Integrand size = 21, antiderivative size = 155 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx=-\frac {a^3 A}{6 x^6}-\frac {a^2 (3 A b+a B)}{5 x^5}-\frac {3 a \left (a b B+A \left (b^2+a c\right )\right )}{4 x^4}-\frac {3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )}{3 x^3}-\frac {b^3 B+3 A b^2 c+6 a b B c+3 a A c^2}{2 x^2}-\frac {3 c \left (b^2 B+A b c+a B c\right )}{x}+B c^3 x+c^2 (3 b B+A c) \log (x) \]
-1/6*a^3*A/x^6-1/5*a^2*(3*A*b+B*a)/x^5-3/4*a*(a*b*B+A*(a*c+b^2))/x^4+1/3*( -3*a*B*(a*c+b^2)-A*(6*a*b*c+b^3))/x^3+1/2*(-3*A*a*c^2-3*A*b^2*c-6*B*a*b*c- B*b^3)/x^2-3*c*(A*b*c+B*a*c+B*b^2)/x+B*c^3*x+c^2*(A*c+3*B*b)*ln(x)
Time = 0.06 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx=-\frac {2 a^3 (5 A+6 B x)+3 a^2 x (5 B x (3 b+4 c x)+3 A (4 b+5 c x))+15 a x^2 \left (4 B x \left (b^2+3 b c x+3 c^2 x^2\right )+A \left (3 b^2+8 b c x+6 c^2 x^2\right )\right )+10 x^3 \left (A b \left (2 b^2+9 b c x+18 c^2 x^2\right )+3 B x \left (b^3+6 b^2 c x-2 c^3 x^3\right )\right )-60 c^2 (3 b B+A c) x^6 \log (x)}{60 x^6} \]
-1/60*(2*a^3*(5*A + 6*B*x) + 3*a^2*x*(5*B*x*(3*b + 4*c*x) + 3*A*(4*b + 5*c *x)) + 15*a*x^2*(4*B*x*(b^2 + 3*b*c*x + 3*c^2*x^2) + A*(3*b^2 + 8*b*c*x + 6*c^2*x^2)) + 10*x^3*(A*b*(2*b^2 + 9*b*c*x + 18*c^2*x^2) + 3*B*x*(b^3 + 6* b^2*c*x - 2*c^3*x^3)) - 60*c^2*(3*b*B + A*c)*x^6*Log[x])/x^6
Time = 0.32 (sec) , antiderivative size = 155, 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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^3 A}{x^7}+\frac {a^2 (a B+3 A b)}{x^6}+\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x^5}+\frac {3 c \left (a B c+A b c+b^2 B\right )}{x^2}+\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{x^3}+\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{x^4}+\frac {c^2 (A c+3 b B)}{x}+B c^3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 A}{6 x^6}-\frac {a^2 (a B+3 A b)}{5 x^5}-\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{4 x^4}-\frac {3 c \left (a B c+A b c+b^2 B\right )}{x}-\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{2 x^2}-\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{3 x^3}+c^2 \log (x) (A c+3 b B)+B c^3 x\) |
-1/6*(a^3*A)/x^6 - (a^2*(3*A*b + a*B))/(5*x^5) - (3*a*(a*b*B + A*(b^2 + a* c)))/(4*x^4) - (3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))/(3*x^3) - (b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)/(2*x^2) - (3*c*(b^2*B + A*b*c + a*B*c)) /x + B*c^3*x + c^2*(3*b*B + A*c)*Log[x]
3.9.76.3.1 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 = 149, normalized size of antiderivative = 0.96
method | result | size |
default | \(B \,c^{3} x -\frac {a^{3} A}{6 x^{6}}-\frac {3 a \left (A a c +A \,b^{2}+a b B \right )}{4 x^{4}}-\frac {a^{2} \left (3 A b +B a \right )}{5 x^{5}}+c^{2} \left (A c +3 B b \right ) \ln \left (x \right )-\frac {3 A a \,c^{2}+3 A \,b^{2} c +6 B a b c +B \,b^{3}}{2 x^{2}}-\frac {3 c \left (A b c +B a c +B \,b^{2}\right )}{x}-\frac {6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}}{3 x^{3}}\) | \(149\) |
