Risch’s algorithm for integration

 We want to integrate

 f  :=−e x

− x+ ln( x) x+ ln( x) xe x

 x(e x + x)2

 The elementary fi eld we obtain is Q( x)(θ1,θ2) with θ1 = e x, θ2 = ln( x), and

so the integrand becomes

−θ1− x+θ2 x+θ2 xθ1

 x(θ1 + x)2= −

1

 x(θ1 + x)+θ2

1+θ1

(θ1 + x)2

 We set A0 = −1

 x(θ1 + x)and A1 =

1+θ1

(θ1 + x)2and using

 A0 + A1θ2 = B0 + B1θ2 + B2θ

2

2

obtain by differentiation the equations:

0 = B2

 A1 = B1 +2 B2θ

2 = B1 +2 B2

1

 x

 A0 = B0 + B1θ

2

 Thus B2 is a constant. Integrating the second equation gives us that  

A1d x−2 B2θ2 = B1−b1

 where b1 is a constant. By a recursive call we fi nd that  

A1 = 

1+θ1

(θ1 + x)2=

 1+e x

(e x + x)2= −

1

e x + x= −

1

θ1 + x

 As no θ2-term is involved we fi nd that B2 = 0, and we set B1 = ¯ B1 +b1 with

¯ B1 = −1

θ1 + x, and b1 still an unevaluated constant.

 Now we integrate the third equation. We get: 

A0− B1θ

2d x =

 A0−

¯ B1θ

2d x−b1θ2 = B0 + c

 

(where c is a constant we will ignore – its the integration constant). Substi-

tuting values, we get 

 A0−¯ B1θ

2 = −1

 x(θ1 + x)− (−

1

θ1 + x)1

 x= 0

 This shows that b1 = 0 and also B0 = 0. The integral thus is 

f  = θ2 B1 = −θ21

θ1 + x= − ln( x)

1

e x + x

 Now for a more complicated example. We want to integrate the function

 f  = x

1

2

1

( x+ 1

2) ln( x+ 1

2)

+1+2ln( x)

 x

2

+(12ln( x+ 1

2)− x) ln( x)2− x

4

( x+ 1

2)2 ln( x+ 1

2)2

+(( x2 + x+1) ln( x+ 1

2) + x2−1) ln( x)

( x+ 1

2)2

+( x− 1

 x) ln( x+ 1

2)

 x+ 1

2

Our fi eld extension now is Q( x)(θ1,θ2) with θ1 = ln( x + 1

2) and θ2 = ln( x).

Substituting yields:

 x

1

2( x+ 1

2)θ1

+1+2θ2

 x

2

+(θ12− x)θ2

2−

x4

( x+ 1

2)2θ2

1

+(( x2 + x+1)θ1 + x2−1)θ2

( x+ 1

2)2

+( x− 1

 x)θ1

 x+ 1

2

 We collect this as a polynomial in θ2 and obtain A2θ2

2+ A1θ2 + A0 with

 A2 :=θ12− x

( x+ 1

2)2θ2

1

+4

 x, A1 :=

2

( x+ 1

2)θ1

+4

 x+

( x2 + x+1)θ1 + x2−1

( x+ 1

2)2

,

 A0 := x

1

2( x+ 1

2)θ1

+1

 x

2

+( x− 1

 x)θ1

 x+ 1

2

− x

4( x+ 1

2)2θ2

1

.

 This gives us the equations:

0 = B3

 A2 = B2 +3 B3θ

2 = B1 +2 B2

1

 x A1 = B1 +2 B2θ

2

 A0 = B0 + B1θ

2

 

 We integrate recursively, and obtain (see below) A2 = 4θ2 +

x

( x+ 1

2)θ1

 As (with B3 constant):  A2 = B2 +3 B3θ2

 we get that B3 =4

3and B2 = ¯ B2 +b2 with ¯ B2 =

x

( x+ 1

2)θ1

. The equation for A1

thus gives after integration: A1−2 ¯ B2

1

 xd x = 2b2θ2 + B1

 The integral on the left hand side is evaluated recursively again: 

4

 x+

( x2 + x+1) ln( x+ 1

2) + x2−1

( x+ 1

2)2

d x =x2−1

 x+ 1

2

ln( x+1

2) +4ln( x)

 The only term involving θ2 is the second summand, thus we get that  b2 =4/2 = 2 and B1 = ¯ B1 +b1 with a constant b1 and

¯ B1 =( x2−1)

 x+ 1

2

θ1

Finally we integrate the last equation and get: A0−

¯ B1

1

 xd x = b1θ2 + B0

 The integral is 

2 x+2 x ln( x+ 1

2) + ln( x+ 1

2)

 x ln( x+ 1

2)(2 x+1)

d x = ln( x) + ln(ln( x+1

2))

and therefore b1 = 1 and B0 = ln(θ1) + c, where again c is the constant of 

integration. The result thus is f  =

4

3θ3

2+ ( ¯ B2 +2)θ22 + ( ¯ B1 +1)θ2 + B0

=4

3ln( x+

1

2)3 +

x

( x+ 1

2) ln( x+ 1

2)

+2

ln( x+1

2)2

+

( x2−1) ln( x+ 1

2)

 x+ 1

2

+1

ln( x+1

2) + ln(ln( x+

1

2)) + c.

 

It remains to solve the recursive integrals. The fi rst one was the integral

of 

 A2 =θ12− x

( x+ 1

2)2θ2

1

+4

 x=

4

 x+

2θ1−4 x

(2 x+1)2θ21

 The polynomial part of this integrates easily to 4ln( x).

 The fractional part simplifi es to:

1

2θ1

 x+ 1

2

2−

 x

 x+ 1

2

2θ1

2

 We deal with the quadratic term in θ1 fi rst. Setting

u =x

 x+ 1

2

and v =1

θ1

 we have (chain rule for θ1

!)

u =1

2( x+ 1

2)2

and v = −1

θ21

( x+ 1

2)

and we get by partial integration (this is essentially equation (∗∗) ): −

 x

 x+ 1

2

2θ1

2=

 uv = uv−

 uv =

x

( x+ 1

2)θ1

−

 1

2( x+ 1

2)2θ1

 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

 x+

( x2 + x+1)θ1 + x2−1

( x+ 1

2)2

=x2 + x+1

( x+ 1

2)2

θ1 +4

 x+

x2−1

( x+1/2)2

 We thus have A1 =x2 + x+1

( x+ 1

2)2

and A0 =4

 x+

x2−1

( x+1/2)2.

Recursively we evaluate (using partial fractions): 

A1 = x−3

2(2 x+1)

 

and thus B2 = 0, B1 = ¯ B1 +b1, ¯ B1 = x− 3

2(2 x+1).

 The next integral is 

A0−¯ B1

1

 x+ 1

2

= 

7 x+4

 x(2 x+1)= 4ln( x)−

1

2ln( x+

1

2)

 Thus b1 = −1/2 and B0 = 4ln( x). This gives the claimed integral

( ¯ B1 +b1)θ1 + B0 =

 x−3

2(2 x+1)−

1

2

θ1 +4ln( x) =x2−1

 x+ 1

2

θ1 +4ln( x)

 The third integral to be evaluated recursively is that of 

2 x+2 xθ1 +θ1

 xθ1 (2 x+1)=

1

 x+

1

θ1( x+ 1

2

 The polynomial part yields the integral ln( x), the rational part has numerator 1

 x+ 1

2

and denominator θ1.

 We set 

a :=

1

 x+1

2

θ1

= 1

and get that a = 0, so a is constant. Thus the integral of the rational part is

a · ln(θ1) = ln(ln( x+1

2))

and we are done.