\(\int \frac {(A+B x) (a+b x+c x^2)^3}{x^3} \, dx\) [30]

Optimal result

Integrand size = 21, antiderivative size = 153 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=-\frac {a^3 A}{2 x^2}-\frac {a^2 (3 A b+a B)}{x}+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x+\frac {1}{2} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^2+c \left (b^2 B+A b c+a B c\right ) x^3+\frac {1}{4} c^2 (3 b B+A c) x^4+\frac {1}{5} B c^3 x^5+3 a \left (a b B+A \left (b^2+a c\right )\right ) \log (x) \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=-\frac {a^3 A}{2 x^2}-\frac {a^2 (3 A b+a B)}{x}+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x+\frac {1}{2} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^2+c \left (b^2 B+A b c+a B c\right ) x^3+\frac {1}{4} c^2 (3 b B+A c) x^4+\frac {1}{5} B c^3 x^5+3 a \left (a b B+A \left (b^2+a c\right )\right ) \log (x) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 153, 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.

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

\(\Big \downarrow \) 1195

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(a^3*A)/x^2 - (a^2*(3*A*b + a*B))/x + (3*a*B*(b^2 + a*c) + A*(b^3 + 6 
*a*b*c))*x + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^2)/2 + c*(b^2* 
B + A*b*c + a*B*c)*x^3 + (c^2*(3*b*B + A*c)*x^4)/4 + (B*c^3*x^5)/5 + 3*a*( 
a*b*B + A*(b^2 + a*c))*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 = 0.99 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.08

