt{a-b} \sqrt{b-c} (r-p) \log \left(-a (-2 c+p+t)+2 \left(\sqrt{a-p} \sqrt{t-a} \sqrt{c-p} \sqrt{t-c}+p t\right)-c (p+t)\right)\right)}{\sqrt{a-b} \sqrt{a-p} \sqrt{b-c} (p-b) \sqrt{c-p} \sqrt{(t-a) (t-c)}}$$
That is what it is. I don't know how you can simplify it without having values for your constants.