I found the default dealing of the terms often displays a negative sign at the beginning, including those beneath a square-root sign, which is a little bit annoying since for most cases, there exists at least one positive-signed term in the result and the negative sign can be adjusted by directly being moved backward after the positive-signed term or being multiplied into one difference term, swapping the two numbers in the difference term and thus avoiding a negative sign at the beginning. How can I remove the negative sign to make sure the term with positive sign comes first(if there is any)?
Example1:
-(a+b)(c-d)
Current result:
(-a - b) (c - d)
What I want:
(a+b)(d-c)
Example2:
Integrate[2 g y Sqrt[2 r y - y^2] \[Rho], {y, 0, h},
Assumptions -> (r | h | g) \[Element] PositiveReals &&
h <= 2 r] // FullSimply
Current Result:
ConditionalExpression[
1/3 g \[Rho] (2 Sqrt[-h^5 (h - 2 r)] -
r (Sqrt[-h^3 (h - 2 r)] + 3 Sqrt[-h (h - 2 r)] r) +
6 r^3 ArcCsc[Sqrt[2] Sqrt[r/h]]), h < 2 r]
i.e.
$\frac{1}{3} g \rho \left(2 \sqrt{-h^5 (h-2 r)}-r \left(\sqrt{-h^3 (h-2 r)}+3 r \sqrt{-h (h-2 r)}\right)+6 r^3 \text{arccsc}\left(\sqrt{2} \sqrt{\frac{r}{h}}\right)\right)\text{ if }h<2 r$
The result I want:
ConditionalExpression[
1/3 g \[Rho] (2 Sqrt[h^5 (2 r-h)] -
r (Sqrt[h^3 (2 r-h)] + 3 Sqrt[h (2 r-h)] r) +
6 r^3 ArcCsc[Sqrt[2] Sqrt[r/h]]), h < 2 r]
i.e.
$\frac{1}{3} g \rho \left(2 \sqrt{h^5 (2 r-h)}-r \left(\sqrt{h^3 (2 r-h)}+3 r \sqrt{h (2 r-h)}\right)+6 r^3 \text{arccsc}\left(\sqrt{2} \sqrt{\frac{r}{h}}\right)\right)\text{ if }h<2 r$
Example 3:
`Integrate[Sin[x]^2, {x, a, b}]`
Current result:
1/2 (-a + b + Cos[a] Sin[a] - Cos[b] Sin[b])
What I want:
1/2 (b -a + Cos[a] Sin[a] - Cos[b] Sin[b])
Example 4:
D[1/2 Log[1 - x] - 1/2 Log[1 + x], x] // Simplify
Current result:
1/(-1 + x^2)
What I want:
1/(x^2 - 1)
Example 5:
`Integrate[1/(x^4 - a^4), x] // FullSimplify`
Current result:
`-((2 ArcTan[x/a] - Log[a - x] + Log[a + x])/(4 a^3))`
what I want:
(Log[a - x]-2 ArcTan[x/a] + Log[a + x])/(4 a^3)
