\(\int \frac {(A+B x) (a+b x+c x^2)^3}{x^2} \, dx\) [29]

Optimal result

Integrand size = 21, antiderivative size = 156 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=-\frac {a^3 A}{x}+3 a \left (a b B+A \left (b^2+a c\right )\right ) x+\frac {1}{2} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^2+\frac {1}{3} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^3+\frac {3}{4} c \left (b^2 B+A b c+a B c\right ) x^4+\frac {1}{5} c^2 (3 b B+A c) x^5+\frac {1}{6} B c^3 x^6+a^2 (3 A b+a B) \log (x) \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=-\frac {a^3 A}{x}+3 a \left (a b B+A \left (b^2+a c\right )\right ) x+\frac {1}{2} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^2+\frac {1}{3} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^3+\frac {3}{4} c \left (b^2 B+A b c+a B c\right ) x^4+\frac {1}{5} c^2 (3 b B+A c) x^5+\frac {1}{6} B c^3 x^6+a^2 (3 A b+a B) \log (x) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 156, 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.

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

\(\Big \downarrow \) 1195

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^3 A}{x}+a^2 \log (x) (a B+3 A b)+\frac {3}{4} c x^4 \left (a B c+A b c+b^2 B\right )+3 a x \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{3} x^3 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {1}{2} x^2 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{5} c^2 x^5 (A c+3 b B)+\frac {1}{6} B c^3 x^6\)

Input:

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

Output:

-((a^3*A)/x) + 3*a*(a*b*B + A*(b^2 + a*c))*x + ((3*a*B*(b^2 + a*c) + A*(b^ 
3 + 6*a*b*c))*x^2)/2 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^3)/3 
 + (3*c*(b^2*B + A*b*c + a*B*c)*x^4)/4 + (c^2*(3*b*B + A*c)*x^5)/5 + (B*c^ 
3*x^6)/6 + a^2*(3*A*b + a*B)*Log[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 = 1.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08

method	result	size
norman	\(\frac {\left (\frac {1}{5} A \,c^{3}+\frac {3}{5} B b \,c^{2}\right ) x^{6}+\left (\frac {3}{4} A b \,c^{2}+\frac {3}{4} B a \,c^{2}+\frac {3}{4} B \,b^{2} c \right ) x^{5}+\left (A a \,c^{2}+A \,b^{2} c +2 B a b c +\frac {1}{3} B \,b^{3}\right ) x^{4}+\left (3 A a b c +\frac {1}{2} A \,b^{3}+\frac {3}{2} B \,a^{2} c +\frac {3}{2} B a \,b^{2}\right ) x^{3}+\left (3 a^{2} A c +3 A a \,b^{2}+3 B \,a^{2} b \right ) x^{2}-a^{3} A +\frac {B \,c^{3} x^{7}}{6}}{x}+\left (3 A \,a^{2} b +B \,a^{3}\right ) \ln \left (x \right )\)	\(168\)
default	\(\frac {B \,c^{3} x^{6}}{6}+\frac {A \,c^{3} x^{5}}{5}+\frac {3 B b \,c^{2} x^{5}}{5}+\frac {3 A b \,c^{2} x^{4}}{4}+\frac {3 B a \,c^{2} x^{4}}{4}+\frac {3 x^{4} B \,b^{2} c}{4}+A a \,c^{2} x^{3}+A \,b^{2} c \,x^{3}+2 B a b c \,x^{3}+\frac {x^{3} B \,b^{3}}{3}+3 A a b c \,x^{2}+\frac {A \,b^{3} x^{2}}{2}+\frac {3 B \,a^{2} c \,x^{2}}{2}+\frac {3 B a \,b^{2} x^{2}}{2}+3 a^{2} A c x +3 A a \,b^{2} x +3 B \,a^{2} b x +a^{2} \left (3 A b +B a \right ) \ln \left (x \right )-\frac {a^{3} A}{x}\)	\(181\)
risch	\(\frac {B \,c^{3} x^{6}}{6}+\frac {A \,c^{3} x^{5}}{5}+\frac {3 B b \,c^{2} x^{5}}{5}+\frac {3 A b \,c^{2} x^{4}}{4}+\frac {3 B a \,c^{2} x^{4}}{4}+\frac {3 x^{4} B \,b^{2} c}{4}+A a \,c^{2} x^{3}+A \,b^{2} c \,x^{3}+2 B a b c \,x^{3}+\frac {x^{3} B \,b^{3}}{3}+3 A a b c \,x^{2}+\frac {A \,b^{3} x^{2}}{2}+\frac {3 B \,a^{2} c \,x^{2}}{2}+\frac {3 B a \,b^{2} x^{2}}{2}+3 a^{2} A c x +3 A a \,b^{2} x +3 B \,a^{2} b x -\frac {a^{3} A}{x}+3 A \ln \left (x \right ) a^{2} b +B \ln \left (x \right ) a^{3}\)	\(183\)
parallelrisch	\(\frac {10 B \,c^{3} x^{7}+12 A \,c^{3} x^{6}+36 B b \,c^{2} x^{6}+45 A b \,c^{2} x^{5}+45 B a \,c^{2} x^{5}+45 B \,b^{2} c \,x^{5}+60 A a \,c^{2} x^{4}+60 A \,b^{2} c \,x^{4}+120 B a b c \,x^{4}+20 B \,b^{3} x^{4}+180 A a b c \,x^{3}+30 A \,b^{3} x^{3}+90 B \,a^{2} c \,x^{3}+90 B a \,b^{2} x^{3}+180 A \ln \left (x \right ) x \,a^{2} b +180 A \,a^{2} c \,x^{2}+180 A a \,b^{2} x^{2}+60 B \ln \left (x \right ) x \,a^{3}+180 B \,a^{2} b \,x^{2}-60 a^{3} A}{60 x}\)	\(196\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=\frac {10 \, B c^{3} x^{7} + 12 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 45 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 20 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 60 \, A a^{3} + 30 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 180 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 60 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x \log \left (x\right )}{60 \, x} \] Input:

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

Output:

1/60*(10*B*c^3*x^7 + 12*(3*B*b*c^2 + A*c^3)*x^6 + 45*(B*b^2*c + (B*a + A*b 
)*c^2)*x^5 + 20*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 - 60*A*a^3 
 + 30*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 180*(B*a^2*b + A*a 
*b^2 + A*a^2*c)*x^2 + 60*(B*a^3 + 3*A*a^2*b)*x*log(x))/x
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=- \frac {A a^{3}}{x} + \frac {B c^{3} x^{6}}{6} + a^{2} \cdot \left (3 A b + B a\right ) \log {\left (x \right )} + x^{5} \left (\frac {A c^{3}}{5} + \frac {3 B b c^{2}}{5}\right ) + x^{4} \cdot \left (\frac {3 A b c^{2}}{4} + \frac {3 B a c^{2}}{4} + \frac {3 B b^{2} c}{4}\right ) + x^{3} \left (A a c^{2} + A b^{2} c + 2 B a b c + \frac {B b^{3}}{3}\right ) + x^{2} \cdot \left (3 A a b c + \frac {A b^{3}}{2} + \frac {3 B a^{2} c}{2} + \frac {3 B a b^{2}}{2}\right ) + x \left (3 A a^{2} c + 3 A a b^{2} + 3 B a^{2} b\right ) \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=\frac {1}{6} \, B c^{3} x^{6} + \frac {1}{5} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{5} + \frac {3}{4} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{3} - \frac {A a^{3}}{x} + \frac {1}{2} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{2} + 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x + {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left (x\right ) \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=\frac {1}{6} \, B c^{3} x^{6} + \frac {3}{5} \, B b c^{2} x^{5} + \frac {1}{5} \, A c^{3} x^{5} + \frac {3}{4} \, B b^{2} c x^{4} + \frac {3}{4} \, B a c^{2} x^{4} + \frac {3}{4} \, A b c^{2} x^{4} + \frac {1}{3} \, B b^{3} x^{3} + 2 \, B a b c x^{3} + A b^{2} c x^{3} + A a c^{2} x^{3} + \frac {3}{2} \, B a b^{2} x^{2} + \frac {1}{2} \, A b^{3} x^{2} + \frac {3}{2} \, B a^{2} c x^{2} + 3 \, A a b c x^{2} + 3 \, B a^{2} b x + 3 \, A a b^{2} x + 3 \, A a^{2} c x - \frac {A a^{3}}{x} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | x \right |}\right ) \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 10.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=x^2\,\left (\frac {3\,B\,c\,a^2}{2}+\frac {3\,B\,a\,b^2}{2}+3\,A\,c\,a\,b+\frac {A\,b^3}{2}\right )+x^3\,\left (\frac {B\,b^3}{3}+A\,b^2\,c+2\,B\,a\,b\,c+A\,a\,c^2\right )+x\,\left (3\,B\,a^2\,b+3\,A\,c\,a^2+3\,A\,a\,b^2\right )+x^5\,\left (\frac {A\,c^3}{5}+\frac {3\,B\,b\,c^2}{5}\right )+\ln \left (x\right )\,\left (B\,a^3+3\,A\,b\,a^2\right )+x^4\,\left (\frac {3\,B\,b^2\,c}{4}+\frac {3\,A\,b\,c^2}{4}+\frac {3\,B\,a\,c^2}{4}\right )-\frac {A\,a^3}{x}+\frac {B\,c^3\,x^6}{6} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^2} \, dx=\frac {240 \,\mathrm {log}\left (x \right ) a^{3} b x -60 a^{4}+180 a^{3} c \,x^{2}+360 a^{2} b^{2} x^{2}+270 a^{2} b c \,x^{3}+60 a^{2} c^{2} x^{4}+120 a \,b^{3} x^{3}+180 a \,b^{2} c \,x^{4}+90 a b \,c^{2} x^{5}+12 a \,c^{3} x^{6}+20 b^{4} x^{4}+45 b^{3} c \,x^{5}+36 b^{2} c^{2} x^{6}+10 b \,c^{3} x^{7}}{60 x} \] Input:

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

Output:

(240*log(x)*a**3*b*x - 60*a**4 + 180*a**3*c*x**2 + 360*a**2*b**2*x**2 + 27 
0*a**2*b*c*x**3 + 60*a**2*c**2*x**4 + 120*a*b**3*x**3 + 180*a*b**2*c*x**4 
+ 90*a*b*c**2*x**5 + 12*a*c**3*x**6 + 20*b**4*x**4 + 45*b**3*c*x**5 + 36*b 
**2*c**2*x**6 + 10*b*c**3*x**7)/(60*x)