Integrand size = 21, antiderivative size = 154 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx=-\frac {a^3 A}{5 x^5}-\frac {a^2 (3 A b+a B)}{4 x^4}-\frac {a \left (a b B+A \left (b^2+a c\right )\right )}{x^3}-\frac {3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )}{2 x^2}-\frac {b^3 B+3 A b^2 c+6 a b B c+3 a A c^2}{x}+c^2 (3 b B+A c) x+\frac {1}{2} B c^3 x^2+3 c \left (b^2 B+A b c+a B c\right ) \log (x) \]
-1/5*a^3*A/x^5-1/4*a^2*(3*A*b+B*a)/x^4-a*(a*b*B+A*(a*c+b^2))/x^3+1/2*(-3*a *B*(a*c+b^2)-A*(6*a*b*c+b^3))/x^2+(-3*A*a*c^2-3*A*b^2*c-6*B*a*b*c-B*b^3)/x +c^2*(A*c+3*B*b)*x+1/2*B*c^3*x^2+3*c*(A*b*c+B*a*c+B*b^2)*ln(x)
Time = 0.06 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx=-\frac {a^3 (4 A+5 B x)+5 a^2 x \left (3 A b+4 b B x+4 A c x+6 B c x^2\right )+10 a x^2 \left (3 b B x (b+4 c x)+2 A \left (b^2+3 b c x+3 c^2 x^2\right )\right )+10 x^3 \left (A \left (b^3+6 b^2 c x-2 c^3 x^3\right )-B x \left (-2 b^3+6 b c^2 x^2+c^3 x^3\right )\right )-60 c \left (b^2 B+A b c+a B c\right ) x^5 \log (x)}{20 x^5} \]
-1/20*(a^3*(4*A + 5*B*x) + 5*a^2*x*(3*A*b + 4*b*B*x + 4*A*c*x + 6*B*c*x^2) + 10*a*x^2*(3*b*B*x*(b + 4*c*x) + 2*A*(b^2 + 3*b*c*x + 3*c^2*x^2)) + 10*x ^3*(A*(b^3 + 6*b^2*c*x - 2*c^3*x^3) - B*x*(-2*b^3 + 6*b*c^2*x^2 + c^3*x^3) ) - 60*c*(b^2*B + A*b*c + a*B*c)*x^5*Log[x])/x^5
Time = 0.33 (sec) , antiderivative size = 154, 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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^3 A}{x^6}+\frac {a^2 (a B+3 A b)}{x^5}+\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x^4}+\frac {3 c \left (a B c+A b c+b^2 B\right )}{x}+\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{x^2}+\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{x^3}+c^2 (A c+3 b B)+B c^3 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 A}{5 x^5}-\frac {a^2 (a B+3 A b)}{4 x^4}-\frac {a \left (A \left (a c+b^2\right )+a b B\right )}{x^3}+3 c \log (x) \left (a B c+A b c+b^2 B\right )-\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{x}-\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{2 x^2}+c^2 x (A c+3 b B)+\frac {1}{2} B c^3 x^2\) |
-1/5*(a^3*A)/x^5 - (a^2*(3*A*b + a*B))/(4*x^4) - (a*(a*b*B + A*(b^2 + a*c) ))/x^3 - (3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))/(2*x^2) - (b^3*B + 3*A*b^ 2*c + 6*a*b*B*c + 3*a*A*c^2)/x + c^2*(3*b*B + A*c)*x + (B*c^3*x^2)/2 + 3*c *(b^2*B + A*b*c + a*B*c)*Log[x]
3.9.75.3.1 Defintions of rubi rules used
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 = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {B \,c^{3} x^{2}}{2}+A \,c^{3} x +3 B b \,c^{2} x -\frac {a^{2} \left (3 A b +B a \right )}{4 x^{4}}-\frac {a^{3} A}{5 x^{5}}+3 c \left (A b c +B a c +B \,b^{2}\right ) \ln \left (x \right )-\frac {6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}}{2 x^{2}}-\frac {3 A a \,c^{2}+3 A \,b^{2} c +6 B a b c +B \,b^{3}}{x}-\frac {a \left (A a c +A \,b^{2}+a b B \right )}{x^{3}}\) | \(151\) |
