Integrand size = 21, antiderivative size = 95 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=-\frac {a^2 A}{5 x^5}-\frac {a (2 A b+a B)}{4 x^4}-\frac {2 a b B+A \left (b^2+2 a c\right )}{3 x^3}-\frac {b^2 B+2 A b c+2 a B c}{2 x^2}-\frac {c (2 b B+A c)}{x}+B c^2 \log (x) \] Output:
-1/5*a^2*A/x^5-1/4*a*(2*A*b+B*a)/x^4-1/3*(2*a*b*B+A*(2*a*c+b^2))/x^3-1/2*( 2*A*b*c+2*B*a*c+B*b^2)/x^2-c*(A*c+2*B*b)/x+B*c^2*ln(x)
Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=-\frac {3 a^2 (4 A+5 B x)+10 a x \left (3 A b+4 b B x+4 A c x+6 B c x^2\right )+10 x^2 \left (3 b B x (b+4 c x)+2 A \left (b^2+3 b c x+3 c^2 x^2\right )\right )}{60 x^5}+B c^2 \log (x) \] Input:
Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x]
Output:
-1/60*(3*a^2*(4*A + 5*B*x) + 10*a*x*(3*A*b + 4*b*B*x + 4*A*c*x + 6*B*c*x^2 ) + 10*x^2*(3*b*B*x*(b + 4*c*x) + 2*A*(b^2 + 3*b*c*x + 3*c^2*x^2)))/x^5 + B*c^2*Log[x]
Time = 0.25 (sec) , antiderivative size = 95, 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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^2 A}{x^6}+\frac {A \left (2 a c+b^2\right )+2 a b B}{x^4}+\frac {2 a B c+2 A b c+b^2 B}{x^3}+\frac {a (a B+2 A b)}{x^5}+\frac {c (A c+2 b B)}{x^2}+\frac {B c^2}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 A}{5 x^5}-\frac {A \left (2 a c+b^2\right )+2 a b B}{3 x^3}-\frac {2 a B c+2 A b c+b^2 B}{2 x^2}-\frac {a (a B+2 A b)}{4 x^4}-\frac {c (A c+2 b B)}{x}+B c^2 \log (x)\) |
Input:
Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x]
Output:
-1/5*(a^2*A)/x^5 - (a*(2*A*b + a*B))/(4*x^4) - (2*a*b*B + A*(b^2 + 2*a*c)) /(3*x^3) - (b^2*B + 2*A*b*c + 2*a*B*c)/(2*x^2) - (c*(2*b*B + A*c))/x + B*c ^2*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 = 1.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {a^{2} A}{5 x^{5}}-\frac {2 A a c +b^{2} A +2 a b B}{3 x^{3}}-\frac {2 A b c +2 a B c +B \,b^{2}}{2 x^{2}}-\frac {a \left (2 A b +B a \right )}{4 x^{4}}+B \,c^{2} \ln \left (x \right )-\frac {c \left (A c +2 B b \right )}{x}\) | \(88\) |
norman | \(\frac {\left (-\frac {1}{2} a b A -\frac {1}{4} a^{2} B \right ) x +\left (-\frac {2}{3} A a c -\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) x^{2}+\left (-A b c -a B c -\frac {1}{2} B \,b^{2}\right ) x^{3}+\left (-A \,c^{2}-2 B b c \right ) x^{4}-\frac {a^{2} A}{5}}{x^{5}}+B \,c^{2} \ln \left (x \right )\) | \(92\) |
risch | \(\frac {\left (-\frac {1}{2} a b A -\frac {1}{4} a^{2} B \right ) x +\left (-\frac {2}{3} A a c -\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) x^{2}+\left (-A b c -a B c -\frac {1}{2} B \,b^{2}\right ) x^{3}+\left (-A \,c^{2}-2 B b c \right ) x^{4}-\frac {a^{2} A}{5}}{x^{5}}+B \,c^{2} \ln \left (x \right )\) | \(92\) |
parallelrisch | \(-\frac {-60 B \,c^{2} \ln \left (x \right ) x^{5}+60 x^{4} A \,c^{2}+120 x^{4} B b c +60 x^{3} A b c +60 B a c \,x^{3}+30 x^{3} B \,b^{2}+40 A a c \,x^{2}+20 x^{2} b^{2} A +40 B a \,x^{2} b +30 a b A x +15 a^{2} B x +12 a^{2} A}{60 x^{5}}\) | \(104\) |
Input:
int((B*x+A)*(c*x^2+b*x+a)^2/x^6,x,method=_RETURNVERBOSE)
Output:
-1/5*a^2*A/x^5-1/3*(2*A*a*c+A*b^2+2*B*a*b)/x^3-1/2*(2*A*b*c+2*B*a*c+B*b^2) /x^2-1/4*a*(2*A*b+B*a)/x^4+B*c^2*ln(x)-c*(A*c+2*B*b)/x
