Integrand size = 21, antiderivative size = 92 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=a (2 A b+a B) x+\frac {1}{2} \left (2 a b B+A \left (b^2+2 a c\right )\right ) x^2+\frac {1}{3} \left (b^2 B+2 A b c+2 a B c\right ) x^3+\frac {1}{4} c (2 b B+A c) x^4+\frac {1}{5} B c^2 x^5+a^2 A \log (x) \] Output:
a*(2*A*b+B*a)*x+1/2*(2*a*b*B+A*(2*a*c+b^2))*x^2+1/3*(2*A*b*c+2*B*a*c+B*b^2 )*x^3+1/4*c*(A*c+2*B*b)*x^4+1/5*B*c^2*x^5+a^2*A*ln(x)
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=a (2 A b+a B) x+\frac {1}{2} \left (A b^2+2 a b B+2 a A c\right ) x^2+\frac {1}{3} \left (b^2 B+2 A b c+2 a B c\right ) x^3+\frac {1}{4} c (2 b B+A c) x^4+\frac {1}{5} B c^2 x^5+a^2 A \log (x) \] Input:
Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x,x]
Output:
a*(2*A*b + a*B)*x + ((A*b^2 + 2*a*b*B + 2*a*A*c)*x^2)/2 + ((b^2*B + 2*A*b* c + 2*a*B*c)*x^3)/3 + (c*(2*b*B + A*c)*x^4)/4 + (B*c^2*x^5)/5 + a^2*A*Log[ x]
Time = 0.25 (sec) , antiderivative size = 92, 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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^2 A}{x}+x^2 \left (2 a B c+2 A b c+b^2 B\right )+x \left (A \left (2 a c+b^2\right )+2 a b B\right )+a (a B+2 A b)+c x^3 (A c+2 b B)+B c^2 x^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^2 A \log (x)+\frac {1}{3} x^3 \left (2 a B c+2 A b c+b^2 B\right )+\frac {1}{2} x^2 \left (A \left (2 a c+b^2\right )+2 a b B\right )+a x (a B+2 A b)+\frac {1}{4} c x^4 (A c+2 b B)+\frac {1}{5} B c^2 x^5\) |
Input:
Int[((A + B*x)*(a + b*x + c*x^2)^2)/x,x]
Output:
a*(2*A*b + a*B)*x + ((2*a*b*B + A*(b^2 + 2*a*c))*x^2)/2 + ((b^2*B + 2*A*b* c + 2*a*B*c)*x^3)/3 + (c*(2*b*B + A*c)*x^4)/4 + (B*c^2*x^5)/5 + a^2*A*Log[ x]
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]
Time = 0.93 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95
method | result | size |
norman | \(\left (\frac {1}{4} A \,c^{2}+\frac {1}{2} B b c \right ) x^{4}+\left (A a c +\frac {1}{2} b^{2} A +a b B \right ) x^{2}+\left (\frac {2}{3} A b c +\frac {2}{3} a B c +\frac {1}{3} B \,b^{2}\right ) x^{3}+\left (2 a b A +a^{2} B \right ) x +\frac {B \,c^{2} x^{5}}{5}+a^{2} A \ln \left (x \right )\) | \(87\) |
default | \(\frac {B \,c^{2} x^{5}}{5}+\frac {x^{4} A \,c^{2}}{4}+\frac {x^{4} B b c}{2}+\frac {2 x^{3} A b c}{3}+\frac {2 B a c \,x^{3}}{3}+\frac {x^{3} B \,b^{2}}{3}+A a c \,x^{2}+\frac {x^{2} b^{2} A}{2}+B a \,x^{2} b +2 a b A x +a^{2} B x +a^{2} A \ln \left (x \right )\) | \(95\) |
risch | \(\frac {B \,c^{2} x^{5}}{5}+\frac {x^{4} A \,c^{2}}{4}+\frac {x^{4} B b c}{2}+\frac {2 x^{3} A b c}{3}+\frac {2 B a c \,x^{3}}{3}+\frac {x^{3} B \,b^{2}}{3}+A a c \,x^{2}+\frac {x^{2} b^{2} A}{2}+B a \,x^{2} b +2 a b A x +a^{2} B x +a^{2} A \ln \left (x \right )\) | \(95\) |
parallelrisch | \(\frac {B \,c^{2} x^{5}}{5}+\frac {x^{4} A \,c^{2}}{4}+\frac {x^{4} B b c}{2}+\frac {2 x^{3} A b c}{3}+\frac {2 B a c \,x^{3}}{3}+\frac {x^{3} B \,b^{2}}{3}+A a c \,x^{2}+\frac {x^{2} b^{2} A}{2}+B a \,x^{2} b +2 a b A x +a^{2} B x +a^{2} A \ln \left (x \right )\) | \(95\) |
Input:
int((B*x+A)*(c*x^2+b*x+a)^2/x,x,method=_RETURNVERBOSE)
Output:
(1/4*A*c^2+1/2*B*b*c)*x^4+(A*a*c+1/2*b^2*A+a*b*B)*x^2+(2/3*A*b*c+2/3*a*B*c +1/3*B*b^2)*x^3+(2*A*a*b+B*a^2)*x+1/5*B*c^2*x^5+a^2*A*ln(x)
Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=\frac {1}{5} \, B c^{2} x^{5} + \frac {1}{4} \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + A a^{2} \log \left (x\right ) + \frac {1}{2} \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + {\left (B a^{2} + 2 \, A a b\right )} x \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x,x, algorithm="fricas")
Output:
1/5*B*c^2*x^5 + 1/4*(2*B*b*c + A*c^2)*x^4 + 1/3*(B*b^2 + 2*(B*a + A*b)*c)* x^3 + A*a^2*log(x) + 1/2*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + (B*a^2 + 2*A*a* b)*x
Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=A a^{2} \log {\left (x \right )} + \frac {B c^{2} x^{5}}{5} + x^{4} \left (\frac {A c^{2}}{4} + \frac {B b c}{2}\right ) + x^{3} \cdot \left (\frac {2 A b c}{3} + \frac {2 B a c}{3} + \frac {B b^{2}}{3}\right ) + x^{2} \left (A a c + \frac {A b^{2}}{2} + B a b\right ) + x \left (2 A a b + B a^{2}\right ) \] Input:
integrate((B*x+A)*(c*x**2+b*x+a)**2/x,x)
Output:
A*a**2*log(x) + B*c**2*x**5/5 + x**4*(A*c**2/4 + B*b*c/2) + x**3*(2*A*b*c/ 3 + 2*B*a*c/3 + B*b**2/3) + x**2*(A*a*c + A*b**2/2 + B*a*b) + x*(2*A*a*b + B*a**2)
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=\frac {1}{5} \, B c^{2} x^{5} + \frac {1}{4} \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + A a^{2} \log \left (x\right ) + \frac {1}{2} \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + {\left (B a^{2} + 2 \, A a b\right )} x \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x,x, algorithm="maxima")
Output:
1/5*B*c^2*x^5 + 1/4*(2*B*b*c + A*c^2)*x^4 + 1/3*(B*b^2 + 2*(B*a + A*b)*c)* x^3 + A*a^2*log(x) + 1/2*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + (B*a^2 + 2*A*a* b)*x
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=\frac {1}{5} \, B c^{2} x^{5} + \frac {1}{2} \, B b c x^{4} + \frac {1}{4} \, A c^{2} x^{4} + \frac {1}{3} \, B b^{2} x^{3} + \frac {2}{3} \, B a c x^{3} + \frac {2}{3} \, A b c x^{3} + B a b x^{2} + \frac {1}{2} \, A b^{2} x^{2} + A a c x^{2} + B a^{2} x + 2 \, A a b x + A a^{2} \log \left ({\left | x \right |}\right ) \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x,x, algorithm="giac")
Output:
1/5*B*c^2*x^5 + 1/2*B*b*c*x^4 + 1/4*A*c^2*x^4 + 1/3*B*b^2*x^3 + 2/3*B*a*c* x^3 + 2/3*A*b*c*x^3 + B*a*b*x^2 + 1/2*A*b^2*x^2 + A*a*c*x^2 + B*a^2*x + 2* A*a*b*x + A*a^2*log(abs(x))
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=x^4\,\left (\frac {A\,c^2}{4}+\frac {B\,b\,c}{2}\right )+x^2\,\left (\frac {A\,b^2}{2}+B\,a\,b+A\,a\,c\right )+x^3\,\left (\frac {B\,b^2}{3}+\frac {2\,A\,c\,b}{3}+\frac {2\,B\,a\,c}{3}\right )+x\,\left (B\,a^2+2\,A\,b\,a\right )+\frac {B\,c^2\,x^5}{5}+A\,a^2\,\ln \left (x\right ) \] Input:
int(((A + B*x)*(a + b*x + c*x^2)^2)/x,x)
Output:
x^4*((A*c^2)/4 + (B*b*c)/2) + x^2*((A*b^2)/2 + A*a*c + B*a*b) + x^3*((B*b^ 2)/3 + (2*A*b*c)/3 + (2*B*a*c)/3) + x*(B*a^2 + 2*A*a*b) + (B*c^2*x^5)/5 + A*a^2*log(x)
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx=\mathrm {log}\left (x \right ) a^{3}+3 a^{2} b x +a^{2} c \,x^{2}+\frac {3 a \,b^{2} x^{2}}{2}+\frac {4 a b c \,x^{3}}{3}+\frac {a \,c^{2} x^{4}}{4}+\frac {b^{3} x^{3}}{3}+\frac {b^{2} c \,x^{4}}{2}+\frac {b \,c^{2} x^{5}}{5} \] Input:
int((B*x+A)*(c*x^2+b*x+a)^2/x,x)
Output:
(60*log(x)*a**3 + 180*a**2*b*x + 60*a**2*c*x**2 + 90*a*b**2*x**2 + 80*a*b* c*x**3 + 15*a*c**2*x**4 + 20*b**3*x**3 + 30*b**2*c*x**4 + 12*b*c**2*x**5)/ 60