\(\int \frac {(A+B x) (a+b x+c x^2)^3}{x^9} \, dx\) [36]

Optimal result

Integrand size = 21, antiderivative size = 162 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=-\frac {a^3 A}{8 x^8}-\frac {a^2 (3 A b+a B)}{7 x^7}-\frac {a \left (a b B+A \left (b^2+a c\right )\right )}{2 x^6}-\frac {3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )}{5 x^5}-\frac {b^3 B+3 A b^2 c+6 a b B c+3 a A c^2}{4 x^4}-\frac {c \left (b^2 B+A b c+a B c\right )}{x^3}-\frac {c^2 (3 b B+A c)}{2 x^2}-\frac {B c^3}{x} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=-\frac {5 a^3 (7 A+8 B x)+4 a^2 x (7 B x (5 b+6 c x)+5 A (6 b+7 c x))+14 a x^2 \left (2 B x \left (6 b^2+15 b c x+10 c^2 x^2\right )+A \left (10 b^2+24 b c x+15 c^2 x^2\right )\right )+14 x^3 \left (5 B x \left (b^3+4 b^2 c x+6 b c^2 x^2+4 c^3 x^3\right )+A \left (4 b^3+15 b^2 c x+20 b c^2 x^2+10 c^3 x^3\right )\right )}{280 x^8} \] Input:

Integrate[((A + B*x)*(a + b*x + c*x^2)^3)/x^9,x]
 

Output:

-1/280*(5*a^3*(7*A + 8*B*x) + 4*a^2*x*(7*B*x*(5*b + 6*c*x) + 5*A*(6*b + 7* 
c*x)) + 14*a*x^2*(2*B*x*(6*b^2 + 15*b*c*x + 10*c^2*x^2) + A*(10*b^2 + 24*b 
*c*x + 15*c^2*x^2)) + 14*x^3*(5*B*x*(b^3 + 4*b^2*c*x + 6*b*c^2*x^2 + 4*c^3 
*x^3) + A*(4*b^3 + 15*b^2*c*x + 20*b*c^2*x^2 + 10*c^3*x^3)))/x^8
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1195, 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.

\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx\)

\(\Big \downarrow \) 1195

\(\displaystyle \int \left (\frac {a^3 A}{x^9}+\frac {a^2 (a B+3 A b)}{x^8}+\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x^7}+\frac {3 c \left (a B c+A b c+b^2 B\right )}{x^4}+\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{x^5}+\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{x^6}+\frac {c^2 (A c+3 b B)}{x^3}+\frac {B c^3}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^3 A}{8 x^8}-\frac {a^2 (a B+3 A b)}{7 x^7}-\frac {a \left (A \left (a c+b^2\right )+a b B\right )}{2 x^6}-\frac {c \left (a B c+A b c+b^2 B\right )}{x^3}-\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{4 x^4}-\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{5 x^5}-\frac {c^2 (A c+3 b B)}{2 x^2}-\frac {B c^3}{x}\)

Input:

Int[((A + B*x)*(a + b*x + c*x^2)^3)/x^9,x]
 

Output:

