Integrand size = 21, antiderivative size = 156 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx=-\frac {a^3 A}{4 x^4}-\frac {a^2 (3 A b+a B)}{3 x^3}-\frac {3 a \left (a b B+A \left (b^2+a c\right )\right )}{2 x^2}-\frac {3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )}{x}+3 c \left (b^2 B+A b c+a B c\right ) x+\frac {1}{2} c^2 (3 b B+A c) x^2+\frac {1}{3} B c^3 x^3+\left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) \log (x) \]
-1/4*a^3*A/x^4-1/3*a^2*(3*A*b+B*a)/x^3-3/2*a*(a*b*B+A*(a*c+b^2))/x^2+(-3*a *B*(a*c+b^2)-A*(6*a*b*c+b^3))/x+3*c*(A*b*c+B*a*c+B*b^2)*x+1/2*c^2*(A*c+3*B *b)*x^2+1/3*B*c^3*x^3+(3*A*a*c^2+3*A*b^2*c+6*B*a*b*c+B*b^3)*ln(x)
Time = 0.04 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx=\frac {-3 a^3 A-4 a^2 (3 A b+a B) x-18 a \left (a b B+A \left (b^2+a c\right )\right ) x^2-12 \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^3+36 c \left (b^2 B+A b c+a B c\right ) x^5+6 c^2 (3 b B+A c) x^6+4 B c^3 x^7+12 \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^4 \log (x)}{12 x^4} \]
(-3*a^3*A - 4*a^2*(3*A*b + a*B)*x - 18*a*(a*b*B + A*(b^2 + a*c))*x^2 - 12* (3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^3 + 36*c*(b^2*B + A*b*c + a*B*c) *x^5 + 6*c^2*(3*b*B + A*c)*x^6 + 4*B*c^3*x^7 + 12*(b^3*B + 3*A*b^2*c + 6*a *b*B*c + 3*a*A*c^2)*x^4*Log[x])/(12*x^4)
Time = 0.34 (sec) , antiderivative size = 156, 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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^3 A}{x^5}+\frac {a^2 (a B+3 A b)}{x^4}+\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x^3}+3 c \left (a B c+A b c+b^2 B\right )+\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{x}+\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{x^2}+c^2 x (A c+3 b B)+B c^3 x^2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 A}{4 x^4}-\frac {a^2 (a B+3 A b)}{3 x^3}-\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{2 x^2}+3 c x \left (a B c+A b c+b^2 B\right )+\log (x) \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )-\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{x}+\frac {1}{2} c^2 x^2 (A c+3 b B)+\frac {1}{3} B c^3 x^3\) |
-1/4*(a^3*A)/x^4 - (a^2*(3*A*b + a*B))/(3*x^3) - (3*a*(a*b*B + A*(b^2 + a* c)))/(2*x^2) - (3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))/x + 3*c*(b^2*B + A* b*c + a*B*c)*x + (c^2*(3*b*B + A*c)*x^2)/2 + (B*c^3*x^3)/3 + (b^3*B + 3*A* b^2*c + 6*a*b*B*c + 3*a*A*c^2)*Log[x]
3.9.74.3.1 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 = 0.17 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.02
method | result | size |
default | \(\frac {B \,c^{3} x^{3}}{3}+\frac {A \,c^{3} x^{2}}{2}+\frac {3 B b \,c^{2} x^{2}}{2}+3 A b \,c^{2} x +3 B a \,c^{2} x +3 B \,b^{2} c x -\frac {a^{3} A}{4 x^{4}}+\left (3 A a \,c^{2}+3 A \,b^{2} c +6 B a b c +B \,b^{3}\right ) \ln \left (x \right )-\frac {3 a \left (A a c +A \,b^{2}+a b B \right )}{2 x^{2}}-\frac {6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}}{x}-\frac {a^{2} \left (3 A b +B a \right )}{3 x^{3}}\) | \(159\) |
