\(\int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx\) [40]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 276 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx=\frac {-A b^3+a \left (b^2 B-a b C+a^2 D\right )}{2 b^2 (b c-a d)^2 (a+b x)^2}-\frac {b^3 (B c-2 A d)-a b^2 (2 c C-B d)+3 a^2 b c D-a^3 d D}{b^2 (b c-a d)^3 (a+b x)}+\frac {c^2 C d-B c d^2+A d^3-c^3 D}{d (b c-a d)^3 (c+d x)}+\frac {\left (b \left (c^2 C-2 B c d+3 A d^2\right )+a \left (2 c C d-B d^2-3 c^2 D\right )\right ) \log (a+b x)}{(b c-a d)^4}-\frac {\left (b \left (c^2 C-2 B c d+3 A d^2\right )+a \left (2 c C d-B d^2-3 c^2 D\right )\right ) \log (c+d x)}{(b c-a d)^4} \] Output:

1/2*(-A*b^3+a*(B*b^2-C*a*b+D*a^2))/b^2/(-a*d+b*c)^2/(b*x+a)^2-(b^3*(-2*A*d 
+B*c)-a*b^2*(-B*d+2*C*c)+3*a^2*b*c*D-a^3*d*D)/b^2/(-a*d+b*c)^3/(b*x+a)+(A* 
d^3-B*c*d^2+C*c^2*d-D*c^3)/d/(-a*d+b*c)^3/(d*x+c)+(b*(3*A*d^2-2*B*c*d+C*c^ 
2)+a*(-B*d^2+2*C*c*d-3*D*c^2))*ln(b*x+a)/(-a*d+b*c)^4-(b*(3*A*d^2-2*B*c*d+ 
C*c^2)+a*(-B*d^2+2*C*c*d-3*D*c^2))*ln(d*x+c)/(-a*d+b*c)^4
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx=\frac {-A b^3+a \left (b^2 B-a b C+a^2 D\right )}{2 b^2 (b c-a d)^2 (a+b x)^2}+\frac {b^3 (-B c+2 A d)+a b^2 (2 c C-B d)-3 a^2 b c D+a^3 d D}{b^2 (b c-a d)^3 (a+b x)}+\frac {-c^2 C d+B c d^2-A d^3+c^3 D}{d (-b c+a d)^3 (c+d x)}+\frac {\left (b \left (c^2 C-2 B c d+3 A d^2\right )-a \left (-2 c C d+B d^2+3 c^2 D\right )\right ) \log (a+b x)}{(b c-a d)^4}+\frac {\left (-b \left (c^2 C-2 B c d+3 A d^2\right )+a \left (-2 c C d+B d^2+3 c^2 D\right )\right ) \log (c+d x)}{(b c-a d)^4} \] Input:

Integrate[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^3*(c + d*x)^2),x]
 

Output:

(-(A*b^3) + a*(b^2*B - a*b*C + a^2*D))/(2*b^2*(b*c - a*d)^2*(a + b*x)^2) + 
 (b^3*(-(B*c) + 2*A*d) + a*b^2*(2*c*C - B*d) - 3*a^2*b*c*D + a^3*d*D)/(b^2 
*(b*c - a*d)^3*(a + b*x)) + (-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D)/(d*(-(b 
*c) + a*d)^3*(c + d*x)) + ((b*(c^2*C - 2*B*c*d + 3*A*d^2) - a*(-2*c*C*d + 
B*d^2 + 3*c^2*D))*Log[a + b*x])/(b*c - a*d)^4 + ((-(b*(c^2*C - 2*B*c*d + 3 
*A*d^2)) + a*(-2*c*C*d + B*d^2 + 3*c^2*D))*Log[c + d*x])/(b*c - a*d)^4
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2123, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx\)