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

Defintions of rubi rules used

rule 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 2009

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

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}}{5 x^{5}}-\frac {a \left (A a c +b^{2} A +a b B \right )}{2 x^{6}}-\frac {c \left (A b c +a B c +B \,b^{2}\right )}{x^{3}}-\frac {c^{2} \left (A c +3 B b \right )}{2 x^{2}}-\frac {a^{3} A}{8 x^{8}}-\frac {3 A a \,c^{2}+3 A \,b^{2} c +6 B a b c +B \,b^{3}}{4 x^{4}}-\frac {a^{2} \left (3 A b +B a \right )}{7 x^{7}}-\frac {B \,c^{3}}{x}\)	\(154\)
norman	\(\frac {-B \,c^{3} x^{7}+\left (-\frac {1}{2} A \,c^{3}-\frac {3}{2} B b \,c^{2}\right ) x^{6}+\left (-A b \,c^{2}-B a \,c^{2}-B \,b^{2} c \right ) x^{5}+\left (-\frac {3}{4} A a \,c^{2}-\frac {3}{4} A \,b^{2} c -\frac {3}{2} B a b c -\frac {1}{4} B \,b^{3}\right ) x^{4}+\left (-\frac {6}{5} A a b c -\frac {1}{5} A \,b^{3}-\frac {3}{5} B \,a^{2} c -\frac {3}{5} B a \,b^{2}\right ) x^{3}+\left (-\frac {1}{2} a^{2} A c -\frac {1}{2} A a \,b^{2}-\frac {1}{2} B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{7} A \,a^{2} b -\frac {1}{7} B \,a^{3}\right ) x -\frac {a^{3} A}{8}}{x^{8}}\)	\(169\)
risch	\(\frac {-B \,c^{3} x^{7}+\left (-\frac {1}{2} A \,c^{3}-\frac {3}{2} B b \,c^{2}\right ) x^{6}+\left (-A b \,c^{2}-B a \,c^{2}-B \,b^{2} c \right ) x^{5}+\left (-\frac {3}{4} A a \,c^{2}-\frac {3}{4} A \,b^{2} c -\frac {3}{2} B a b c -\frac {1}{4} B \,b^{3}\right ) x^{4}+\left (-\frac {6}{5} A a b c -\frac {1}{5} A \,b^{3}-\frac {3}{5} B \,a^{2} c -\frac {3}{5} B a \,b^{2}\right ) x^{3}+\left (-\frac {1}{2} a^{2} A c -\frac {1}{2} A a \,b^{2}-\frac {1}{2} B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{7} A \,a^{2} b -\frac {1}{7} B \,a^{3}\right ) x -\frac {a^{3} A}{8}}{x^{8}}\)	\(169\)
gosper	\(-\frac {280 B \,c^{3} x^{7}+140 A \,c^{3} x^{6}+420 B b \,c^{2} x^{6}+280 A b \,c^{2} x^{5}+280 B a \,c^{2} x^{5}+280 B \,b^{2} c \,x^{5}+210 A a \,c^{2} x^{4}+210 A \,b^{2} c \,x^{4}+420 B a b c \,x^{4}+70 B \,b^{3} x^{4}+336 A a b c \,x^{3}+56 A \,b^{3} x^{3}+168 B \,a^{2} c \,x^{3}+168 B a \,b^{2} x^{3}+140 A \,a^{2} c \,x^{2}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 A \,a^{2} b x +40 B \,a^{3} x +35 a^{3} A}{280 x^{8}}\)	\(192\)
parallelrisch	\(-\frac {280 B \,c^{3} x^{7}+140 A \,c^{3} x^{6}+420 B b \,c^{2} x^{6}+280 A b \,c^{2} x^{5}+280 B a \,c^{2} x^{5}+280 B \,b^{2} c \,x^{5}+210 A a \,c^{2} x^{4}+210 A \,b^{2} c \,x^{4}+420 B a b c \,x^{4}+70 B \,b^{3} x^{4}+336 A a b c \,x^{3}+56 A \,b^{3} x^{3}+168 B \,a^{2} c \,x^{3}+168 B a \,b^{2} x^{3}+140 A \,a^{2} c \,x^{2}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 A \,a^{2} b x +40 B \,a^{3} x +35 a^{3} A}{280 x^{8}}\)	\(192\)
orering	\(-\frac {280 B \,c^{3} x^{7}+140 A \,c^{3} x^{6}+420 B b \,c^{2} x^{6}+280 A b \,c^{2} x^{5}+280 B a \,c^{2} x^{5}+280 B \,b^{2} c \,x^{5}+210 A a \,c^{2} x^{4}+210 A \,b^{2} c \,x^{4}+420 B a b c \,x^{4}+70 B \,b^{3} x^{4}+336 A a b c \,x^{3}+56 A \,b^{3} x^{3}+168 B \,a^{2} c \,x^{3}+168 B a \,b^{2} x^{3}+140 A \,a^{2} c \,x^{2}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 A \,a^{2} b x +40 B \,a^{3} x +35 a^{3} A}{280 x^{8}}\)	\(192\)

