Risch’s algorithm for integration
We want to integrate
−ex − x + ln(x)x + ln(x)xex
f :=
x(ex + x)2
The elementary field we obtain is Q(x)(θ1 , θ2 ) with θ1 = ex , θ2 = ln(x), and
so the integrand becomes
1
1 + θ1
−θ1 − x + θ2 x + θ2 xθ1
=
−
+
θ
2
x(θ1 + x)
x(θ1 + x)2
(θ1 + x)2
We set A0 = −
1
1 + θ1
and A1 =
and using
x(θ1 + x)
(θ1 + x)2
Z
A0 + A1 θ2 = B0 + B1 θ2 + B2 θ22
obtain by differentiation the equations:
0 = B02
A1 = B01 + 2B2 θ02 = B01 + 2B2
A0 = B00 + B1 θ02
1
x
Thus B2 is a constant. Integrating the second equation gives us that
Z
A1 dx − 2B2 θ2 = B1 − b1
where b1 is a constant. By a recursive call we find that
Z
Z
A1 =
1 + θ1
=
(θ1 + x)2
Z
1 + ex
1
1
=
−
=
−
ex + x
θ1 + x
(ex + x)2
As no θ2 -term is involved we find that B2 = 0, and we set B1 = B̄1 + b1 with
1
B̄1 = −
, and b1 still an unevaluated constant.
θ1 + x
Now we integrate the third equation. We get:
Z
A0 − B1 θ02 dx =
Z
A0 − B̄1 θ02 dx − b1 θ2 = B0 + c
(where c is a constant we will ignore – its the integration constant). Substituting values, we get
A0 − B̄1 θ02 = −
1
1
1
− (−
) =0
x(θ1 + x)
θ1 + x x
This shows that b1 = 0 and also B0 = 0. The integral thus is
Z
f = θ2 B1 = −θ2
1
1
= − ln(x) x
θ1 + x
e +x
Now for a more complicated example. We want to integrate the function
!2
( 12 ln(x + 12 ) − x) ln(x)2 − 4x
1
1
1 + 2 ln(x)
f = x
+
+
2 (x + 12 ) ln(x + 12 )
x
(x + 12 )2 ln(x + 12 )2
+
((x2 + x + 1) ln(x + 12 ) + x2 − 1) ln(x)
(x + 12 )2
+
(x − 1x ) ln(x + 12 )
x + 12
Our field extension now is Q(x)(θ1 , θ2 ) with θ1 = ln(x + 12 ) and θ2 = ln(x).
Substituting yields:
!2
( θ21 − x)θ22 − 4x ((x2 + x + 1)θ1 + x2 − 1)θ2 (x − 1x )θ1
1
1 + 2θ2
x
+
+
+
+
x
2(x + 12 )θ1
(x + 12 )2 θ21
(x + 12 )2
x + 12
We collect this as a polynomial in θ2 and obtain A2 θ22 + A1 θ2 + A0 with
A2 :=
θ1
2
2
4 (x2 + x + 1)θ1 + x2 − 1
+
+
,
(x + 12 )θ1 x
(x + 12 )2
!2
(x − 1x )θ1
1
1
x
A0 := x
+
+
−
.