\(\Big \downarrow \) 2123

\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b (a+b x)^3 (b c-a d)^2}+\frac {a^3 (-d) D+3 a^2 b c D-a b^2 (2 c C-B d)+b^3 (B c-2 A d)}{b (a+b x)^2 (b c-a d)^3}+\frac {b \left (a \left (-B d^2-3 c^2 D+2 c C d\right )+b \left (3 A d^2-2 B c d+c^2 C\right )\right )}{(a+b x) (b c-a d)^4}+\frac {d \left (-a \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (3 A d^2-2 B c d+c^2 C\right )\right )}{(c+d x) (b c-a d)^4}+\frac {-A d^3+B c d^2+c^3 D-c^2 C d}{(c+d x)^2 (b c-a d)^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^2 (a+b x)^2 (b c-a d)^2}-\frac {a^3 (-d) D+3 a^2 b c D-a b^2 (2 c C-B d)+b^3 (B c-2 A d)}{b^2 (a+b x) (b c-a d)^3}+\frac {\log (a+b x) \left (a \left (-B d^2-3 c^2 D+2 c C d\right )+b \left (3 A d^2-2 B c d+c^2 C\right )\right )}{(b c-a d)^4}-\frac {\log (c+d x) \left (a \left (-B d^2-3 c^2 D+2 c C d\right )+b \left (3 A d^2-2 B c d+c^2 C\right )\right )}{(b c-a d)^4}+\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{d (c+d x) (b c-a d)^3}\)

Input:

Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^3*(c + d*x)^2),x]
 

Output:

-1/2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(b^2*(b*c - a*d)^2*(a + b*x)^2) - 
 (b^3*(B*c - 2*A*d) - a*b^2*(2*c*C - B*d) + 3*a^2*b*c*D - a^3*d*D)/(b^2*(b 
*c - a*d)^3*(a + b*x)) + (c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)/(d*(b*c - a*d 
)^3*(c + d*x)) + ((b*(c^2*C - 2*B*c*d + 3*A*d^2) + a*(2*c*C*d - B*d^2 - 3* 
c^2*D))*Log[a + b*x])/(b*c - a*d)^4 - ((b*(c^2*C - 2*B*c*d + 3*A*d^2) + a* 
(2*c*C*d - B*d^2 - 3*c^2*D))*Log[c + d*x])/(b*c - a*d)^4
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2123
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 
Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.01

method result size
default \(-\frac {A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}}{\left (a d -b c \right )^{3} d \left (x d +c \right )}-\frac {\left (3 b \,d^{2} A -B a \,d^{2}-2 B b c d +2 C a c d +C b \,c^{2}-3 a \,c^{2} D\right ) \ln \left (x d +c \right )}{\left (a d -b c \right )^{4}}+\frac {\left (3 b \,d^{2} A -B a \,d^{2}-2 B b c d +2 C a c d +C b \,c^{2}-3 a \,c^{2} D\right ) \ln \left (b x +a \right )}{\left (a d -b c \right )^{4}}-\frac {b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D}{2 b^{2} \left (a d -b c \right )^{2} \left (b x +a \right )^{2}}-\frac {2 b^{3} d A -B a \,b^{2} d -B \,b^{3} c +2 a \,b^{2} c C +a^{3} d D-3 a^{2} b c D}{\left (a d -b c \right )^{3} b^{2} \left (b x +a \right )}\) \(278\)
norman \(\frac {-\frac {2 A \,a^{2} b^{2} d^{3}+5 A a \,b^{3} c \,d^{2}-A \,b^{4} c^{2} d -5 B \,a^{2} b^{2} c \,d^{2}-B a \,b^{3} c^{2} d +C \,a^{3} b c \,d^{2}+5 C \,a^{2} b^{2} c^{2} d +D a^{4} c \,d^{2}-5 D a^{3} b \,c^{2} d -2 D a^{2} b^{2} c^{3}}{2 b^{2} d \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}-\frac {\left (3 A \,b^{3} d^{3}-B a \,b^{2} d^{3}-2 B \,b^{3} c \,d^{2}+2 C a \,b^{2} c \,d^{2}+C \,b^{3} c^{2} d +a^{3} d^{3} D-3 D a^{2} b c \,d^{2}-D b^{3} c^{3}\right ) x^{2}}{b d \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}-\frac {\left (9 A a \,b^{3} d^{3}+3 A \,b^{4} c \,d^{2}-3 B \,a^{2} b^{2} d^{3}-7 B a \,b^{3} c \,d^{2}-2 B \,b^{4} c^{2} d +C \,a^{3} b \,d^{3}+3 C \,a^{2} b^{2} c \,d^{2}+8 C a \,b^{3} c^{2} d +D a^{4} d^{3}-3 D a^{3} b c \,d^{2}-6 D a^{2} b^{2} c^{2} d -4 D a \,b^{3} c^{3}\right ) x}{2 b^{2} d \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}}{\left (b x +a \right )^{2} \left (x d +c \right )}+\frac {\left (3 b \,d^{2} A -B a \,d^{2}-2 B b c d +2 C a c d +C b \,c^{2}-3 a \,c^{2} D\right ) \ln \left (b x +a \right )}{d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}}-\frac {\left (3 b \,d^{2} A -B a \,d^{2}-2 B b c d +2 C a c d +C b \,c^{2}-3 a \,c^{2} D\right ) \ln \left (x d +c \right )}{d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}}\) \(677\)
parallelrisch \(\text {Expression too large to display}\) \(1927\)

