\(\int \frac {(a+b x) (A+B x+C x^2+D x^3)}{c+d x} \, dx\) [28]

Optimal result

Integrand size = 28, antiderivative size = 178 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=-\frac {\left (a d \left (c C d-B d^2-c^2 D\right )-b \left (c^2 C d-B c d^2+A d^3-c^3 D\right )\right ) x}{d^4}+\frac {\left (a d (C d-c D)-b \left (c C d-B d^2-c^2 D\right )\right ) x^2}{2 d^3}+\frac {(b C d-b c D+a d D) x^3}{3 d^2}+\frac {b D x^4}{4 d}-\frac {(b c-a d) \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \log (c+d x)}{d^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {d x \left (2 a d \left (6 c^2 D-3 c d (2 C+D x)+d^2 \left (6 B+3 C x+2 D x^2\right )\right )+b \left (-12 c^3 D+6 c^2 d (2 C+D x)-2 c d^2 \left (6 B+3 C x+2 D x^2\right )+d^3 \left (12 A+6 B x+4 C x^2+3 D x^3\right )\right )\right )+12 (b c-a d) \left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \log (c+d x)}{12 d^5} \] Input:

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

Output:

(d*x*(2*a*d*(6*c^2*D - 3*c*d*(2*C + D*x) + d^2*(6*B + 3*C*x + 2*D*x^2)) + 
b*(-12*c^3*D + 6*c^2*d*(2*C + D*x) - 2*c*d^2*(6*B + 3*C*x + 2*D*x^2) + d^3 
*(12*A + 6*B*x + 4*C*x^2 + 3*D*x^3))) + 12*(b*c - a*d)*(-(c^2*C*d) + B*c*d 
^2 - A*d^3 + c^3*D)*Log[c + d*x])/(12*d^5)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 178, 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.

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

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-(((a*d*(c*C*d - B*d^2 - c^2*D) - b*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))*x 
)/d^4) + ((a*d*(C*d - c*D) - b*(c*C*d - B*d^2 - c^2*D))*x^2)/(2*d^3) + ((b 
*C*d - b*c*D + a*d*D)*x^3)/(3*d^2) + (b*D*x^4)/(4*d) - ((b*c - a*d)*(c^2*C 
*d - B*c*d^2 + A*d^3 - 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 = 201, normalized size of antiderivative = 1.13

