\(\int \frac {(a+b x) (A+B x+C x^2+D x^3)}{(c+d x)^3} \, dx\) [43]

Optimal result

Integrand size = 28, antiderivative size = 184 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {(b C d-3 b c D+a d D) x}{d^4}+\frac {b D x^2}{2 d^3}+\frac {(b c-a d) \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{2 d^5 (c+d x)^2}+\frac {a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )}{d^5 (c+d x)}+\frac {\left (a d (C d-3 c D)-b \left (3 c C d-B d^2-6 c^2 D\right )\right ) \log (c+d x)}{d^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {2 d (b C d-3 b c D+a d D) x+b d^2 D x^2-\frac {(b c-a d) \left (-c^2 C d+B c d^2-A d^3+c^3 D\right )}{(c+d x)^2}-\frac {2 \left (a d \left (-2 c C d+B d^2+3 c^2 D\right )+b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )\right )}{c+d x}+2 \left (a d (C d-3 c D)+b \left (-3 c C d+B d^2+6 c^2 D\right )\right ) \log (c+d x)}{2 d^5} \] Input:

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

Output:

(2*d*(b*C*d - 3*b*c*D + a*d*D)*x + b*d^2*D*x^2 - ((b*c - a*d)*(-(c^2*C*d) 
+ B*c*d^2 - A*d^3 + c^3*D))/(c + d*x)^2 - (2*(a*d*(-2*c*C*d + B*d^2 + 3*c^ 
2*D) + b*(3*c^2*C*d - 2*B*c*d^2 + A*d^3 - 4*c^3*D)))/(c + d*x) + 2*(a*d*(C 
*d - 3*c*D) + b*(-3*c*C*d + B*d^2 + 6*c^2*D))*Log[c + d*x])/(2*d^5)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, 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.

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

\(\Big \downarrow \) 2123

\(\displaystyle \int \left (\frac {(a d-b c) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^4 (c+d x)^3}+\frac {b \left (A d^3-2 B c d^2-4 c^3 D+3 c^2 C d\right )-a d \left (-B d^2-3 c^2 D+2 c C d\right )}{d^4 (c+d x)^2}+\frac {a d (C d-3 c D)-b \left (-B d^2-6 c^2 D+3 c C d\right )}{d^4 (c+d x)}+\frac {a d D-3 b c D+b C d}{d^4}+\frac {b D x}{d^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (A d^3-2 B c d^2-4 c^3 D+3 c^2 C d\right )}{d^5 (c+d x)}+\frac {(b c-a d) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 d^5 (c+d x)^2}+\frac {\log (c+d x) \left (a d (C d-3 c D)-b \left (-B d^2-6 c^2 D+3 c C d\right )\right )}{d^5}+\frac {x (a d D-3 b c D+b C d)}{d^4}+\frac {b D x^2}{2 d^3}\)

Input:

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

Output:

((b*C*d - 3*b*c*D + a*d*D)*x)/d^4 + (b*D*x^2)/(2*d^3) + ((b*c - a*d)*(c^2* 
C*d - B*c*d^2 + A*d^3 - c^3*D))/(2*d^5*(c + d*x)^2) + (a*d*(2*c*C*d - B*d^ 
2 - 3*c^2*D) - b*(3*c^2*C*d - 2*B*c*d^2 + A*d^3 - 4*c^3*D))/(d^5*(c + d*x) 
) + ((a*d*(C*d - 3*c*D) - b*(3*c*C*d - B*d^2 - 6*c^2*D))*Log[c + d*x])/d^5
 

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 = 0.33 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.15

