RUBI and integration as term re-writing

Abstract

We describe the development of a term-rewriting system for indenite integration; it is also called a rule-based evaluation system. The development is separated into modules, and we describe the module for a wide class of integrands containing the tangent function.

Full text

⃝
f(x)
F(x)
•F′(x) = f(x)
•
•
•Æ
•
•
√−2 tan x dx =−1
2ln 2−2 tan x−2√−2 tan x
−arctan √−2 tan x−1
+1
2ln 2−2 tan x+ 2√−2 tan x
−arctan √−2 tan x+ 1,

√−2 tan x dx =
−[x]
2[x]−21−√2[x]
+2 1 + √2[x]
+1−√2[x] + [x]
−1 + √2[x] + [x] ,
= ln cos x+ ln(1 + √−2 tan x−tan x)
+ arctan 1 + tan x
√−2 tan x,
=−arctan √−2 tan x
1 + tan x+ arctanh √−2 tan x
1−tan x,
= arctan 1 + tan x
√−2 tan x+ arctanh 1−tan x
√−2 tan x.
f F
F′=f
dx
x=ln x ,
ln |x|,
(nπ, nπ +π/2) n∈Z
F′=f
√−2 tan x x =nπ +π/2
x=nπ
x= (n+
1/2)π
tan x=±1
x=nπ +π/2
x2+ 2
x4−3x2+ 4 dx = arctan x
2−x2,
= arctan 2x+√7+ arctan 2x−√7.
Æ

√2 tan x dx =√tan xcos xarccos (cos x−sin x)
√cos xsin x
−ln cos x+√2√tan xcos x+ sin x.
x→ −x
f(ax +b)dx
y=ax +b y
(m+n+ 1)
m+n+ 1 ̸= 0
T1= tan(c+dx)m m −1
m≥1
m+n+ 1 = 0
C= 0
tanm(c+dx)(a+btan(c+dx))n
∗(A+Btan(c+dx) + Ctan2(c+dx)) ,
a, b, c, d, A, B, C ∈Cm, n ∈R
m n
B=C= 0
a2+b2= 0
u= tan(c+dx)
u
u= tan((c+dx)/2)

tan(1 + i+x)∗
4−12 tan(1 + i+x) + 9 tan(1 + i+x)2
(2 −3 tan(1 + i+x))3/2dx .
m= 1 n=−3/2n≤ −1
Ab2−abB +
a2C= 0
−1
13 tan(1 + i+x)(−26 + 39 tan(1 + i+x))
2−3 tan(1 + i+x)dx .
tan(1 + i+x)2−3 tan(1 + i+x)dx .
dn T1Tn
2dx −→ Tn
2−dn Tn−1
2T3(b, −a, 0) dx
3 + 2 tan(1 + i+x)
2−3 tan(1 + i+x)dx + 22−3 tan(1 + i+x).
A2+B2̸= 0 a2+b2̸= 0
A+Btan(c+dx)
a+btan(c+dx)dx −→
1
2(A−Bi)1 + itan(c+dx)
a+btan(c+dx)dx+
1
2(A+Bi)1−itan(c+dx)
a+btan(c+dx)dx
A2+B2= 0 bA +aB ̸= 0
A+Btan(c+dx)
a+btan(c+dx)dx −→
−
2Barctanh √a+btan(c+dx)
√a+bA
B
da+bA
B
+ 22−3 tan(1 + i+x)
−√2−3iarctanh 2−3 tan(1 + i+x)
√2−3i
−√2 + 3iarctanh 2−3 tan(1 + i+x)
√2+3i
(sin(c+dx))−1
csc(c+dx) tan(c+dx)−1
cot(c+dx)

