I was pleased to see the following concise result from Rubi

In[8203]:= Int[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8203]= ArcTan[u^2/(Sqrt[2] (1 + u))]/Sqrt[2]

My implementation of Bronstein's Risch algorithm (from Symbolic Integration I) returns two arctangents

In[8206]:= Risch[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8206]= (ArcTan[(-1 + u)/Sqrt[2]] - ArcTan[(2 + 2 u + (-1 + u) u^2)/Sqrt[2]])/Sqrt[2]

and Mathematica returns the naive form

In[8207]:= Integrate[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8207]= 1/4 RootSum[
2 + 4 #1 + 2 #1^2 + #1^4 &, (2 Log[u - #1] #1 + Log[u - #1] #1^2)/(
1 + #1 + #1^3) &]

FriCAS also returns two arctangents

(8) -> integrate((2*u + u^2)/(2 + 4*u + 2*u^2 + u^4), u)

+-+ 3 2
(u - 1)\|2 u - u + 2 u + 2
atan(-----------) - atan(-----------------)
2 +-+
\|2
(8) -------------------------------------------
+-+
\|2
Type: Union(Expression(Integer),...)

Perhaps there is still some room for improvement in the log to arctan conversions in symbolic integrators...

Sam

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On 10/10/2023 11:06 PM, Sam Blake wrote:

I was pleased to see the following concise result from Rubi

In[8203]:= Int[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8203]= ArcTan[u^2/(Sqrt[2] (1 + u))]/Sqrt[2]

Fyi, the rules used are (3 rules)

TraceScan[Print[#/. Rubi`Private`ShowStep[i_,___]:>i]&,Int[(2 u+u^2)/(2+4 u+2 u^2+u^4),u],Rubi`Private`ShowStep[i_,___]]
1607
2119
209

Rule 1607 is

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol]
:> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; FreeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

2119 is

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_Symbol] :>
Dist[A^2*((m - n + 1)/(m + 1)), Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x], x, x^(m + 1)/(A*(m - n + 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x]
&& EqQ[n2, 2*n] && EqQ[k, 2*(m + 1)] && EqQ[a*B^2*(m + 1)^2 -
A^2*d*(m - n + 1)^2, 0] && EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]

Rule 209 is

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :>
Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

My implementation of Bronstein's Risch algorithm (from Symbolic Integration I) returns two arctangents

In[8206]:= Risch[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8206]= (ArcTan[(-1 + u)/Sqrt[2]] - ArcTan[(2 + 2 u + (-1 + u) u^2)/Sqrt[2]])/Sqrt[2]

and Mathematica returns the naive form

In[8207]:= Integrate[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8207]= 1/4 RootSum[
2 + 4 #1 + 2 #1^2 + #1^4 &, (2 Log[u - #1] #1 + Log[u - #1] #1^2)/(
1 + #1 + #1^3) &]

FriCAS also returns two arctangents

(8) -> integrate((2*u + u^2)/(2 + 4*u + 2*u^2 + u^4), u)

+-+ 3 2
(u - 1)\|2 u - u + 2 u + 2
atan(-----------) - atan(-----------------)
2 +-+
\|2
(8) -------------------------------------------
+-+
\|2
Type: Union(Expression(Integer),...)

Perhaps there is still some room for improvement in the log to arctan conversions in symbolic integrators...

Sam

--Nasser

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Sam Blake schrieb:

I was pleased to see the following concise result from Rubi

In[8203]:= Int[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8203]= ArcTan[u^2/(Sqrt[2] (1 + u))]/Sqrt[2]

My implementation of Bronstein's Risch algorithm (from Symbolic
Integration I) returns two arctangents

In[8206]:= Risch[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8206]= (ArcTan[(-1 + u)/Sqrt[2]] - ArcTan[(2 + 2 u + (-1 + u) u^2)/Sqrt[2]])/Sqrt[2]

and Mathematica returns the naive form

In[8207]:= Integrate[(2 u + u^2)/(2 + 4 u + 2 u^2 + u^4), u]
Out[8207]= 1/4 RootSum[
2 + 4 #1 + 2 #1^2 + #1^4 &, (2 Log[u - #1] #1 + Log[u - #1] #1^2)/(
1 + #1 + #1^3) &]

FriCAS also returns two arctangents

(8) -> integrate((2*u + u^2)/(2 + 4*u + 2*u^2 + u^4), u)

+-+ 3 2
(u - 1)\|2 u - u + 2 u + 2
atan(-----------) - atan(-----------------)
2 +-+
\|2
(8) -------------------------------------------
+-+
\|2
Type: Union(Expression(Integer),...)

Perhaps there is still some room for improvement in the log to arctan conversions in symbolic integrators...

Bronstein's Risch algorithm presumably involves Rioboo's decomposition
of arc tangents to ensure continuity on the real axis (see chapter 2.8
in his Symbolic Integration I). Indeed, ATAN(u^2/(SQRT(2)*(1 + u)))/
SQRT(2) is discontinuous at x = -1. However, a user can easily change
this antiderivative to -ATAN(SQRT(2)*(1 + u)/u^2)/SQRT(2), but he then
needs a limit to evaluate it at x = 0.

Martin.

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