method	result	size
norman	\(\frac {\frac {\left (C b d +D a d -2 D b c \right ) x^{3}}{d^{2}}-\frac {A a \,d^{4}+A c \,d^{3} b +B c \,d^{3} a -3 B b \,c^{2} d^{2}-3 C a \,c^{2} d^{2}+9 C b \,c^{3} d +9 D a \,c^{3} d -18 D b \,c^{4}}{2 d^{5}}-\frac {\left (A b \,d^{3}+B a \,d^{3}-2 B b c \,d^{2}-2 C a c \,d^{2}+6 C b \,c^{2} d +6 D a \,c^{2} d -12 D b \,c^{3}\right ) x}{d^{4}}+\frac {b D x^{4}}{2 d}}{\left (x d +c \right )^{2}}+\frac {\left (b B \,d^{2}+C a \,d^{2}-3 C b c d -3 D a c d +6 D b \,c^{2}\right ) \ln \left (x d +c \right )}{d^{5}}\)	\(211\)
default	\(\frac {\frac {1}{2} d D x^{2} b +C b d x +D a d x -3 D b c x}{d^{4}}-\frac {A b \,d^{3}+B a \,d^{3}-2 B b c \,d^{2}-2 C a c \,d^{2}+3 C b \,c^{2} d +3 D a \,c^{2} d -4 D b \,c^{3}}{d^{5} \left (x d +c \right )}+\frac {\left (b B \,d^{2}+C a \,d^{2}-3 C b c d -3 D a c d +6 D b \,c^{2}\right ) \ln \left (x d +c \right )}{d^{5}}-\frac {A a \,d^{4}-A c \,d^{3} b -B c \,d^{3} a +B b \,c^{2} d^{2}+C a \,c^{2} d^{2}-C b \,c^{3} d -D a \,c^{3} d +D b \,c^{4}}{2 d^{5} \left (x d +c \right )^{2}}\)	\(212\)
parallelrisch	\(-\frac {-4 C \ln \left (x d +c \right ) x a c \,d^{3}+12 C \ln \left (x d +c \right ) x b \,c^{2} d^{2}+12 D \ln \left (x d +c \right ) x a \,c^{2} d^{2}-24 D \ln \left (x d +c \right ) x b \,c^{3} d -3 B b \,c^{2} d^{2}-4 B \ln \left (x d +c \right ) x b c \,d^{3}-3 C a \,c^{2} d^{2}+9 D a \,c^{3} d +6 D \ln \left (x d +c \right ) a \,c^{3} d -D b \,x^{4} d^{4}+2 B x a \,d^{4}-12 D \ln \left (x d +c \right ) b \,c^{4}-2 C \,x^{3} b \,d^{4}-2 D x^{3} a \,d^{4}+2 A x b \,d^{4}+A c \,d^{3} b +B c \,d^{3} a +9 C b \,c^{3} d -2 B \ln \left (x d +c \right ) x^{2} b \,d^{4}-2 C \ln \left (x d +c \right ) x^{2} a \,d^{4}+12 D x a \,c^{2} d^{2}-24 D x b \,c^{3} d +4 D x^{3} b c \,d^{3}-4 B x b c \,d^{3}-4 C x a c \,d^{3}+12 C x b \,c^{2} d^{2}-2 B \ln \left (x d +c \right ) b \,c^{2} d^{2}-2 C \ln \left (x d +c \right ) a \,c^{2} d^{2}+6 C \ln \left (x d +c \right ) b \,c^{3} d +6 C \ln \left (x d +c \right ) x^{2} b c \,d^{3}+6 D \ln \left (x d +c \right ) x^{2} a c \,d^{3}-12 D \ln \left (x d +c \right ) x^{2} b \,c^{2} d^{2}-18 D b \,c^{4}+A a \,d^{4}}{2 d^{5} \left (x d +c \right )^{2}}\)	\(420\)

Input:

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

Output:

((C*b*d+D*a*d-2*D*b*c)/d^2*x^3-1/2*(A*a*d^4+A*b*c*d^3+B*a*c*d^3-3*B*b*c^2* 
d^2-3*C*a*c^2*d^2+9*C*b*c^3*d+9*D*a*c^3*d-18*D*b*c^4)/d^5-(A*b*d^3+B*a*d^3 
-2*B*b*c*d^2-2*C*a*c*d^2+6*C*b*c^2*d+6*D*a*c^2*d-12*D*b*c^3)/d^4*x+1/2*b*D 
*x^4/d)/(d*x+c)^2+1/d^5*(B*b*d^2+C*a*d^2-3*C*b*c*d-3*D*a*c*d+6*D*b*c^2)*ln 
(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.73 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {D b d^{4} x^{4} + 7 \, D b c^{4} - A a d^{4} - 5 \, {\left (D a + C b\right )} c^{3} d + 3 \, {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3} - 2 \, {\left (2 \, D b c d^{3} - {\left (D a + C b\right )} d^{4}\right )} x^{3} - {\left (11 \, D b c^{2} d^{2} - 4 \, {\left (D a + C b\right )} c d^{3}\right )} x^{2} + 2 \, {\left (D b c^{3} d - 2 \, {\left (D a + C b\right )} c^{2} d^{2} + 2 \, {\left (C a + B b\right )} c d^{3} - {\left (B a + A b\right )} d^{4}\right )} x + 2 \, {\left (6 \, D b c^{4} - 3 \, {\left (D a + C b\right )} c^{3} d + {\left (C a + B b\right )} c^{2} d^{2} + {\left (6 \, D b c^{2} d^{2} - 3 \, {\left (D a + C b\right )} c d^{3} + {\left (C a + B b\right )} d^{4}\right )} x^{2} + 2 \, {\left (6 \, D b c^{3} d - 3 \, {\left (D a + C b\right )} c^{2} d^{2} + {\left (C a + B b\right )} c d^{3}\right )} x\right )} \log \left (d x + c\right )}{2 \, {\left (d^{7} x^{2} + 2 \, c d^{6} x + c^{2} d^{5}\right )}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 6.02 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.41 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {D b x^{2}}{2 d^{3}} + x \left (\frac {C b}{d^{3}} + \frac {D a}{d^{3}} - \frac {3 D b c}{d^{4}}\right ) + \frac {- A a d^{4} - A b c d^{3} - B a c d^{3} + 3 B b c^{2} d^{2} + 3 C a c^{2} d^{2} - 5 C b c^{3} d - 5 D a c^{3} d + 7 D b c^{4} + x \left (- 2 A b d^{4} - 2 B a d^{4} + 4 B b c d^{3} + 4 C a c d^{3} - 6 C b c^{2} d^{2} - 6 D a c^{2} d^{2} + 8 D b c^{3} d\right )}{2 c^{2} d^{5} + 4 c d^{6} x + 2 d^{7} x^{2}} - \frac {\left (- B b d^{2} - C a d^{2} + 3 C b c d + 3 D a c d - 6 D b c^{2}\right ) \log {\left (c + d x \right )}}{d^{5}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {7 \, D b c^{4} - A a d^{4} - 5 \, {\left (D a + C b\right )} c^{3} d + 3 \, {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3} + 2 \, {\left (4 \, D b c^{3} d - 3 \, {\left (D a + C b\right )} c^{2} d^{2} + 2 \, {\left (C a + B b\right )} c d^{3} - {\left (B a + A b\right )} d^{4}\right )} x}{2 \, {\left (d^{7} x^{2} + 2 \, c d^{6} x + c^{2} d^{5}\right )}} + \frac {D b d x^{2} - 2 \, {\left (3 \, D b c - {\left (D a + C b\right )} d\right )} x}{2 \, d^{4}} + \frac {{\left (6 \, D b c^{2} - 3 \, {\left (D a + C b\right )} c d + {\left (C a + B b\right )} d^{2}\right )} \log \left (d x + c\right )}{d^{5}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.22 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {{\left (6 \, D b c^{2} - 3 \, D a c d - 3 \, C b c d + C a d^{2} + B b d^{2}\right )} \log \left ({\left | d x + c \right |}\right )}{d^{5}} + \frac {D b d^{3} x^{2} - 6 \, D b c d^{2} x + 2 \, D a d^{3} x + 2 \, C b d^{3} x}{2 \, d^{6}} + \frac {7 \, D b c^{4} - 5 \, D a c^{3} d - 5 \, C b c^{3} d + 3 \, C a c^{2} d^{2} + 3 \, B b c^{2} d^{2} - B a c d^{3} - A b c d^{3} - A a d^{4} + 2 \, {\left (4 \, D b c^{3} d - 3 \, D a c^{2} d^{2} - 3 \, C b c^{2} d^{2} + 2 \, C a c d^{3} + 2 \, B b c d^{3} - B a d^{4} - A b d^{4}\right )} x}{2 \, {\left (d x + c\right )}^{2} d^{5}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.52 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.88 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {\frac {3\,B\,b\,c^2}{2\,d^3}+\frac {2\,B\,b\,c\,x}{d^2}}{c^2+2\,c\,d\,x+d^2\,x^2}+\frac {\frac {3\,C\,a\,c^2}{2\,d^3}+\frac {2\,C\,a\,c\,x}{d^2}}{c^2+2\,c\,d\,x+d^2\,x^2}-\frac {\frac {A\,b\,c}{2\,d^2}+\frac {A\,b\,x}{d}}{c^2+2\,c\,d\,x+d^2\,x^2}-\frac {\frac {B\,a\,c}{2\,d^2}+\frac {B\,a\,x}{d}}{c^2+2\,c\,d\,x+d^2\,x^2}+\frac {B\,b\,\ln \left (c+d\,x\right )}{d^3}+\frac {C\,a\,\ln \left (c+d\,x\right )}{d^3}+\frac {b\,D\,\left (\frac {{\left (c+d\,x\right )}^2}{2}+\frac {4\,c^3}{c+d\,x}-\frac {c^4}{2\,{\left (c+d\,x\right )}^2}+6\,c^2\,\ln \left (c+d\,x\right )-4\,c\,d\,x\right )}{d^5}-\frac {C\,b\,\left (3\,c\,\ln \left (c+d\,x\right )-d\,x+\frac {3\,c^2}{c+d\,x}-\frac {c^3}{2\,{\left (c+d\,x\right )}^2}\right )}{d^4}-\frac {A\,a}{2\,d\,\left (c^2+2\,c\,d\,x+d^2\,x^2\right )}-\frac {a\,D\,\left (3\,c\,\ln \left (c+d\,x\right )-d\,x+\frac {3\,c^2}{c+d\,x}-\frac {c^3}{2\,{\left (c+d\,x\right )}^2}\right )}{d^4} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {-4 \,\mathrm {log}\left (d x +c \right ) a \,c^{4} d -8 \,\mathrm {log}\left (d x +c \right ) a \,c^{3} d^{2} x -4 \,\mathrm {log}\left (d x +c \right ) a \,c^{2} d^{3} x^{2}+2 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{3} d +4 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{2} d^{2} x +2 \,\mathrm {log}\left (d x +c \right ) b^{2} c \,d^{3} x^{2}+6 \,\mathrm {log}\left (d x +c \right ) b \,c^{5}+12 \,\mathrm {log}\left (d x +c \right ) b \,c^{4} d x +6 \,\mathrm {log}\left (d x +c \right ) b \,c^{3} d^{2} x^{2}-a^{2} c \,d^{3}+2 a b \,d^{4} x^{2}-2 a \,c^{4} d +4 a \,c^{2} d^{3} x^{2}+2 a c \,d^{4} x^{3}+b^{2} c^{3} d -2 b^{2} c \,d^{3} x^{2}+3 b \,c^{5}-6 b \,c^{3} d^{2} x^{2}-2 b \,c^{2} d^{3} x^{3}+b c \,d^{4} x^{4}}{2 c \,d^{4} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

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

Output:

( - 4*log(c + d*x)*a*c**4*d - 8*log(c + d*x)*a*c**3*d**2*x - 4*log(c + d*x 
)*a*c**2*d**3*x**2 + 2*log(c + d*x)*b**2*c**3*d + 4*log(c + d*x)*b**2*c**2 
*d**2*x + 2*log(c + d*x)*b**2*c*d**3*x**2 + 6*log(c + d*x)*b*c**5 + 12*log 
(c + d*x)*b*c**4*d*x + 6*log(c + d*x)*b*c**3*d**2*x**2 - a**2*c*d**3 + 2*a 
*b*d**4*x**2 - 2*a*c**4*d + 4*a*c**2*d**3*x**2 + 2*a*c*d**4*x**3 + b**2*c* 
*3*d - 2*b**2*c*d**3*x**2 + 3*b*c**5 - 6*b*c**3*d**2*x**2 - 2*b*c**2*d**3* 
x**3 + b*c*d**4*x**4)/(2*c*d**4*(c**2 + 2*c*d*x + d**2*x**2))