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=\frac {60 \, B c^{2} x^{5} \log \left (x\right ) - 60 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} - 30 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} - 12 \, A a^{2} - 20 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x^6,x, algorithm="fricas")
Output:
1/60*(60*B*c^2*x^5*log(x) - 60*(2*B*b*c + A*c^2)*x^4 - 30*(B*b^2 + 2*(B*a + A*b)*c)*x^3 - 12*A*a^2 - 20*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 - 15*(B*a^2 + 2*A*a*b)*x)/x^5
Time = 3.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=B c^{2} \log {\left (x \right )} + \frac {- 12 A a^{2} + x^{4} \left (- 60 A c^{2} - 120 B b c\right ) + x^{3} \left (- 60 A b c - 60 B a c - 30 B b^{2}\right ) + x^{2} \left (- 40 A a c - 20 A b^{2} - 40 B a b\right ) + x \left (- 30 A a b - 15 B a^{2}\right )}{60 x^{5}} \] Input:
integrate((B*x+A)*(c*x**2+b*x+a)**2/x**6,x)
Output:
B*c**2*log(x) + (-12*A*a**2 + x**4*(-60*A*c**2 - 120*B*b*c) + x**3*(-60*A* b*c - 60*B*a*c - 30*B*b**2) + x**2*(-40*A*a*c - 20*A*b**2 - 40*B*a*b) + x* (-30*A*a*b - 15*B*a**2))/(60*x**5)
Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=B c^{2} \log \left (x\right ) - \frac {60 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 30 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + 12 \, A a^{2} + 20 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x^6,x, algorithm="maxima")
Output:
B*c^2*log(x) - 1/60*(60*(2*B*b*c + A*c^2)*x^4 + 30*(B*b^2 + 2*(B*a + A*b)* c)*x^3 + 12*A*a^2 + 20*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + 15*(B*a^2 + 2*A*a *b)*x)/x^5
Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=B c^{2} \log \left ({\left | x \right |}\right ) - \frac {60 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 30 \, {\left (B b^{2} + 2 \, B a c + 2 \, A b c\right )} x^{3} + 12 \, A a^{2} + 20 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x^6,x, algorithm="giac")
Output:
B*c^2*log(abs(x)) - 1/60*(60*(2*B*b*c + A*c^2)*x^4 + 30*(B*b^2 + 2*B*a*c + 2*A*b*c)*x^3 + 12*A*a^2 + 20*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/x^5
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=B\,c^2\,\ln \left (x\right )-\frac {x^4\,\left (A\,c^2+2\,B\,b\,c\right )+\frac {A\,a^2}{5}+x^2\,\left (\frac {A\,b^2}{3}+\frac {2\,B\,a\,b}{3}+\frac {2\,A\,a\,c}{3}\right )+x^3\,\left (\frac {B\,b^2}{2}+A\,c\,b+B\,a\,c\right )+x\,\left (\frac {B\,a^2}{4}+\frac {A\,b\,a}{2}\right )}{x^5} \] Input:
int(((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x)
Output:
B*c^2*log(x) - (x^4*(A*c^2 + 2*B*b*c) + (A*a^2)/5 + x^2*((A*b^2)/3 + (2*A* a*c)/3 + (2*B*a*b)/3) + x^3*((B*b^2)/2 + A*b*c + B*a*c) + x*((B*a^2)/4 + ( A*a*b)/2))/x^5
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx=\frac {60 \,\mathrm {log}\left (x \right ) b \,c^{2} x^{5}-12 a^{3}-45 a^{2} b x -40 a^{2} c \,x^{2}-60 a \,b^{2} x^{2}-120 a b c \,x^{3}-60 a \,c^{2} x^{4}-30 b^{3} x^{3}-120 b^{2} c \,x^{4}}{60 x^{5}} \] Input:
int((B*x+A)*(c*x^2+b*x+a)^2/x^6,x)
Output:
(60*log(x)*b*c**2*x**5 - 12*a**3 - 45*a**2*b*x - 40*a**2*c*x**2 - 60*a*b** 2*x**2 - 120*a*b*c*x**3 - 60*a*c**2*x**4 - 30*b**3*x**3 - 120*b**2*c*x**4) /(60*x**5)