\(\int \frac {(a+b x) (A+B x+C x^2+D x^3)}{(c+d x)^2} \, dx\) [36]

Optimal result

Integrand size = 28, antiderivative size = 181 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {\left (a d (C d-2 c D)-b \left (2 c C d-B d^2-3 c^2 D\right )\right ) x}{d^4}+\frac {(b C d-2 b c D+a d D) x^2}{2 d^3}+\frac {b D x^3}{3 d^2}+\frac {(b c-a d) \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{d^5 (c+d x)}-\frac {\left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )\right ) \log (c+d x)}{d^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {6 d \left (a d (C d-2 c D)+b \left (-2 c C d+B d^2+3 c^2 D\right )\right ) x+3 d^2 (b C d-2 b c D+a d D) x^2+2 b d^3 D x^3-\frac {6 (b c-a d) \left (-c^2 C d+B c d^2-A d^3+c^3 D\right )}{c+d x}+6 \left (a d \left (-2 c C d+B d^2+3 c^2 D\right )+b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )\right ) \log (c+d x)}{6 d^5} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 181, 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.

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

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((a*d*(C*d - 2*c*D) - b*(2*c*C*d - B*d^2 - 3*c^2*D))*x)/d^4 + ((b*C*d - 2* 
b*c*D + a*d*D)*x^2)/(2*d^3) + (b*D*x^3)/(3*d^2) + ((b*c - a*d)*(c^2*C*d - 
B*c*d^2 + A*d^3 - c^3*D))/(d^5*(c + d*x)) - ((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))*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 = 0.34 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\frac {1}{3} b D x^{3} d^{2}+\frac {1}{2} C b \,d^{2} x^{2}+\frac {1}{2} D a \,d^{2} x^{2}-D b c d \,x^{2}+b B \,d^{2} x +C a \,d^{2} x -2 C b c d x -2 D a c d x +3 D b \,c^{2} x}{d^{4}}-\frac {A a \,d^{4}-A c \,d^{3} b -B c \,d^{3} a +B b \,c^{2} d^{2}+C a \,c^{2} d^{2}-C b \,c^{3} d -D a \,c^{3} d +D b \,c^{4}}{d^{5} \left (x d +c \right )}+\frac {\left (A b \,d^{3}+B a \,d^{3}-2 B b c \,d^{2}-2 C a c \,d^{2}+3 C b \,c^{2} d +3 D a \,c^{2} d -4 D b \,c^{3}\right ) \ln \left (x d +c \right )}{d^{5}}\)	\(219\)
norman	\(\frac {\frac {\left (A a \,d^{4}-A c \,d^{3} b -B c \,d^{3} a +2 B b \,c^{2} d^{2}+2 C a \,c^{2} d^{2}-3 C b \,c^{3} d -3 D a \,c^{3} d +4 D b \,c^{4}\right ) x}{d^{4} c}+\frac {\left (3 C b d +3 D a d -4 D b c \right ) x^{3}}{6 d^{2}}+\frac {\left (2 b B \,d^{2}+2 C a \,d^{2}-3 C b c d -3 D a c d +4 D b \,c^{2}\right ) x^{2}}{2 d^{3}}+\frac {b D x^{4}}{3 d}}{x d +c}+\frac {\left (A b \,d^{3}+B a \,d^{3}-2 B b c \,d^{2}-2 C a c \,d^{2}+3 C b \,c^{2} d +3 D a \,c^{2} d -4 D b \,c^{3}\right ) \ln \left (x d +c \right )}{d^{5}}\)	\(223\)
parallelrisch	\(\frac {-12 C \ln \left (x d +c \right ) x a c \,d^{3}+18 C \ln \left (x d +c \right ) x b \,c^{2} d^{2}+18 D \ln \left (x d +c \right ) x a \,c^{2} d^{2}-24 D \ln \left (x d +c \right ) x b \,c^{3} d -12 B b \,c^{2} d^{2}-12 B \ln \left (x d +c \right ) x b c \,d^{3}-12 C a \,c^{2} d^{2}+18 D a \,c^{3} d +18 D \ln \left (x d +c \right ) a \,c^{3} d +2 D b \,x^{4} d^{4}-24 D \ln \left (x d +c \right ) b \,c^{4}+3 C \,x^{3} b \,d^{4}+3 D x^{3} a \,d^{4}+6 B \,x^{2} b \,d^{4}+6 C \,x^{2} a \,d^{4}+6 A c \,d^{3} b +6 B c \,d^{3} a +6 A \ln \left (x d +c \right ) x b \,d^{4}+6 B \ln \left (x d +c \right ) x a \,d^{4}+18 C b \,c^{3} d +6 A \ln \left (x d +c \right ) b c \,d^{3}-4 D x^{3} b c \,d^{3}-9 C \,x^{2} b c \,d^{3}-9 D x^{2} a c \,d^{3}+12 D x^{2} b \,c^{2} d^{2}+6 B \ln \left (x