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

Optimal result

Integrand size = 21, antiderivative size = 157 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=a^2 (3 A b+a B) x+\frac {3}{2} a \left (a b B+A \left (b^2+a c\right )\right ) x^2+\frac {1}{3} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^3+\frac {1}{4} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^4+\frac {3}{5} c \left (b^2 B+A b c+a B c\right ) x^5+\frac {1}{6} c^2 (3 b B+A c) x^6+\frac {1}{7} B c^3 x^7+a^3 A \log (x) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=a^2 (3 A b+a B) x+\frac {3}{2} a \left (a b B+A \left (b^2+a c\right )\right ) x^2+\frac {1}{3} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^3+\frac {1}{4} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^4+\frac {3}{5} c \left (b^2 B+A b c+a B c\right ) x^5+\frac {1}{6} c^2 (3 b B+A c) x^6+\frac {1}{7} B c^3 x^7+a^3 A \log (x) \] Input:

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

Output:

a^2*(3*A*b + a*B)*x + (3*a*(a*b*B + A*(b^2 + a*c))*x^2)/2 + ((3*a*B*(b^2 + 
 a*c) + A*(b^3 + 6*a*b*c))*x^3)/3 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a* 
A*c^2)*x^4)/4 + (3*c*(b^2*B + A*b*c + a*B*c)*x^5)/5 + (c^2*(3*b*B + A*c)*x 
^6)/6 + (B*c^3*x^7)/7 + a^3*A*Log[x]
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 157, 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.

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

\(\Big \downarrow \) 1195

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

\(\Big \downarrow \) 2009

\(\displaystyle a^3 A \log (x)+a^2 x (a B+3 A b)+\frac {3}{5} c x^5 \left (a B c+A b c+b^2 B\right )+\frac {3}{2} a x^2 \left (A \left (a c+b^2\right )+a b B\right )+\frac {1}{4} x^4 \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {1}{3} x^3 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{6} c^2 x^6 (A c+3 b B)+\frac {1}{7} B c^3 x^7\)

Input:

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

Output:

a^2*(3*A*b + a*B)*x + (3*a*(a*b*B + A*(b^2 + a*c))*x^2)/2 + ((3*a*B*(b^2 + 
 a*c) + A*(b^3 + 6*a*b*c))*x^3)/3 + ((b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a* 
A*c^2)*x^4)/4 + (3*c*(b^2*B + A*b*c + a*B*c)*x^5)/5 + (c^2*(3*b*B + A*c)*x 
^6)/6 + (B*c^3*x^7)/7 + a^3*A*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 = 0.98 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.04