Input:

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

-(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/(a*d-b*c)^3/d/(d*x+c)-(3*A*b*d^2-B*a*d^2-2* 
B*b*c*d+2*C*a*c*d+C*b*c^2-3*D*a*c^2)/(a*d-b*c)^4*ln(d*x+c)+(3*A*b*d^2-B*a* 
d^2-2*B*b*c*d+2*C*a*c*d+C*b*c^2-3*D*a*c^2)/(a*d-b*c)^4*ln(b*x+a)-1/2*(A*b^ 
3-B*a*b^2+C*a^2*b-D*a^3)/b^2/(a*d-b*c)^2/(b*x+a)^2-(2*A*b^3*d-B*a*b^2*d-B* 
b^3*c+2*C*a*b^2*c+D*a^3*d-3*D*a^2*b*c)/(a*d-b*c)^3/b^2/(b*x+a)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1420 vs. \(2 (275) = 550\).

Time = 0.12 (sec) , antiderivative size = 1420, normalized size of antiderivative = 5.14 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx=\text {Too large to display} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^2,x, algorithm="fricas")
 

Output:

-1/2*(2*D*a^2*b^3*c^4 + 2*A*a^3*b^2*d^4 + (3*D*a^3*b^2 - 5*C*a^2*b^3 + B*a 
*b^4 + A*b^5)*c^3*d - 2*(3*D*a^4*b - 2*C*a^3*b^2 - 2*B*a^2*b^3 + 3*A*a*b^4 
)*c^2*d^2 + (D*a^5 + C*a^4*b - 5*B*a^3*b^2 + 3*A*a^2*b^3)*c*d^3 + 2*(D*b^5 
*c^4 - (D*a*b^4 + C*b^5)*c^3*d + (3*D*a^2*b^3 - C*a*b^4 + 2*B*b^5)*c^2*d^2 
 - (4*D*a^3*b^2 - 2*C*a^2*b^3 + B*a*b^4 + 3*A*b^5)*c*d^3 + (D*a^4*b - B*a^ 
2*b^3 + 3*A*a*b^4)*d^4)*x^2 + (4*D*a*b^4*c^4 + 2*(D*a^2*b^3 - 4*C*a*b^4 + 
B*b^5)*c^3*d - (3*D*a^3*b^2 - 5*C*a^2*b^3 - 5*B*a*b^4 + 3*A*b^5)*c^2*d^2 - 
 2*(2*D*a^4*b - C*a^3*b^2 + 2*B*a^2*b^3 + 3*A*a*b^4)*c*d^3 + (D*a^5 + C*a^ 
4*b - 3*B*a^3*b^2 + 9*A*a^2*b^3)*d^4)*x + 2*((3*D*a^3*b^2 - C*a^2*b^3)*c^3 
*d - 2*(C*a^3*b^2 - B*a^2*b^3)*c^2*d^2 + (B*a^3*b^2 - 3*A*a^2*b^3)*c*d^3 + 
 ((3*D*a*b^4 - C*b^5)*c^2*d^2 - 2*(C*a*b^4 - B*b^5)*c*d^3 + (B*a*b^4 - 3*A 
*b^5)*d^4)*x^3 + ((3*D*a*b^4 - C*b^5)*c^3*d + 2*(3*D*a^2*b^3 - 2*C*a*b^4 + 
 B*b^5)*c^2*d^2 - (4*C*a^2*b^3 - 5*B*a*b^4 + 3*A*b^5)*c*d^3 + 2*(B*a^2*b^3 
 - 3*A*a*b^4)*d^4)*x^2 + (2*(3*D*a^2*b^3 - C*a*b^4)*c^3*d + (3*D*a^3*b^2 - 
 5*C*a^2*b^3 + 4*B*a*b^4)*c^2*d^2 - 2*(C*a^3*b^2 - 2*B*a^2*b^3 + 3*A*a*b^4 
)*c*d^3 + (B*a^3*b^2 - 3*A*a^2*b^3)*d^4)*x)*log(b*x + a) - 2*((3*D*a^3*b^2 
 - C*a^2*b^3)*c^3*d - 2*(C*a^3*b^2 - B*a^2*b^3)*c^2*d^2 + (B*a^3*b^2 - 3*A 
*a^2*b^3)*c*d^3 + ((3*D*a*b^4 - C*b^5)*c^2*d^2 - 2*(C*a*b^4 - B*b^5)*c*d^3 
 + (B*a*b^4 - 3*A*b^5)*d^4)*x^3 + ((3*D*a*b^4 - C*b^5)*c^3*d + 2*(3*D*a^2* 
b^3 - 2*C*a*b^4 + B*b^5)*c^2*d^2 - (4*C*a^2*b^3 - 5*B*a*b^4 + 3*A*b^5)*...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1912 vs. \(2 (253) = 506\).