d +c \right ) a c \,d^{3}-12 B \ln \left (x d +c \right ) b \,c^{2} d^{2}-12 C \ln \left (x d +c \right ) a \,c^{2} d^{2}+18 C \ln \left (x d +c \right ) b \,c^{3} d -24 D b \,c^{4}-6 A a \,d^{4}}{6 d^{5} \left (x d +c \right )}\)	\(384\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {2 \, D b d^{4} x^{4} - 6 \, D b c^{4} - 6 \, A a d^{4} + 6 \, {\left (D a + C b\right )} c^{3} d - 6 \, {\left (C a + B b\right )} c^{2} d^{2} + 6 \, {\left (B a + A b\right )} c d^{3} - {\left (4 \, D b c d^{3} - 3 \, {\left (D a + C b\right )} d^{4}\right )} x^{3} + 3 \, {\left (4 \, D b c^{2} d^{2} - 3 \, {\left (D a + C b\right )} c d^{3} + 2 \, {\left (C a + B b\right )} d^{4}\right )} x^{2} + 6 \, {\left (3 \, D b c^{3} d - 2 \, {\left (D a + C b\right )} c^{2} d^{2} + {\left (C a + B b\right )} c d^{3}\right )} x - 6 \, {\left (4 \, D b c^{4} - 3 \, {\left (D a + C b\right )} c^{3} d + 2 \, {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3} + {\left (4 \, D b c^{3} d - 3 \, {\left (D a + C b\right )} c^{2} d^{2} + 2 \, {\left (C a + B b\right )} c d^{3} - {\left (B a + A b\right )} d^{4}\right )} x\right )} \log \left (d x + c\right )}{6 \, {\left (d^{6} x + c d^{5}\right )}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {D b x^{3}}{3 d^{2}} + x^{2} \left (\frac {C b}{2 d^{2}} + \frac {D a}{2 d^{2}} - \frac {D b c}{d^{3}}\right ) + x \left (\frac {B b}{d^{2}} + \frac {C a}{d^{2}} - \frac {2 C b c}{d^{3}} - \frac {2 D a c}{d^{3}} + \frac {3 D b c^{2}}{d^{4}}\right ) + \frac {- A a d^{4} + A b c d^{3} + B a c d^{3} - B b c^{2} d^{2} - C a c^{2} d^{2} + C b c^{3} d + D a c^{3} d - D b c^{4}}{c d^{5} + d^{6} x} + \frac {\left (A b d^{3} + B a d^{3} - 2 B b c d^{2} - 2 C a c d^{2} + 3 C b c^{2} d + 3 D a c^{2} d - 4 D b c^{3}\right ) \log {\left (c + d x \right )}}{d^{5}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=-\frac {D b c^{4} + A a d^{4} - {\left (D a + C b\right )} c^{3} d + {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3}}{d^{6} x + c d^{5}} + \frac {2 \, D b d^{2} x^{3} - 3 \, {\left (2 \, D b c d - {\left (D a + C b\right )} d^{2}\right )} x^{2} + 6 \, {\left (3 \, D b c^{2} - 2 \, {\left (D a + C b\right )} c d + {\left (C a + B b\right )} d^{2}\right )} x}{6 \, d^{4}} - \frac {{\left (4 \, D b c^{3} - 3 \, {\left (D a + C b\right )} c^{2} d + 2 \, {\left (C a + B b\right )} c d^{2} - {\left (B a + A b\right )} d^{3}\right )} \log \left (d x + c\right )}{d^{5}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.71 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {{\left (2 \, D b - \frac {3 \, {\left (4 \, D b c d - D a d^{2} - C b d^{2}\right )}}{{\left (d x + c\right )} d} + \frac {6 \, {\left (6 \, D b c^{2} d^{2} - 3 \, D a c d^{3} - 3 \, C b c d^{3} + C a d^{4} + B b d^{4}\right )}}{{\left (d x + c\right )}^{2} d^{2}}\right )} {\left (d x + c\right )}^{3}}{6 \, d^{5}} + \frac {{\left (4 \, D b c^{3} - 3 \, D a c^{2} d - 3 \, C b c^{2} d + 2 \, C a c d^{2} + 2 \, B b c d^{2} - B a d^{3} - A b d^{3}\right )} \log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{d^{5}} - \frac {\frac {D b c^{4} d^{3}}{d x + c} - \frac {D a c^{3} d^{4}}{d x + c} - \frac {C b c^{3} d^{4}}{d x + c} + \frac {C a c^{2} d^{5}}{d x + c} + \frac {B b c^{2} d^{5}}{d x + c} - \frac {B a c d^{6}}{d x + c} - \frac {A b c d^{6}}{d x + c} + \frac {A a d^{7}}{d x + c}}{d^{8}} \] Input:

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

Output:

1/6*(2*D*b - 3*(4*D*b*c*d - D*a*d^2 - C*b*d^2)/((d*x + c)*d) + 6*(6*D*b*c^ 
2*d^2 - 3*D*a*c*d^3 - 3*C*b*c*d^3 + C*a*d^4 + B*b*d^4)/((d*x + c)^2*d^2))* 
(d*x + c)^3/d^5 + (4*D*b*c^3 - 3*D*a*c^2*d - 3*C*b*c^2*d + 2*C*a*c*d^2 + 2 
*B*b*c*d^2 - B*a*d^3 - A*b*d^3)*log(abs(d*x + c)/((d*x + c)^2*abs(d)))/d^5 
 - (D*b*c^4*d^3/(d*x + c) - D*a*c^3*d^4/(d*x + c) - C*b*c^3*d^4/(d*x + c) 
+ C*a*c^2*d^5/(d*x + c) + B*b*c^2*d^5/(d*x + c) - B*a*c*d^6/(d*x + c) - A* 
b*c*d^6/(d*x + c) + A*a*d^7/(d*x + c))/d^8
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {12 \,\mathrm {log}\left (d x +c \right ) a b \,c^{2} d^{2}+12 \,\mathrm {log}\left (d x +c \right ) a b c \,d^{3} x +6 \,\mathrm {log}\left (d x +c \right ) a \,c^{4} d +6 \,\mathrm {log}\left (d x +c \right ) a \,c^{3} d^{2} x -12 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{3} d -12 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{2} d^{2} x -6 \,\mathrm {log}\left (d x +c \right ) b \,c^{5}-6 \,\mathrm {log}\left (d x +c \right ) b \,c^{4} d x +6 a^{2} d^{4} x -12 a b c \,d^{3} x -6 a \,c^{3} d^{2} x -3 a \,c^{2} d^{3} x^{2}+3 a c \,d^{4} x^{3}+12 b^{2} c^{2} d^{2} x +6 b^{2} c \,d^{3} x^{2}+6 b \,c^{4} d x +3 b \,c^{3} d^{2} x^{2}-b \,c^{2} d^{3} x^{3}+2 b c \,d^{4} x^{4}}{6 c \,d^{4} \left (d x +c \right )} \] Input:

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

Output:

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