# Remove the negative sign so that positive-signed term comes first

Posted 3 months ago
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 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)
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Posted 3 months ago
 This seems to work in a couple of cases: fnct = (LeafCount[#] + 3 StringCount[ToString[InputForm[#]], "-"] &); -(a + b) (c - d) FullSimplify[%, ComplexityFunction -> fnct] Integrate[2 g y Sqrt[2 r y - y^2] \[Rho], {y, 0, h}, Assumptions -> (r | h | g) \[Element] PositiveReals && h <= 2 r] Simplify[%, ComplexityFunction -> fnct] 
Posted 3 months ago
 Thanks for your insight, it does work to some degree. But for the first example, it gives-((a + b) (c - d)).Moreover, for the second example, it givesConditionalExpression[ 2 g \[Rho] (1/3 Sqrt[h^5 (-h + 2 r)] - 1/6 r (3 r Sqrt[h (-h + 2 r)] + Sqrt[h^3 (-h + 2 r)]) + r^3 ArcCsc[Sqrt[2] Sqrt[r/h]]), h < 2 r] where the term (-h+2 r) emerges multiple times, which would be better if displayed as (2r-h) especially as I have made the assumption h<=2r.Additionally, for the last example, it gives(-2 ArcTan[x/a] + Log[a - x] - Log[a + x])/(4 a^3), although being moved to the numerator, the negative sign still comes first.
 On my system I get this: fnct = (LeafCount[#] + 3 StringCount[ToString[InputForm[#]], "-"] &); FullSimplify[-(a + b) (c - d), ComplexityFunction -> fnct] // TraditionalForm (a+b) (d-c) TraditionalForm is another trick you can try. In general, getting algebraic expressions in exactly the way you want them can be rather complicated.
 Thanks. The TraditionalForm trick is useful though imperfect. I agree on you that rearranging expressions in exactly the way I want is truly complicated.