method	result	size
norman	\(\frac {\left (\frac {1}{4} A \,c^{3}+\frac {3}{4} B b \,c^{2}\right ) x^{6}+\left (\frac {3}{2} A a \,c^{2}+\frac {3}{2} A \,b^{2} c +3 B a b c +\frac {1}{2} B \,b^{3}\right ) x^{4}+\left (-3 A \,a^{2} b -B \,a^{3}\right ) x +\left (A b \,c^{2}+B a \,c^{2}+B \,b^{2} c \right ) x^{5}+\left (6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}\right ) x^{3}-\frac {a^{3} A}{2}+\frac {B \,c^{3} x^{7}}{5}}{x^{2}}+\left (3 a^{2} A c +3 A a \,b^{2}+3 B \,a^{2} b \right ) \ln \left (x \right )\)	\(165\)
default	\(\frac {B \,c^{3} x^{5}}{5}+\frac {A \,c^{3} x^{4}}{4}+\frac {3 B b \,c^{2} x^{4}}{4}+A b \,c^{2} x^{3}+B a \,c^{2} x^{3}+B \,b^{2} c \,x^{3}+\frac {3 A a \,c^{2} x^{2}}{2}+\frac {3 A \,b^{2} c \,x^{2}}{2}+3 B a b c \,x^{2}+\frac {x^{2} B \,b^{3}}{2}+6 A a b c x +A \,b^{3} x +3 B \,a^{2} c x +3 B a \,b^{2} x -\frac {a^{3} A}{2 x^{2}}+3 a \left (A a c +b^{2} A +a b B \right ) \ln \left (x \right )-\frac {a^{2} \left (3 A b +B a \right )}{x}\)	\(168\)
risch	\(\frac {B \,c^{3} x^{5}}{5}+\frac {A \,c^{3} x^{4}}{4}+\frac {3 B b \,c^{2} x^{4}}{4}+A b \,c^{2} x^{3}+B a \,c^{2} x^{3}+B \,b^{2} c \,x^{3}+\frac {3 A a \,c^{2} x^{2}}{2}+\frac {3 A \,b^{2} c \,x^{2}}{2}+3 B a b c \,x^{2}+\frac {x^{2} B \,b^{3}}{2}+6 A a b c x +A \,b^{3} x +3 B \,a^{2} c x +3 B a \,b^{2} x +\frac {\left (-3 A \,a^{2} b -B \,a^{3}\right ) x -\frac {a^{3} A}{2}}{x^{2}}+3 A \ln \left (x \right ) a^{2} c +3 A \ln \left (x \right ) a \,b^{2}+3 B \ln \left (x \right ) a^{2} b\)	\(178\)
parallelrisch	\(\frac {4 B \,c^{3} x^{7}+5 A \,c^{3} x^{6}+15 B b \,c^{2} x^{6}+20 A b \,c^{2} x^{5}+20 B a \,c^{2} x^{5}+20 B \,b^{2} c \,x^{5}+30 A a \,c^{2} x^{4}+30 A \,b^{2} c \,x^{4}+60 B a b c \,x^{4}+10 B \,b^{3} x^{4}+60 A \ln \left (x \right ) x^{2} a^{2} c +60 A \ln \left (x \right ) x^{2} a \,b^{2}+120 A a b c \,x^{3}+20 A \,b^{3} x^{3}+60 B \ln \left (x \right ) x^{2} a^{2} b +60 B \,a^{2} c \,x^{3}+60 B a \,b^{2} x^{3}-60 A \,a^{2} b x -20 B \,a^{3} x -10 a^{3} A}{20 x^{2}}\)	\(198\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=\frac {4 \, B c^{3} x^{7} + 5 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 20 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 10 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 10 \, A a^{3} + 20 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 60 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} \log \left (x\right ) - 20 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{2}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=\frac {B c^{3} x^{5}}{5} + 3 a \left (A a c + A b^{2} + B a b\right ) \log {\left (x \right )} + x^{4} \left (\frac {A c^{3}}{4} + \frac {3 B b c^{2}}{4}\right ) + x^{3} \left (A b c^{2} + B a c^{2} + B b^{2} c\right ) + x^{2} \cdot \left (\frac {3 A a c^{2}}{2} + \frac {3 A b^{2} c}{2} + 3 B a b c + \frac {B b^{3}}{2}\right ) + x \left (6 A a b c + A b^{3} + 3 B a^{2} c + 3 B a b^{2}\right ) + \frac {- A a^{3} + x \left (- 6 A a^{2} b - 2 B a^{3}\right )}{2 x^{2}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=\frac {1}{5} \, B c^{3} x^{5} + \frac {1}{4} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{3} + \frac {1}{2} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{2} + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x + 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} \log \left (x\right ) - \frac {A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2 \, x^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=\frac {1}{5} \, B c^{3} x^{5} + \frac {3}{4} \, B b c^{2} x^{4} + \frac {1}{4} \, A c^{3} x^{4} + B b^{2} c x^{3} + B a c^{2} x^{3} + A b c^{2} x^{3} + \frac {1}{2} \, B b^{3} x^{2} + 3 \, B a b c x^{2} + \frac {3}{2} \, A b^{2} c x^{2} + \frac {3}{2} \, A a c^{2} x^{2} + 3 \, B a b^{2} x + A b^{3} x + 3 \, B a^{2} c x + 6 \, A a b c x + 3 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} \log \left ({\left | x \right |}\right ) - \frac {A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2 \, x^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=x^2\,\left (\frac {B\,b^3}{2}+\frac {3\,A\,b^2\,c}{2}+3\,B\,a\,b\,c+\frac {3\,A\,a\,c^2}{2}\right )-\frac {x\,\left (B\,a^3+3\,A\,b\,a^2\right )+\frac {A\,a^3}{2}}{x^2}+x^4\,\left (\frac {A\,c^3}{4}+\frac {3\,B\,b\,c^2}{4}\right )+x^3\,\left (B\,b^2\,c+A\,b\,c^2+B\,a\,c^2\right )+x\,\left (3\,B\,c\,a^2+3\,B\,a\,b^2+6\,A\,c\,a\,b+A\,b^3\right )+\ln \left (x\right )\,\left (3\,B\,a^2\,b+3\,A\,c\,a^2+3\,A\,a\,b^2\right )+\frac {B\,c^3\,x^5}{5} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^3} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) a^{3} c \,x^{2}+120 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{2}-10 a^{4}-80 a^{3} b x +180 a^{2} b c \,x^{3}+30 a^{2} c^{2} x^{4}+80 a \,b^{3} x^{3}+90 a \,b^{2} c \,x^{4}+40 a b \,c^{2} x^{5}+5 a \,c^{3} x^{6}+10 b^{4} x^{4}+20 b^{3} c \,x^{5}+15 b^{2} c^{2} x^{6}+4 b \,c^{3} x^{7}}{20 x^{2}} \] Input:

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

Output:

(60*log(x)*a**3*c*x**2 + 120*log(x)*a**2*b**2*x**2 - 10*a**4 - 80*a**3*b*x 
 + 180*a**2*b*c*x**3 + 30*a**2*c**2*x**4 + 80*a*b**3*x**3 + 90*a*b**2*c*x* 
*4 + 40*a*b*c**2*x**5 + 5*a*c**3*x**6 + 10*b**4*x**4 + 20*b**3*c*x**5 + 15 
*b**2*c**2*x**6 + 4*b*c**3*x**7)/(20*x**2)