I need the solution to an Integral of type f[z]*Log[z]/z

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$$\int dz \frac{\sqrt{\frac{1}{4} (x+1)^2 (z-1)^2-x} \log (z)}{z}$$

All values are real. The domain for z and x are both [0,1], with also the constraint that the Sqrt is real ( which means z is really from ##[0, 1-\frac{2 \sqrt{x}}{x+1}]##. I'm just trying to get the anti-derivative, and not apply the limits yet. No complex values for x or z.

I've tried many ways, and I can't seem to get it. Mathematica doesn't give a solution, and I've included the assumptions.

I think it'll be a combination of Log[z], Log[z]^2, and PolyLog[2,z] though I can't get it into a form that I can do this. Its possible due to its nature that it could be solved in terms of HyperGeometric functions too.

Does anyone have any ideas? Is there a method to solving these?

Thanks!
-Hepth
 
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Hepth said:
$$\int dz \frac{\sqrt{\frac{1}{4} (x+1)^2 (z-1)^2-x} \log (z)}{z}$$

All values are real. The domain for z and x are both [0,1], with also the constraint that the Sqrt is real ( which means z is really from ##[0, 1-\frac{2 \sqrt{x}}{x+1}]##. I'm just trying to get the anti-derivative, and not apply the limits yet. No complex values for x or z.

I've tried many ways, and I can't seem to get it. Mathematica doesn't give a solution, and I've included the assumptions.

I think it'll be a combination of Log[z], Log[z]^2, and PolyLog[2,z] though I can't get it into a form that I can do this. Its possible due to its nature that it could be solved in terms of HyperGeometric functions too.

Does anyone have any ideas? Is there a method to solving these?

Thanks!
-Hepth
Is there any relationship between x and z?
 
No. They are both independent at this point. In the end, x will also be integrated from 0..1, but combined with some other integrals it'll be convergent.
 
You can see that at x = 0 the solution is
##\frac{z}{2}+\frac{\log^2(z)}{4}-\frac{1}{2} z \log (z)##,

and at x=1 the solution is $$
\frac{1}{\sqrt{4-2 z}}i \sqrt{\frac{2}{z}-1} \sqrt{z} \left(\sqrt{2} (\log (z)-1) \left(\sqrt{-(z-2) z}+2 \sin ^{-1}\left(\frac{\sqrt{z}}{\sqrt{2}}\right)\right)-4 \sqrt{z} \, 3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};\frac{z}{2}\right)\right)$$
 
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