norman | \(\frac {\left (\frac {1}{2} A \,c^{3}+\frac {3}{2} B b \,c^{2}\right ) x^{6}+\left (-A \,a^{2} b -\frac {1}{3} B \,a^{3}\right ) x +\left (-\frac {3}{2} A \,a^{2} c -\frac {3}{2} A a \,b^{2}-\frac {3}{2} B b \,a^{2}\right ) x^{2}+\left (3 A b \,c^{2}+3 B a \,c^{2}+3 B \,b^{2} c \right ) x^{5}+\left (-6 A a b c -A \,b^{3}-3 B \,a^{2} c -3 B a \,b^{2}\right ) x^{3}-\frac {A \,a^{3}}{4}+\frac {B \,c^{3} x^{7}}{3}}{x^{4}}+\left (3 A a \,c^{2}+3 A \,b^{2} c +6 B a b c +B \,b^{3}\right ) \ln \left (x \right )\) | \(168\) |
risch | \(\frac {B \,c^{3} x^{3}}{3}+\frac {A \,c^{3} x^{2}}{2}+\frac {3 B b \,c^{2} x^{2}}{2}+3 A b \,c^{2} x +3 B a \,c^{2} x +3 B \,b^{2} c x +\frac {\left (-6 A a b c -A \,b^{3}-3 B \,a^{2} c -3 B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{2} A \,a^{2} c -\frac {3}{2} A a \,b^{2}-\frac {3}{2} B b \,a^{2}\right ) x^{2}+\left (-A \,a^{2} b -\frac {1}{3} B \,a^{3}\right ) x -\frac {A \,a^{3}}{4}}{x^{4}}+3 A \ln \left (x \right ) a \,c^{2}+3 A \ln \left (x \right ) b^{2} c +6 B \ln \left (x \right ) a b c +B \,b^{3} \ln \left (x \right )\) | \(171\) |
parallelrisch | \(\frac {4 B \,c^{3} x^{7}+6 A \,c^{3} x^{6}+18 B b \,c^{2} x^{6}+36 A \ln \left (x \right ) x^{4} a \,c^{2}+36 A \ln \left (x \right ) x^{4} b^{2} c +36 A b \,c^{2} x^{5}+72 B \ln \left (x \right ) x^{4} a b c +12 B \,b^{3} \ln \left (x \right ) x^{4}+36 a B \,c^{2} x^{5}+36 B \,b^{2} c \,x^{5}-72 A a b c \,x^{3}-12 A \,b^{3} x^{3}-36 a^{2} B c \,x^{3}-36 B a \,b^{2} x^{3}-18 a^{2} A c \,x^{2}-18 A a \,b^{2} x^{2}-18 B \,a^{2} b \,x^{2}-12 A \,a^{2} b x -4 a^{3} B x -3 A \,a^{3}}{12 x^{4}}\) | \(200\) |
1/3*B*c^3*x^3+1/2*A*c^3*x^2+3/2*B*b*c^2*x^2+3*A*b*c^2*x+3*B*a*c^2*x+3*B*b^ 2*c*x-1/4*a^3*A/x^4+(3*A*a*c^2+3*A*b^2*c+6*B*a*b*c+B*b^3)*ln(x)-3/2*a*(A*a *c+A*b^2+B*a*b)/x^2-(6*A*a*b*c+A*b^3+3*B*a^2*c+3*B*a*b^2)/x-1/3*a^2*(3*A*b +B*a)/x^3
Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx=\frac {4 \, B c^{3} x^{7} + 6 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 36 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 12 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} \log \left (x\right ) - 3 \, A a^{3} - 12 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} - 18 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \]
1/12*(4*B*c^3*x^7 + 6*(3*B*b*c^2 + A*c^3)*x^6 + 36*(B*b^2*c + (B*a + A*b)* c^2)*x^5 + 12*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4*log(x) - 3*A *a^3 - 12*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 - 18*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 - 4*(B*a^3 + 3*A*a^2*b)*x)/x^4