risch | \(\frac {B \,c^{3} x^{2}}{2}+A \,c^{3} x +3 B b \,c^{2} x +\frac {\left (-3 A a \,c^{2}-3 A \,b^{2} c -6 B a b c -B \,b^{3}\right ) x^{4}+\left (-3 A a b c -\frac {1}{2} A \,b^{3}-\frac {3}{2} B \,a^{2} c -\frac {3}{2} B a \,b^{2}\right ) x^{3}+\left (-A \,a^{2} c -A a \,b^{2}-B b \,a^{2}\right ) x^{2}+\left (-\frac {3}{4} A \,a^{2} b -\frac {1}{4} B \,a^{3}\right ) x -\frac {A \,a^{3}}{5}}{x^{5}}+3 A \ln \left (x \right ) b \,c^{2}+3 B \ln \left (x \right ) a \,c^{2}+3 B \ln \left (x \right ) b^{2} c\) | \(167\) |
norman | \(\frac {\left (-\frac {3}{4} A \,a^{2} b -\frac {1}{4} B \,a^{3}\right ) x +\left (-3 A a b c -\frac {1}{2} A \,b^{3}-\frac {3}{2} B \,a^{2} c -\frac {3}{2} B a \,b^{2}\right ) x^{3}+\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{6}+\left (-A \,a^{2} c -A a \,b^{2}-B b \,a^{2}\right ) x^{2}+\left (-3 A a \,c^{2}-3 A \,b^{2} c -6 B a b c -B \,b^{3}\right ) x^{4}-\frac {A \,a^{3}}{5}+\frac {B \,c^{3} x^{7}}{2}}{x^{5}}+\left (3 A b \,c^{2}+3 B a \,c^{2}+3 B \,b^{2} c \right ) \ln \left (x \right )\) | \(168\) |
parallelrisch | \(\frac {10 B \,c^{3} x^{7}+60 A \ln \left (x \right ) x^{5} b \,c^{2}+20 A \,c^{3} x^{6}+60 B \ln \left (x \right ) x^{5} a \,c^{2}+60 B \ln \left (x \right ) x^{5} b^{2} c +60 B b \,c^{2} x^{6}-60 a A \,c^{2} x^{4}-60 A \,b^{2} c \,x^{4}-120 B a b c \,x^{4}-20 x^{4} B \,b^{3}-60 A a b c \,x^{3}-10 A \,b^{3} x^{3}-30 a^{2} B c \,x^{3}-30 B a \,b^{2} x^{3}-20 a^{2} A c \,x^{2}-20 A a \,b^{2} x^{2}-20 B \,a^{2} b \,x^{2}-15 A \,a^{2} b x -5 a^{3} B x -4 A \,a^{3}}{20 x^{5}}\) | \(198\) |
1/2*B*c^3*x^2+A*c^3*x+3*B*b*c^2*x-1/4*a^2*(3*A*b+B*a)/x^4-1/5*a^3*A/x^5+3* c*(A*b*c+B*a*c+B*b^2)*ln(x)-1/2*(6*A*a*b*c+A*b^3+3*B*a^2*c+3*B*a*b^2)/x^2- (3*A*a*c^2+3*A*b^2*c+6*B*a*b*c+B*b^3)/x-a*(A*a*c+A*b^2+B*a*b)/x^3
Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx=\frac {10 \, B c^{3} x^{7} + 20 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 60 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} \log \left (x\right ) - 20 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 4 \, A a^{3} - 10 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} - 20 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \]
1/20*(10*B*c^3*x^7 + 20*(3*B*b*c^2 + A*c^3)*x^6 + 60*(B*b^2*c + (B*a + A*b )*c^2)*x^5*log(x) - 20*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 - 4 *A*a^3 - 10*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 - 20*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 - 5*(B*a^3 + 3*A*a^2*b)*x)/x^5