method	result	size
norman	\(\frac {\left (A b \,d^{3}+B a \,d^{3}-B b c \,d^{2}-C a c \,d^{2}+C b \,c^{2} d +D a \,c^{2} d -D b \,c^{3}\right ) x}{d^{4}}+\frac {\left (b B \,d^{2}+C a \,d^{2}-C b c d -D a c d +D b \,c^{2}\right ) x^{2}}{2 d^{3}}+\frac {\left (C b d +D a d -D b c \right ) x^{3}}{3 d^{2}}+\frac {b D x^{4}}{4 d}+\frac {\left (A a \,d^{4}-A b c \,d^{3}-B a c \,d^{3}+B b \,c^{2} d^{2}+C a \,c^{2} d^{2}-C b \,c^{3} d -D a \,c^{3} d +D b \,c^{4}\right ) \ln \left (x d +c \right )}{d^{5}}\)	\(201\)
default	\(\frac {\frac {1}{4} D b \,x^{4} d^{3}+\frac {1}{3} C b \,d^{3} x^{3}+\frac {1}{3} D a \,d^{3} x^{3}-\frac {1}{3} D b c \,d^{2} x^{3}+\frac {1}{2} B b \,d^{3} x^{2}+\frac {1}{2} C a \,d^{3} x^{2}-\frac {1}{2} C b c \,d^{2} x^{2}-\frac {1}{2} D a c \,d^{2} x^{2}+\frac {1}{2} D b \,c^{2} d \,x^{2}+A b \,d^{3} x +B a \,d^{3} x -B b c \,d^{2} x -C a c \,d^{2} x +C b \,c^{2} d x +D a \,c^{2} d x -D b \,c^{3} x}{d^{4}}+\frac {\left (A a \,d^{4}-A b c \,d^{3}-B a c \,d^{3}+B b \,c^{2} d^{2}+C a \,c^{2} d^{2}-C b \,c^{3} d -D a \,c^{3} d +D b \,c^{4}\right ) \ln \left (x d +c \right )}{d^{5}}\)	\(230\)
parallelrisch	\(\frac {-12 D \ln \left (x d +c \right ) a \,c^{3} d +3 D b \,x^{4} d^{4}+12 B x a \,d^{4}+12 A \ln \left (x d +c \right ) a \,d^{4}+12 D \ln \left (x d +c \right ) b \,c^{4}+4 C \,x^{3} b \,d^{4}+4 D x^{3} a \,d^{4}+6 B \,x^{2} b \,d^{4}+6 C \,x^{2} a \,d^{4}+12 A x b \,d^{4}+12 D x a \,c^{2} d^{2}-12 D x b \,c^{3} d -12 A \ln \left (x d +c \right ) b c \,d^{3}-4 D x^{3} b c \,d^{3}-6 C \,x^{2} b c \,d^{3}-6 D x^{2} a c \,d^{3}+6 D x^{2} b \,c^{2} d^{2}-12 B x b c \,d^{3}-12 C x a c \,d^{3}+12 C x b \,c^{2} d^{2}-12 B \ln \left (x d +c \right ) a c \,d^{3}+12 B \ln \left (x d +c \right ) b \,c^{2} d^{2}+12 C \ln \left (x d +c \right ) a \,c^{2} d^{2}-12 C \ln \left (x d +c \right ) b \,c^{3} d}{12 d^{5}}\)	\(282\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {3 \, D b d^{4} x^{4} - 4 \, {\left (D b c d^{3} - {\left (D a + C b\right )} d^{4}\right )} x^{3} + 6 \, {\left (D b c^{2} d^{2} - {\left (D a + C b\right )} c d^{3} + {\left (C a + B b\right )} d^{4}\right )} x^{2} - 12 \, {\left (D b c^{3} d - {\left (D a + C b\right )} c^{2} d^{2} + {\left (C a + B b\right )} c d^{3} - {\left (B a + A b\right )} d^{4}\right )} x + 12 \, {\left (D b c^{4} + A a d^{4} - {\left (D a + C b\right )} c^{3} d + {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3}\right )} \log \left (d x + c\right )}{12 \, d^{5}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {D b x^{4}}{4 d} + x^{3} \left (\frac {C b}{3 d} + \frac {D a}{3 d} - \frac {D b c}{3 d^{2}}\right ) + x^{2} \left (\frac {B b}{2 d} + \frac {C a}{2 d} - \frac {C b c}{2 d^{2}} - \frac {D a c}{2 d^{2}} + \frac {D b c^{2}}{2 d^{3}}\right ) + x \left (\frac {A b}{d} + \frac {B a}{d} - \frac {B b c}{d^{2}} - \frac {C a c}{d^{2}} + \frac {C b c^{2}}{d^{3}} + \frac {D a c^{2}}{d^{3}} - \frac {D b c^{3}}{d^{4}}\right ) - \frac {\left (a d - b c\right ) \left (- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}\right ) \log {\left (c + d x \right )}}{d^{5}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {3 \, D b d^{3} x^{4} - 4 \, {\left (D b c d^{2} - {\left (D a + C b\right )} d^{3}\right )} x^{3} + 6 \, {\left (D b c^{2} d - {\left (D a + C b\right )} c d^{2} + {\left (C a + B b\right )} d^{3}\right )} x^{2} - 12 \, {\left (D b c^{3} - {\left (D a + C b\right )} c^{2} d + {\left (C a + B b\right )} c d^{2} - {\left (B a + A b\right )} d^{3}\right )} x}{12 \, d^{4}} + \frac {{\left (D b c^{4} + A a d^{4} - {\left (D a + C b\right )} c^{3} d + {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3}\right )} \log \left (d x + c\right )}{d^{5}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {3 \, D b d^{3} x^{4} - 4 \, D b c d^{2} x^{3} + 4 \, D a d^{3} x^{3} + 4 \, C b d^{3} x^{3} + 6 \, D b c^{2} d x^{2} - 6 \, D a c d^{2} x^{2} - 6 \, C b c d^{2} x^{2} + 6 \, C a d^{3} x^{2} + 6 \, B b d^{3} x^{2} - 12 \, D b c^{3} x + 12 \, D a c^{2} d x + 12 \, C b c^{2} d x - 12 \, C a c d^{2} x - 12 \, B b c d^{2} x + 12 \, B a d^{3} x + 12 \, A b d^{3} x}{12 \, d^{4}} + \frac {{\left (D b c^{4} - D a c^{3} d - C b c^{3} d + C a c^{2} d^{2} + B b c^{2} d^{2} - B a c d^{3} - A b c d^{3} + A a d^{4}\right )} \log \left ({\left | d x + c \right |}\right )}{d^{5}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.51 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {12 \,\mathrm {log}\left (d x +c \right ) a^{2} d^{2}-24 \,\mathrm {log}\left (d x +c \right ) a b c d +12 \,\mathrm {log}\left (d x +c \right ) b^{2} c^{2}+24 a b \,d^{2} x +4 a \,d^{3} x^{3}-12 b^{2} c d x +6 b^{2} d^{2} x^{2}+3 b \,d^{3} x^{4}}{12 d^{3}} \] Input:

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

Output:

(12*log(c + d*x)*a**2*d**2 - 24*log(c + d*x)*a*b*c*d + 12*log(c + d*x)*b** 
2*c**2 + 24*a*b*d**2*x + 4*a*d**3*x**3 - 12*b**2*c*d*x + 6*b**2*d**2*x**2 
+ 3*b*d**3*x**4)/(12*d**3)