Time = 42.77 (sec) , antiderivative size = 1912, normalized size of antiderivative = 6.93 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx=\text {Too large to display} \] Input:

integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**3/(d*x+c)**2,x)
 

Output:

(-2*A*a**2*b**2*d**3 - 5*A*a*b**3*c*d**2 + A*b**4*c**2*d + 5*B*a**2*b**2*c 
*d**2 + B*a*b**3*c**2*d - C*a**3*b*c*d**2 - 5*C*a**2*b**2*c**2*d - D*a**4* 
c*d**2 + 5*D*a**3*b*c**2*d + 2*D*a**2*b**2*c**3 + x**2*(-6*A*b**4*d**3 + 2 
*B*a*b**3*d**3 + 4*B*b**4*c*d**2 - 4*C*a*b**3*c*d**2 - 2*C*b**4*c**2*d - 2 
*D*a**3*b*d**3 + 6*D*a**2*b**2*c*d**2 + 2*D*b**4*c**3) + x*(-9*A*a*b**3*d* 
*3 - 3*A*b**4*c*d**2 + 3*B*a**2*b**2*d**3 + 7*B*a*b**3*c*d**2 + 2*B*b**4*c 
**2*d - C*a**3*b*d**3 - 3*C*a**2*b**2*c*d**2 - 8*C*a*b**3*c**2*d - D*a**4* 
d**3 + 3*D*a**3*b*c*d**2 + 6*D*a**2*b**2*c**2*d + 4*D*a*b**3*c**3))/(2*a** 
5*b**2*c*d**4 - 6*a**4*b**3*c**2*d**3 + 6*a**3*b**4*c**3*d**2 - 2*a**2*b** 
5*c**4*d + x**3*(2*a**3*b**4*d**5 - 6*a**2*b**5*c*d**4 + 6*a*b**6*c**2*d** 
3 - 2*b**7*c**3*d**2) + x**2*(4*a**4*b**3*d**5 - 10*a**3*b**4*c*d**4 + 6*a 
**2*b**5*c**2*d**3 + 2*a*b**6*c**3*d**2 - 2*b**7*c**4*d) + x*(2*a**5*b**2* 
d**5 - 2*a**4*b**3*c*d**4 - 6*a**3*b**4*c**2*d**3 + 10*a**2*b**5*c**3*d**2 
 - 4*a*b**6*c**4*d)) + (-3*A*b*d**2 + B*a*d**2 + 2*B*b*c*d - 2*C*a*c*d - C 
*b*c**2 + 3*D*a*c**2)*log(x + (-3*A*a*b*d**3 - 3*A*b**2*c*d**2 + B*a**2*d* 
*3 + 3*B*a*b*c*d**2 + 2*B*b**2*c**2*d - 2*C*a**2*c*d**2 - 3*C*a*b*c**2*d - 
 C*b**2*c**3 + 3*D*a**2*c**2*d + 3*D*a*b*c**3 - a**5*d**5*(-3*A*b*d**2 + B 
*a*d**2 + 2*B*b*c*d - 2*C*a*c*d - C*b*c**2 + 3*D*a*c**2)/(a*d - b*c)**4 + 
5*a**4*b*c*d**4*(-3*A*b*d**2 + B*a*d**2 + 2*B*b*c*d - 2*C*a*c*d - C*b*c**2 
 + 3*D*a*c**2)/(a*d - b*c)**4 - 10*a**3*b**2*c**2*d**3*(-3*A*b*d**2 + B...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (275) = 550\).