Time = 6.04 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx=\frac {B c^{3} x^{2}}{2} + 3 c \left (A b c + B a c + B b^{2}\right ) \log {\left (x \right )} + x \left (A c^{3} + 3 B b c^{2}\right ) + \frac {- 4 A a^{3} + x^{4} \left (- 60 A a c^{2} - 60 A b^{2} c - 120 B a b c - 20 B b^{3}\right ) + x^{3} \left (- 60 A a b c - 10 A b^{3} - 30 B a^{2} c - 30 B a b^{2}\right ) + x^{2} \left (- 20 A a^{2} c - 20 A a b^{2} - 20 B a^{2} b\right ) + x \left (- 15 A a^{2} b - 5 B a^{3}\right )}{20 x^{5}} \]
B*c**3*x**2/2 + 3*c*(A*b*c + B*a*c + B*b**2)*log(x) + x*(A*c**3 + 3*B*b*c* *2) + (-4*A*a**3 + x**4*(-60*A*a*c**2 - 60*A*b**2*c - 120*B*a*b*c - 20*B*b **3) + x**3*(-60*A*a*b*c - 10*A*b**3 - 30*B*a**2*c - 30*B*a*b**2) + x**2*( -20*A*a**2*c - 20*A*a*b**2 - 20*B*a**2*b) + x*(-15*A*a**2*b - 5*B*a**3))/( 20*x**5)
Time = 0.20 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx=\frac {1}{2} \, B c^{3} x^{2} + {\left (3 \, B b c^{2} + A c^{3}\right )} x + 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} \log \left (x\right ) - \frac {20 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + 4 \, A a^{3} + 10 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 20 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \]
1/2*B*c^3*x^2 + (3*B*b*c^2 + A*c^3)*x + 3*(B*b^2*c + (B*a + A*b)*c^2)*log( x) - 1/20*(20*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 + 4*A*a^3 + 10*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 20*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 5*(B*a^3 + 3*A*a^2*b)*x)/x^5
Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx=\frac {1}{2} \, B c^{3} x^{2} + 3 \, B b c^{2} x + A c^{3} x + 3 \, {\left (B b^{2} c + B a c^{2} + A b c^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {20 \, {\left (B b^{3} + 6 \, B a b c + 3 \, A b^{2} c + 3 \, A a c^{2}\right )} x^{4} + 4 \, A a^{3} + 10 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, B a^{2} c + 6 \, A a b c\right )} x^{3} + 20 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \]
1/2*B*c^3*x^2 + 3*B*b*c^2*x + A*c^3*x + 3*(B*b^2*c + B*a*c^2 + A*b*c^2)*lo g(abs(x)) - 1/20*(20*(B*b^3 + 6*B*a*b*c + 3*A*b^2*c + 3*A*a*c^2)*x^4 + 4*A *a^3 + 10*(3*B*a*b^2 + A*b^3 + 3*B*a^2*c + 6*A*a*b*c)*x^3 + 20*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 5*(B*a^3 + 3*A*a^2*b)*x)/x^5
Time = 10.06 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^6} \, dx=x\,\left (A\,c^3+3\,B\,b\,c^2\right )-\frac {x^3\,\left (\frac {3\,B\,c\,a^2}{2}+\frac {3\,B\,a\,b^2}{2}+3\,A\,c\,a\,b+\frac {A\,b^3}{2}\right )+x^4\,\left (B\,b^3+3\,A\,b^2\,c+6\,B\,a\,b\,c+3\,A\,a\,c^2\right )+x\,\left (\frac {B\,a^3}{4}+\frac {3\,A\,b\,a^2}{4}\right )+\frac {A\,a^3}{5}+x^2\,\left (B\,a^2\,b+A\,c\,a^2+A\,a\,b^2\right )}{x^5}+\ln \left (x\right )\,\left (3\,B\,b^2\,c+3\,A\,b\,c^2+3\,B\,a\,c^2\right )+\frac {B\,c^3\,x^2}{2} \]