2(x + 12 )θ1 x
x + 12
4(x + 12 )2 θ21
−x
4
+
,
(x + 12 )2 θ21 x
A1 :=
This gives us the equations:
0 = B03
A2 = B02 + 3B3 θ02 = B01 + 2B2
A1 = B01 + 2B2 θ02
A0 = B00 + B1 θ02
1
x
We integrate recursively, and obtain (see below)
Z
x
A2 = 4θ2 +
(x + 12 )θ1
As (with B3 constant):
Z
A2 = B2 + 3B3 θ2
4
x
and B2 = B̄2 + b2 with B̄2 =
. The equation for A1
3
(x + 12 )θ1
thus gives after integration:
Z
1
A1 − 2B̄2 dx = 2b2 θ2 + B1
x
The integral on the left hand side is evaluated recursively again:
we get that B3 =
Z
4 (x2 + x + 1) ln(x + 12 ) + x2 − 1
x2 − 1
1
+
dx
=
ln(x
+
) + 4 ln(x)
x
2
(x + 12 )2
x + 12
The only term involving θ2 is the second summand, thus we get that b2 =
4/2 = 2 and B1 = B̄1 + b1 with a constant b1 and
(x2 − 1)
B̄1 =
θ1
x + 12
Finally we integrate the last equation and get:
Z
1
A0 − B̄1 dx = b1 θ2 + B0
x
The integral is
Z
2x + 2x ln(x + 12 ) + ln(x + 12 )
x ln(x + 12 )(2x + 1)
1
dx = ln(x) + ln(ln(x + ))
2
and therefore b1 = 1 and B0 = ln(θ1 ) + c, where again c is the constant of
integration. The result thus is
Z
4
f = θ32 + (B̄2 + 2)θ22 + (B̄1 + 1)θ2 + B0
3
!
4
1 3
x
1 2
= ln(x + ) +
+
2
ln(x
+
)
3
2
2
(x + 12 ) ln(x + 12 )
!
1
2
(x − 1) ln(x + 2 )
1
1
+
+
1
ln(x
+
)
+
ln(ln(x
+
)) + c.
2
2
x + 12
It remains to solve the recursive integrals. The first one was the integral
of
A2 =
θ1
2
−x
+
2
(x + 12 )2 θ1
2θ1 − 4x
4 4
= +
x x (2x + 1)2 θ21
The polynomial part of this integrates easily to 4 ln(x).
The fractional part simplifies to:
1
x
1 2
−
1 2
2θ1 x + 2
x+ 2
θ1 2
We deal with the quadratic term in θ1 first. Setting
u=
x
x + 12
and
v=
1
θ1
we have (chain rule for θ01 !)
u0 =
1
2(x + 12 )2
v0 = −
and
1
θ21 (x + 12 )
and we get by partial integration (this is essentially equation (∗∗)):
Z
−
x
1 2
x+ 2
Z
θ1 2
=
0
uv = uv −
Z
x
−
uv=
(x + 12 )θ1
0
Z
1
2(x + 12 )2 θ1
Note that the integral subtracted is exactly the negative of the term only
involving θ1 . This gives the claimed antiderivative
The next integral to solve was that of
4 (x2 + x + 1)θ1 + x2 − 1 x2 + x + 1
4
x2 − 1
+
=
θ1 + +
x
x (x + 1/2)2
(x + 12 )2
(x + 12 )2
x2 + x + 1
4
x2 − 1
We thus have A1 =
and A0 = +
.
x (x + 1/2)2
(x + 12 )2
Recursively we evaluate (using partial fractions):
Z
A1 = x −
3
2(2x + 1)
3
and thus B2 = 0, B1 = B̄1 + b1 , B̄1 = x − 2(2x+1)
.
The next integral is
Z
1
=
A0 − B̄1
x + 12
Z
7x + 4
1
1
= 4 ln(x) − ln(x + )
x(2x + 1)
2
2
Thus b1 = −1/2 and B0 = 4 ln(x). This gives the claimed integral
1
x2 − 1
3
−
θ1 + 4 ln(x) =
θ1 + 4 ln(x)
(B̄1 + b1 )θ1 + B0 = x −
2(2x + 1) 2
x + 12
The third integral to be evaluated recursively is that of
2x + 2xθ1 + θ1 1
1
= +
xθ1 (2x + 1)
x θ1 (x + 12
The polynomial part yields the integral ln(x), the rational part has numerator
1
and denominator θ1 .
x + 12
We set
a :=
1
x+ 12
θ01
=1
and get that a0 = 0, so a is constant. Thus the integral of the rational part is
1
a · ln(θ1 ) = ln(ln(x + ))
2
and we are done.