Input:

int((B*x+A)*(c*x^2+b*x+a)^3/x^9,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=-\frac {280 \, B c^{3} x^{7} + 140 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 280 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 70 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + 35 \, A a^{3} + 56 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 140 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 40 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{280 \, x^{8}} \] Input:

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^9,x, algorithm="fricas")
 

Output:

-1/280*(280*B*c^3*x^7 + 140*(3*B*b*c^2 + A*c^3)*x^6 + 280*(B*b^2*c + (B*a 
+ A*b)*c^2)*x^5 + 70*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 + 35* 
A*a^3 + 56*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 140*(B*a^2*b 
+ A*a*b^2 + A*a^2*c)*x^2 + 40*(B*a^3 + 3*A*a^2*b)*x)/x^8
 

Sympy [A] (verification not implemented)

Time = 59.22 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=\frac {- 35 A a^{3} - 280 B c^{3} x^{7} + x^{6} \left (- 140 A c^{3} - 420 B b c^{2}\right ) + x^{5} \left (- 280 A b c^{2} - 280 B a c^{2} - 280 B b^{2} c\right ) + x^{4} \left (- 210 A a c^{2} - 210 A b^{2} c - 420 B a b c - 70 B b^{3}\right ) + x^{3} \left (- 336 A a b c - 56 A b^{3} - 168 B a^{2} c - 168 B a b^{2}\right ) + x^{2} \left (- 140 A a^{2} c - 140 A a b^{2} - 140 B a^{2} b\right ) + x \left (- 120 A a^{2} b - 40 B a^{3}\right )}{280 x^{8}} \] Input:

integrate((B*x+A)*(c*x**2+b*x+a)**3/x**9,x)
 

Output:

(-35*A*a**3 - 280*B*c**3*x**7 + x**6*(-140*A*c**3 - 420*B*b*c**2) + x**5*( 
-280*A*b*c**2 - 280*B*a*c**2 - 280*B*b**2*c) + x**4*(-210*A*a*c**2 - 210*A 
*b**2*c - 420*B*a*b*c - 70*B*b**3) + x**3*(-336*A*a*b*c - 56*A*b**3 - 168* 
B*a**2*c - 168*B*a*b**2) + x**2*(-140*A*a**2*c - 140*A*a*b**2 - 140*B*a**2 
*b) + x*(-120*A*a**2*b - 40*B*a**3))/(280*x**8)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=-\frac {280 \, B c^{3} x^{7} + 140 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 280 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 70 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + 35 \, A a^{3} + 56 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 140 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 40 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{280 \, x^{8}} \] Input:

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^9,x, algorithm="maxima")
 

Output:

-1/280*(280*B*c^3*x^7 + 140*(3*B*b*c^2 + A*c^3)*x^6 + 280*(B*b^2*c + (B*a 
+ A*b)*c^2)*x^5 + 70*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 + 35* 
A*a^3 + 56*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 140*(B*a^2*b 
+ A*a*b^2 + A*a^2*c)*x^2 + 40*(B*a^3 + 3*A*a^2*b)*x)/x^8
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=-\frac {280 \, B c^{3} x^{7} + 420 \, B b c^{2} x^{6} + 140 \, A c^{3} x^{6} + 280 \, B b^{2} c x^{5} + 280 \, B a c^{2} x^{5} + 280 \, A b c^{2} x^{5} + 70 \, B b^{3} x^{4} + 420 \, B a b c x^{4} + 210 \, A b^{2} c x^{4} + 210 \, A a c^{2} x^{4} + 168 \, B a b^{2} x^{3} + 56 \, A b^{3} x^{3} + 168 \, B a^{2} c x^{3} + 336 \, A a b c x^{3} + 140 \, B a^{2} b x^{2} + 140 \, A a b^{2} x^{2} + 140 \, A a^{2} c x^{2} + 40 \, B a^{3} x + 120 \, A a^{2} b x + 35 \, A a^{3}}{280 \, x^{8}} \] Input:

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^9,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-1/280*(280*B*c^3*x^7 + 420*B*b*c^2*x^6 + 140*A*c^3*x^6 + 280*B*b^2*c*x^5 
+ 280*B*a*c^2*x^5 + 280*A*b*c^2*x^5 + 70*B*b^3*x^4 + 420*B*a*b*c*x^4 + 210 
*A*b^2*c*x^4 + 210*A*a*c^2*x^4 + 168*B*a*b^2*x^3 + 56*A*b^3*x^3 + 168*B*a^ 
2*c*x^3 + 336*A*a*b*c*x^3 + 140*B*a^2*b*x^2 + 140*A*a*b^2*x^2 + 140*A*a^2* 
c*x^2 + 40*B*a^3*x + 120*A*a^2*b*x + 35*A*a^3)/x^8
 

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=-\frac {x^3\,\left (\frac {3\,B\,c\,a^2}{5}+\frac {3\,B\,a\,b^2}{5}+\frac {6\,A\,c\,a\,b}{5}+\frac {A\,b^3}{5}\right )+x^4\,\left (\frac {B\,b^3}{4}+\frac {3\,A\,b^2\,c}{4}+\frac {3\,B\,a\,b\,c}{2}+\frac {3\,A\,a\,c^2}{4}\right )+x\,\left (\frac {B\,a^3}{7}+\frac {3\,A\,b\,a^2}{7}\right )+\frac {A\,a^3}{8}+x^6\,\left (\frac {A\,c^3}{2}+\frac {3\,B\,b\,c^2}{2}\right )+x^2\,\left (\frac {B\,a^2\,b}{2}+\frac {A\,c\,a^2}{2}+\frac {A\,a\,b^2}{2}\right )+x^5\,\left (B\,b^2\,c+A\,b\,c^2+B\,a\,c^2\right )+B\,c^3\,x^7}{x^8} \] Input:

int(((A + B*x)*(a + b*x + c*x^2)^3)/x^9,x)
 

Output:

-(x^3*((A*b^3)/5 + (3*B*a*b^2)/5 + (3*B*a^2*c)/5 + (6*A*a*b*c)/5) + x^4*(( 
B*b^3)/4 + (3*A*a*c^2)/4 + (3*A*b^2*c)/4 + (3*B*a*b*c)/2) + x*((B*a^3)/7 + 
 (3*A*a^2*b)/7) + (A*a^3)/8 + x^6*((A*c^3)/2 + (3*B*b*c^2)/2) + x^2*((A*a* 
b^2)/2 + (A*a^2*c)/2 + (B*a^2*b)/2) + x^5*(A*b*c^2 + B*a*c^2 + B*b^2*c) + 
B*c^3*x^7)/x^8
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^9} \, dx=\frac {-280 b \,c^{3} x^{7}-140 a \,c^{3} x^{6}-420 b^{2} c^{2} x^{6}-560 a b \,c^{2} x^{5}-280 b^{3} c \,x^{5}-210 a^{2} c^{2} x^{4}-630 a \,b^{2} c \,x^{4}-70 b^{4} x^{4}-504 a^{2} b c \,x^{3}-224 a \,b^{3} x^{3}-140 a^{3} c \,x^{2}-280 a^{2} b^{2} x^{2}-160 a^{3} b x -35 a^{4}}{280 x^{8}} \] Input:

int((B*x+A)*(c*x^2+b*x+a)^3/x^9,x)
 

Output:

( - 35*a**4 - 160*a**3*b*x - 140*a**3*c*x**2 - 280*a**2*b**2*x**2 - 504*a* 
*2*b*c*x**3 - 210*a**2*c**2*x**4 - 224*a*b**3*x**3 - 630*a*b**2*c*x**4 - 5 
60*a*b*c**2*x**5 - 140*a*c**3*x**6 - 70*b**4*x**4 - 280*b**3*c*x**5 - 420* 
b**2*c**2*x**6 - 280*b*c**3*x**7)/(280*x**8)