Time = 1.82 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx=\frac {B c^{3} x^{3}}{3} + x^{2} \left (\frac {A c^{3}}{2} + \frac {3 B b c^{2}}{2}\right ) + x \left (3 A b c^{2} + 3 B a c^{2} + 3 B b^{2} c\right ) + \left (3 A a c^{2} + 3 A b^{2} c + 6 B a b c + B b^{3}\right ) \log {\left (x \right )} + \frac {- 3 A a^{3} + x^{3} \left (- 72 A a b c - 12 A b^{3} - 36 B a^{2} c - 36 B a b^{2}\right ) + x^{2} \left (- 18 A a^{2} c - 18 A a b^{2} - 18 B a^{2} b\right ) + x \left (- 12 A a^{2} b - 4 B a^{3}\right )}{12 x^{4}} \]
B*c**3*x**3/3 + x**2*(A*c**3/2 + 3*B*b*c**2/2) + x*(3*A*b*c**2 + 3*B*a*c** 2 + 3*B*b**2*c) + (3*A*a*c**2 + 3*A*b**2*c + 6*B*a*b*c + B*b**3)*log(x) + (-3*A*a**3 + x**3*(-72*A*a*b*c - 12*A*b**3 - 36*B*a**2*c - 36*B*a*b**2) + x**2*(-18*A*a**2*c - 18*A*a*b**2 - 18*B*a**2*b) + x*(-12*A*a**2*b - 4*B*a* *3))/(12*x**4)
Time = 0.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx=\frac {1}{3} \, B c^{3} x^{3} + \frac {1}{2} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + 3 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x + {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \log \left (x\right ) - \frac {3 \, A a^{3} + 12 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \]
1/3*B*c^3*x^3 + 1/2*(3*B*b*c^2 + A*c^3)*x^2 + 3*(B*b^2*c + (B*a + A*b)*c^2 )*x + (B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*log(x) - 1/12*(3*A*a^3 + 12*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 18*(B*a^2*b + A*a*b^ 2 + A*a^2*c)*x^2 + 4*(B*a^3 + 3*A*a^2*b)*x)/x^4
Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx=\frac {1}{3} \, B c^{3} x^{3} + \frac {3}{2} \, B b c^{2} x^{2} + \frac {1}{2} \, A c^{3} x^{2} + 3 \, B b^{2} c x + 3 \, B a c^{2} x + 3 \, A b c^{2} x + {\left (B b^{3} + 6 \, B a b c + 3 \, A b^{2} c + 3 \, A a c^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac {3 \, A a^{3} + 12 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, B a^{2} c + 6 \, A a b c\right )} x^{3} + 18 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \]
1/3*B*c^3*x^3 + 3/2*B*b*c^2*x^2 + 1/2*A*c^3*x^2 + 3*B*b^2*c*x + 3*B*a*c^2* x + 3*A*b*c^2*x + (B*b^3 + 6*B*a*b*c + 3*A*b^2*c + 3*A*a*c^2)*log(abs(x)) - 1/12*(3*A*a^3 + 12*(3*B*a*b^2 + A*b^3 + 3*B*a^2*c + 6*A*a*b*c)*x^3 + 18* (B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 4*(B*a^3 + 3*A*a^2*b)*x)/x^4
Time = 10.01 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^5} \, dx=\ln \left (x\right )\,\left (B\,b^3+3\,A\,b^2\,c+6\,B\,a\,b\,c+3\,A\,a\,c^2\right )-\frac {x^3\,\left (3\,B\,c\,a^2+3\,B\,a\,b^2+6\,A\,c\,a\,b+A\,b^3\right )+x\,\left (\frac {B\,a^3}{3}+A\,b\,a^2\right )+\frac {A\,a^3}{4}+x^2\,\left (\frac {3\,B\,a^2\,b}{2}+\frac {3\,A\,c\,a^2}{2}+\frac {3\,A\,a\,b^2}{2}\right )}{x^4}+x\,\left (3\,B\,b^2\,c+3\,A\,b\,c^2+3\,B\,a\,c^2\right )+x^2\,\left (\frac {A\,c^3}{2}+\frac {3\,B\,b\,c^2}{2}\right )+\frac {B\,c^3\,x^3}{3} \]