Time = 0.07 (sec) , antiderivative size = 703, normalized size of antiderivative = 2.55 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx=-\frac {{\left ({\left (3 \, D a - C b\right )} c^{2} - 2 \, {\left (C a - B b\right )} c d + {\left (B a - 3 \, A b\right )} d^{2}\right )} \log \left (b x + a\right )}{b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}} + \frac {{\left ({\left (3 \, D a - C b\right )} c^{2} - 2 \, {\left (C a - B b\right )} c d + {\left (B a - 3 \, A b\right )} d^{2}\right )} \log \left (d x + c\right )}{b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}} - \frac {2 \, D a^{2} b^{2} c^{3} - 2 \, A a^{2} b^{2} d^{3} + {\left (5 \, D a^{3} b - 5 \, C a^{2} b^{2} + B a b^{3} + A b^{4}\right )} c^{2} d - {\left (D a^{4} + C a^{3} b - 5 \, B a^{2} b^{2} + 5 \, A a b^{3}\right )} c d^{2} + 2 \, {\left (D b^{4} c^{3} - C b^{4} c^{2} d + {\left (3 \, D a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4}\right )} c d^{2} - {\left (D a^{3} b - B a b^{3} + 3 \, A b^{4}\right )} d^{3}\right )} x^{2} + {\left (4 \, D a b^{3} c^{3} + 2 \, {\left (3 \, D a^{2} b^{2} - 4 \, C a b^{3} + B b^{4}\right )} c^{2} d + {\left (3 \, D a^{3} b - 3 \, C a^{2} b^{2} + 7 \, B a b^{3} - 3 \, A b^{4}\right )} c d^{2} - {\left (D a^{4} + C a^{3} b - 3 \, B a^{2} b^{2} + 9 \, A a b^{3}\right )} d^{3}\right )} x}{2 \, {\left (a^{2} b^{5} c^{4} d - 3 \, a^{3} b^{4} c^{3} d^{2} + 3 \, a^{4} b^{3} c^{2} d^{3} - a^{5} b^{2} c d^{4} + {\left (b^{7} c^{3} d^{2} - 3 \, a b^{6} c^{2} d^{3} + 3 \, a^{2} b^{5} c d^{4} - a^{3} b^{4} d^{5}\right )} x^{3} + {\left (b^{7} c^{4} d - a b^{6} c^{3} d^{2} - 3 \, a^{2} b^{5} c^{2} d^{3} + 5 \, a^{3} b^{4} c d^{4} - 2 \, a^{4} b^{3} d^{5}\right )} x^{2} + {\left (2 \, a b^{6} c^{4} d - 5 \, a^{2} b^{5} c^{3} d^{2} + 3 \, a^{3} b^{4} c^{2} d^{3} + a^{4} b^{3} c d^{4} - a^{5} b^{2} d^{5}\right )} x\right )}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^2,x, algorithm="maxima")
 

Output:

-((3*D*a - C*b)*c^2 - 2*(C*a - B*b)*c*d + (B*a - 3*A*b)*d^2)*log(b*x + a)/ 
(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4) + 
((3*D*a - C*b)*c^2 - 2*(C*a - B*b)*c*d + (B*a - 3*A*b)*d^2)*log(d*x + c)/( 
b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4) - 1 
/2*(2*D*a^2*b^2*c^3 - 2*A*a^2*b^2*d^3 + (5*D*a^3*b - 5*C*a^2*b^2 + B*a*b^3 
 + A*b^4)*c^2*d - (D*a^4 + C*a^3*b - 5*B*a^2*b^2 + 5*A*a*b^3)*c*d^2 + 2*(D 
*b^4*c^3 - C*b^4*c^2*d + (3*D*a^2*b^2 - 2*C*a*b^3 + 2*B*b^4)*c*d^2 - (D*a^ 
3*b - B*a*b^3 + 3*A*b^4)*d^3)*x^2 + (4*D*a*b^3*c^3 + 2*(3*D*a^2*b^2 - 4*C* 
a*b^3 + B*b^4)*c^2*d + (3*D*a^3*b - 3*C*a^2*b^2 + 7*B*a*b^3 - 3*A*b^4)*c*d 
^2 - (D*a^4 + C*a^3*b - 3*B*a^2*b^2 + 9*A*a*b^3)*d^3)*x)/(a^2*b^5*c^4*d - 
3*a^3*b^4*c^3*d^2 + 3*a^4*b^3*c^2*d^3 - a^5*b^2*c*d^4 + (b^7*c^3*d^2 - 3*a 
*b^6*c^2*d^3 + 3*a^2*b^5*c*d^4 - a^3*b^4*d^5)*x^3 + (b^7*c^4*d - a*b^6*c^3 
*d^2 - 3*a^2*b^5*c^2*d^3 + 5*a^3*b^4*c*d^4 - 2*a^4*b^3*d^5)*x^2 + (2*a*b^6 
*c^4*d - 5*a^2*b^5*c^3*d^2 + 3*a^3*b^4*c^2*d^3 + a^4*b^3*c*d^4 - a^5*b^2*d 
^5)*x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.63 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx=-\frac {{\left (3 \, D a c^{2} d - C b c^{2} d - 2 \, C a c d^{2} + 2 \, B b c d^{2} + B a d^{3} - 3 \, A b d^{3}\right )} \log \left ({\left | b - \frac {b c}{d x + c} + \frac {a d}{d x + c} \right |}\right )}{b^{4} c^{4} d - 4 \, a b^{3} c^{3} d^{2} + 6 \, a^{2} b^{2} c^{2} d^{3} - 4 \, a^{3} b c d^{4} + a^{4} d^{5}} - \frac {\frac {D c^{3} d^{2}}{d x + c} - \frac {C c^{2} d^{3}}{d x + c} + \frac {B c d^{4}}{d x + c} - \frac {A d^{5}}{d x + c}}{b^{3} c^{3} d^{3} - 3 \, a b^{2} c^{2} d^{4} + 3 \, a^{2} b c d^{5} - a^{3} d^{6}} - \frac {6 \, D a^{2} b c d - 4 \, C a b^{2} c d + 2 \, B b^{3} c d - D a^{3} d^{2} - C a^{2} b d^{2} + 3 \, B a b^{2} d^{2} - 5 \, A b^{3} d^{2} - \frac {2 \, {\left (3 \, D a^{2} b c^{2} d^{2} - 2 \, C a b^{2} c^{2} d^{2} + B b^{3} c^{2} d^{2} - 3 \, D a^{3} c d^{3} + C a^{2} b c d^{3} + B a b^{2} c d^{3} - 3 \, A b^{3} c d^{3} + C a^{3} d^{4} - 2 \, B a^{2} b d^{4} + 3 \, A a b^{2} d^{4}\right )}}{{\left (d x + c\right )} d}}{2 \, {\left (b c - a d\right )}^{4} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )}^{2}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^2,x, algorithm="giac")
 

Output:

-(3*D*a*c^2*d - C*b*c^2*d - 2*C*a*c*d^2 + 2*B*b*c*d^2 + B*a*d^3 - 3*A*b*d^ 
3)*log(abs(b - b*c/(d*x + c) + a*d/(d*x + c)))/(b^4*c^4*d - 4*a*b^3*c^3*d^ 
2 + 6*a^2*b^2*c^2*d^3 - 4*a^3*b*c*d^4 + a^4*d^5) - (D*c^3*d^2/(d*x + c) - 
C*c^2*d^3/(d*x + c) + B*c*d^4/(d*x + c) - A*d^5/(d*x + c))/(b^3*c^3*d^3 - 
3*a*b^2*c^2*d^4 + 3*a^2*b*c*d^5 - a^3*d^6) - 1/2*(6*D*a^2*b*c*d - 4*C*a*b^ 
2*c*d + 2*B*b^3*c*d - D*a^3*d^2 - C*a^2*b*d^2 + 3*B*a*b^2*d^2 - 5*A*b^3*d^ 
2 - 2*(3*D*a^2*b*c^2*d^2 - 2*C*a*b^2*c^2*d^2 + B*b^3*c^2*d^2 - 3*D*a^3*c*d 
^3 + C*a^2*b*c*d^3 + B*a*b^2*c*d^3 - 3*A*b^3*c*d^3 + C*a^3*d^4 - 2*B*a^2*b 
*d^4 + 3*A*a*b^2*d^4)/((d*x + c)*d))/((b*c - a*d)^4*(b - b*c/(d*x + c) + a 
*d/(d*x + c))^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^2} \,d x \] Input:

int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^3*(c + d*x)^2),x)
 

Output:

int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^3*(c + d*x)^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 1223, normalized size of antiderivative = 4.43 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^2} \, dx =\text {Too large to display} \] Input:

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^2,x)
 

Output:

(8*log(a + b*x)*a**3*b**2*c*d**2 + 8*log(a + b*x)*a**3*b**2*d**3*x - 4*log 
(a + b*x)*a**3*b*c**3*d - 4*log(a + b*x)*a**3*b*c**2*d**2*x + 4*log(a + b* 
x)*a**2*b**3*c**2*d + 20*log(a + b*x)*a**2*b**3*c*d**2*x + 16*log(a + b*x) 
*a**2*b**3*d**3*x**2 - 2*log(a + b*x)*a**2*b**2*c**4 - 10*log(a + b*x)*a** 
2*b**2*c**3*d*x - 8*log(a + b*x)*a**2*b**2*c**2*d**2*x**2 + 8*log(a + b*x) 
*a*b**4*c**2*d*x + 16*log(a + b*x)*a*b**4*c*d**2*x**2 + 8*log(a + b*x)*a*b 
**4*d**3*x**3 - 4*log(a + b*x)*a*b**3*c**4*x - 8*log(a + b*x)*a*b**3*c**3* 
d*x**2 - 4*log(a + b*x)*a*b**3*c**2*d**2*x**3 + 4*log(a + b*x)*b**5*c**2*d 
*x**2 + 4*log(a + b*x)*b**5*c*d**2*x**3 - 2*log(a + b*x)*b**4*c**4*x**2 - 
2*log(a + b*x)*b**4*c**3*d*x**3 - 8*log(c + d*x)*a**3*b**2*c*d**2 - 8*log( 
c + d*x)*a**3*b**2*d**3*x + 4*log(c + d*x)*a**3*b*c**3*d + 4*log(c + d*x)* 
a**3*b*c**2*d**2*x - 4*log(c + d*x)*a**2*b**3*c**2*d - 20*log(c + d*x)*a** 
2*b**3*c*d**2*x - 16*log(c + d*x)*a**2*b**3*d**3*x**2 + 2*log(c + d*x)*a** 
2*b**2*c**4 + 10*log(c + d*x)*a**2*b**2*c**3*d*x + 8*log(c + d*x)*a**2*b** 
2*c**2*d**2*x**2 - 8*log(c + d*x)*a*b**4*c**2*d*x - 16*log(c + d*x)*a*b**4 
*c*d**2*x**2 - 8*log(c + d*x)*a*b**4*d**3*x**3 + 4*log(c + d*x)*a*b**3*c** 
4*x + 8*log(c + d*x)*a*b**3*c**3*d*x**2 + 4*log(c + d*x)*a*b**3*c**2*d**2* 
x**3 - 4*log(c + d*x)*b**5*c**2*d*x**2 - 4*log(c + d*x)*b**5*c*d**2*x**3 + 
 2*log(c + d*x)*b**4*c**4*x**2 + 2*log(c + d*x)*b**4*c**3*d*x**3 - 4*a**4* 
b*d**3 + a**4*c**2*d**2 + a**4*c*d**3*x + 2*a**3*b**2*c*d**2 - 8*a**3*b...