T1= tan(c+dx), T2=a+btan(c+dx),
T3(A, B, C ) = A+Btan(c+dx) + Ctan2(c+dx).
A, B, C ∈C
d(m+ 1) Tm
1Tn
2T3(A, B, C )dx =
AT m+1
1Tn
2+dTm+1
1Tn−1
2T3(ˆ
A, ˆ
B, ˆ
C)dx ,
ˆ
A=aB(m+ 1) −Abn ,
ˆ
B= (bB −aA +aC)(m+ 1) ,
ˆ
C=bC(m+ 1) −Ab(m+n+ 1) .
d(m+n+ 1) Tm
1Tn
2T3(A, B, C )dx =
CT m+1
1Tn
2+dTm
1Tn−1
2T3(ˆ
A, ˆ
B, ˆ
C)dx ,
ˆ
A=Aa(m+n+ 1) −C(m+ 1)a ,
ˆ
B= (aB +bA −bC)(m+n+ 1) ,
ˆ
C=aCn +bB(m+n+ 1) .
bd(n+ 1) a2+b2Tm
1Tn
2T3dx =
Ab2−abB +a2CTm
1Tn+1
2+dTm−1
1Tn+1
2ˆ
T3dx ,
ˆ
A=−Ab2−abB +a2Cm ,
ˆ
B=b(bB +aA −aC)(n+ 1) ,
ˆ
C= (m+n+ 1)(aB −Ab)b−ma2C+ (n+ 1)b2C .
ad(n+ 1) a2+b2Tm
1Tn
2T3dx =
−Ab2−abB +a2CTm+1
1Tn+1
2+dTm
1Tn+1
2ˆ
T3dx
ˆ
A=Aa2(n+ 1) + b2(m+n+ 2)
−a(bB −aC)(m+ 1) ,ˆ
B=a(aB −bA +bC)(n+ 1),
ˆ
C=Ab2−abB +a2C(m+n+ 2) .
bd(m+n+ 1) Tm
1Tn
2T3(A, B, C )dx =
CT m
1Tn+1
2−dTm−1
1Tn
2ˆ
T3dx ,
ˆ
A=aCm , ˆ
B=b(C−A)(m+n+ 1) ,
ˆ
C=aCm −bB(m+n+ 1) .
ad(m+ 1) Tm
1Tn
2T3(A, B, C )dx =
AT m+1
1Tn+1
2+dTm+1
1Tn
2ˆ
T3dx ,
ˆ
A=aB(m+ 1) −Ab(m+n+ 2) ,
ˆ
B=−a(A−C)(m+ 1) ,ˆ
C=−Ab(m+n+ 2) .
a2+b2= 0
T3
d(m+ 1) Tm
1Tn
2T3(A, B, 0) dx =
AaT m+1
1Tn−1
2−dTm+1
1Tn−1
2T3(ˆ
A, ˆ
B, 0) dx ,
ˆ
A=Ab(n−1) −(Ab +Ba)(m+ 1) ,
ˆ
B=Aa(m+n)−Bb(m+ 1) .
d(m+n)Tm
1Tn
2T3(A, B, 0) dx =
BbT m+1
1Tn−1
2+dTm
1Tn−1
2T3(ˆ
A, ˆ
B, 0) dx ,
ˆ
A=Aa(n+m)−Bb(m+ 1) ,
ˆ
B=Ba(n−1) + (Ab +Ba)(m+n).
2a2nd Tm
1Tn
2T3(A, B, 0) dx =
BbT m
1Tn
2+dTm−1
1Tn+1
2T3(ˆ
A, ˆ
B, 0) dx ,
ˆ
A= (Ab −Ba)m , ˆ
B=Bb(m−n) + Aa(m+n).
2a2nd Tm
1Tn
2T3(A, B, 0) dx =
−a(aA +bB)Tm+1
1Tn
2+dTm
1Tn+1
2ˆ
T3(ˆ
A, ˆ
B, 0) dx ,
ˆ
A=bB(m+ 1) + aA(m+ 2n+ 1) ,
ˆ
B= (aB −Ab)(m+n+ 1) .

ad(m+n)Tm
1Tn
2T3(A, B, 0) dx =
aBT m
1Tn
2+dTm−1
1Tn
2ˆ
T3(ˆ
A, ˆ
B, 0) dx ,
ˆ
A=−aBm , ˆ
B=Aam + (Aa −Bb)n .
ad(m+ 1) Tm
1Tn
2T3(A, B, 0) dx =
aAT m+1
1Tn
2+dTm+1
1Tn
2ˆ
T3(ˆ
A, ˆ
B, 0) dx ,
ˆ
A=Abn −Ba(m+ 1) ,ˆ
B=Aa(m+n+ 1) .