method	result	size
norman	\(\left (\frac {1}{6} A \,c^{3}+\frac {1}{2} B b \,c^{2}\right ) x^{6}+\left (\frac {3}{5} A b \,c^{2}+\frac {3}{5} B a \,c^{2}+\frac {3}{5} B \,b^{2} c \right ) x^{5}+\left (\frac {3}{2} a^{2} A c +\frac {3}{2} A a \,b^{2}+\frac {3}{2} B \,a^{2} b \right ) x^{2}+\left (\frac {3}{4} A a \,c^{2}+\frac {3}{4} A \,b^{2} c +\frac {3}{2} B a b c +\frac {1}{4} B \,b^{3}\right ) x^{4}+\left (2 A a b c +\frac {1}{3} A \,b^{3}+B \,a^{2} c +B a \,b^{2}\right ) x^{3}+\left (3 A \,a^{2} b +B \,a^{3}\right ) x +\frac {B \,c^{3} x^{7}}{7}+a^{3} A \ln \left (x \right )\)	\(163\)
default	\(\frac {B \,c^{3} x^{7}}{7}+\frac {A \,c^{3} x^{6}}{6}+\frac {B b \,c^{2} x^{6}}{2}+\frac {3 A b \,c^{2} x^{5}}{5}+\frac {3 B a \,c^{2} x^{5}}{5}+\frac {3 B \,b^{2} c \,x^{5}}{5}+\frac {3 A a \,c^{2} x^{4}}{4}+\frac {3 A \,b^{2} c \,x^{4}}{4}+\frac {3 B a b c \,x^{4}}{2}+\frac {B \,b^{3} x^{4}}{4}+2 A a b c \,x^{3}+\frac {A \,b^{3} x^{3}}{3}+B \,a^{2} c \,x^{3}+B a \,b^{2} x^{3}+\frac {3 A \,a^{2} c \,x^{2}}{2}+\frac {3 A a \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}+3 A \,a^{2} b x +B \,a^{3} x +a^{3} A \ln \left (x \right )\)	\(185\)
risch	\(\frac {B \,c^{3} x^{7}}{7}+\frac {A \,c^{3} x^{6}}{6}+\frac {B b \,c^{2} x^{6}}{2}+\frac {3 A b \,c^{2} x^{5}}{5}+\frac {3 B a \,c^{2} x^{5}}{5}+\frac {3 B \,b^{2} c \,x^{5}}{5}+\frac {3 A a \,c^{2} x^{4}}{4}+\frac {3 A \,b^{2} c \,x^{4}}{4}+\frac {3 B a b c \,x^{4}}{2}+\frac {B \,b^{3} x^{4}}{4}+2 A a b c \,x^{3}+\frac {A \,b^{3} x^{3}}{3}+B \,a^{2} c \,x^{3}+B a \,b^{2} x^{3}+\frac {3 A \,a^{2} c \,x^{2}}{2}+\frac {3 A a \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}+3 A \,a^{2} b x +B \,a^{3} x +a^{3} A \ln \left (x \right )\)	\(185\)
parallelrisch	\(\frac {B \,c^{3} x^{7}}{7}+\frac {A \,c^{3} x^{6}}{6}+\frac {B b \,c^{2} x^{6}}{2}+\frac {3 A b \,c^{2} x^{5}}{5}+\frac {3 B a \,c^{2} x^{5}}{5}+\frac {3 B \,b^{2} c \,x^{5}}{5}+\frac {3 A a \,c^{2} x^{4}}{4}+\frac {3 A \,b^{2} c \,x^{4}}{4}+\frac {3 B a b c \,x^{4}}{2}+\frac {B \,b^{3} x^{4}}{4}+2 A a b c \,x^{3}+\frac {A \,b^{3} x^{3}}{3}+B \,a^{2} c \,x^{3}+B a \,b^{2} x^{3}+\frac {3 A \,a^{2} c \,x^{2}}{2}+\frac {3 A a \,b^{2} x^{2}}{2}+\frac {3 B \,a^{2} b \,x^{2}}{2}+3 A \,a^{2} b x +B \,a^{3} x +a^{3} A \ln \left (x \right )\)	\(185\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=\frac {1}{7} \, B c^{3} x^{7} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{5} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + A a^{3} \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + \frac {3}{2} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} x \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.22 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=A a^{3} \log {\left (x \right )} + \frac {B c^{3} x^{7}}{7} + x^{6} \left (\frac {A c^{3}}{6} + \frac {B b c^{2}}{2}\right ) + x^{5} \cdot \left (\frac {3 A b c^{2}}{5} + \frac {3 B a c^{2}}{5} + \frac {3 B b^{2} c}{5}\right ) + x^{4} \cdot \left (\frac {3 A a c^{2}}{4} + \frac {3 A b^{2} c}{4} + \frac {3 B a b c}{2} + \frac {B b^{3}}{4}\right ) + x^{3} \cdot \left (2 A a b c + \frac {A b^{3}}{3} + B a^{2} c + B a b^{2}\right ) + x^{2} \cdot \left (\frac {3 A a^{2} c}{2} + \frac {3 A a b^{2}}{2} + \frac {3 B a^{2} b}{2}\right ) + x \left (3 A a^{2} b + B a^{3}\right ) \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=\frac {1}{7} \, B c^{3} x^{7} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{5} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + A a^{3} \log \left (x\right ) + \frac {1}{3} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + \frac {3}{2} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} x \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=\frac {1}{7} \, B c^{3} x^{7} + \frac {1}{2} \, B b c^{2} x^{6} + \frac {1}{6} \, A c^{3} x^{6} + \frac {3}{5} \, B b^{2} c x^{5} + \frac {3}{5} \, B a c^{2} x^{5} + \frac {3}{5} \, A b c^{2} x^{5} + \frac {1}{4} \, B b^{3} x^{4} + \frac {3}{2} \, B a b c x^{4} + \frac {3}{4} \, A b^{2} c x^{4} + \frac {3}{4} \, A a c^{2} x^{4} + B a b^{2} x^{3} + \frac {1}{3} \, A b^{3} x^{3} + B a^{2} c x^{3} + 2 \, A a b c x^{3} + \frac {3}{2} \, B a^{2} b x^{2} + \frac {3}{2} \, A a b^{2} x^{2} + \frac {3}{2} \, A a^{2} c x^{2} + B a^{3} x + 3 \, A a^{2} b x + A a^{3} \log \left ({\left | x \right |}\right ) \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 10.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=x^3\,\left (B\,c\,a^2+B\,a\,b^2+2\,A\,c\,a\,b+\frac {A\,b^3}{3}\right )+x^4\,\left (\frac {B\,b^3}{4}+\frac {3\,A\,b^2\,c}{4}+\frac {3\,B\,a\,b\,c}{2}+\frac {3\,A\,a\,c^2}{4}\right )+x\,\left (B\,a^3+3\,A\,b\,a^2\right )+x^6\,\left (\frac {A\,c^3}{6}+\frac {B\,b\,c^2}{2}\right )+x^2\,\left (\frac {3\,B\,a^2\,b}{2}+\frac {3\,A\,c\,a^2}{2}+\frac {3\,A\,a\,b^2}{2}\right )+x^5\,\left (\frac {3\,B\,b^2\,c}{5}+\frac {3\,A\,b\,c^2}{5}+\frac {3\,B\,a\,c^2}{5}\right )+\frac {B\,c^3\,x^7}{7}+A\,a^3\,\ln \left (x\right ) \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x} \, dx=\mathrm {log}\left (x \right ) a^{4}+4 a^{3} b x +\frac {3 a^{3} c \,x^{2}}{2}+3 a^{2} b^{2} x^{2}+3 a^{2} b c \,x^{3}+\frac {3 a^{2} c^{2} x^{4}}{4}+\frac {4 a \,b^{3} x^{3}}{3}+\frac {9 a \,b^{2} c \,x^{4}}{4}+\frac {6 a b \,c^{2} x^{5}}{5}+\frac {a \,c^{3} x^{6}}{6}+\frac {b^{4} x^{4}}{4}+\frac {3 b^{3} c \,x^{5}}{5}+\frac {b^{2} c^{2} x^{6}}{2}+\frac {b \,c^{3} x^{7}}{7} \] Input:

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

Output:

(420*log(x)*a**4 + 1680*a**3*b*x + 630*a**3*c*x**2 + 1260*a**2*b**2*x**2 + 
 1260*a**2*b*c*x**3 + 315*a**2*c**2*x**4 + 560*a*b**3*x**3 + 945*a*b**2*c* 
x**4 + 504*a*b*c**2*x**5 + 70*a*c**3*x**6 + 105*b**4*x**4 + 252*b**3*c*x** 
5 + 210*b**2*c**2*x**6 + 60*b*c**3*x**7)/420