risch | \(B \,c^{3} x +\frac {\left (-3 A b \,c^{2}-3 B a \,c^{2}-3 B \,b^{2} c \right ) x^{5}+\left (-\frac {3}{2} A a \,c^{2}-\frac {3}{2} A \,b^{2} c -3 B a b c -\frac {1}{2} B \,b^{3}\right ) x^{4}+\left (-2 A a b c -\frac {1}{3} A \,b^{3}-B \,a^{2} c -B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{4} A \,a^{2} c -\frac {3}{4} A a \,b^{2}-\frac {3}{4} B b \,a^{2}\right ) x^{2}+\left (-\frac {3}{5} A \,a^{2} b -\frac {1}{5} B \,a^{3}\right ) x -\frac {A \,a^{3}}{6}}{x^{6}}+A \ln \left (x \right ) c^{3}+3 B \ln \left (x \right ) b \,c^{2}\) | \(165\) |
norman | \(\frac {\left (-\frac {3}{5} A \,a^{2} b -\frac {1}{5} B \,a^{3}\right ) x +\left (-\frac {3}{4} A \,a^{2} c -\frac {3}{4} A a \,b^{2}-\frac {3}{4} B b \,a^{2}\right ) x^{2}+\left (-\frac {3}{2} A a \,c^{2}-\frac {3}{2} A \,b^{2} c -3 B a b c -\frac {1}{2} B \,b^{3}\right ) x^{4}+\left (-2 A a b c -\frac {1}{3} A \,b^{3}-B \,a^{2} c -B a \,b^{2}\right ) x^{3}+\left (-3 A b \,c^{2}-3 B a \,c^{2}-3 B \,b^{2} c \right ) x^{5}+B \,c^{3} x^{7}-\frac {A \,a^{3}}{6}}{x^{6}}+\left (A \,c^{3}+3 B b \,c^{2}\right ) \ln \left (x \right )\) | \(167\) |
parallelrisch | \(\frac {60 A \ln \left (x \right ) x^{6} c^{3}+180 B \ln \left (x \right ) x^{6} b \,c^{2}+60 B \,c^{3} x^{7}-180 A b \,c^{2} x^{5}-180 a B \,c^{2} x^{5}-180 B \,b^{2} c \,x^{5}-90 a A \,c^{2} x^{4}-90 A \,b^{2} c \,x^{4}-180 B a b c \,x^{4}-30 x^{4} B \,b^{3}-120 A a b c \,x^{3}-20 A \,b^{3} x^{3}-60 a^{2} B c \,x^{3}-60 B a \,b^{2} x^{3}-45 a^{2} A c \,x^{2}-45 A a \,b^{2} x^{2}-45 B \,a^{2} b \,x^{2}-36 A \,a^{2} b x -12 a^{3} B x -10 A \,a^{3}}{60 x^{6}}\) | \(196\) |
B*c^3*x-1/6*a^3*A/x^6-3/4*a*(A*a*c+A*b^2+B*a*b)/x^4-1/5*a^2*(3*A*b+B*a)/x^ 5+c^2*(A*c+3*B*b)*ln(x)-1/2*(3*A*a*c^2+3*A*b^2*c+6*B*a*b*c+B*b^3)/x^2-3*c* (A*b*c+B*a*c+B*b^2)/x-1/3*(6*A*a*b*c+A*b^3+3*B*a^2*c+3*B*a*b^2)/x^3
Time = 0.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx=\frac {60 \, B c^{3} x^{7} + 60 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} \log \left (x\right ) - 180 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} - 30 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 10 \, A a^{3} - 20 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} - 45 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 12 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{60 \, x^{6}} \]
1/60*(60*B*c^3*x^7 + 60*(3*B*b*c^2 + A*c^3)*x^6*log(x) - 180*(B*b^2*c + (B *a + A*b)*c^2)*x^5 - 30*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 - 10*A*a^3 - 20*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 - 45*(B*a^2* b + A*a*b^2 + A*a^2*c)*x^2 - 12*(B*a^3 + 3*A*a^2*b)*x)/x^6
Time = 53.87 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx=B c^{3} x + c^{2} \left (A c + 3 B b\right ) \log {\left (x \right )} + \frac {- 10 A a^{3} + x^{5} \left (- 180 A b c^{2} - 180 B a c^{2} - 180 B b^{2} c\right ) + x^{4} \left (- 90 A a c^{2} - 90 A b^{2} c - 180 B a b c - 30 B b^{3}\right ) + x^{3} \left (- 120 A a b c - 20 A b^{3} - 60 B a^{2} c - 60 B a b^{2}\right ) + x^{2} \left (- 45 A a^{2} c - 45 A a b^{2} - 45 B a^{2} b\right ) + x \left (- 36 A a^{2} b - 12 B a^{3}\right )}{60 x^{6}} \]
B*c**3*x + c**2*(A*c + 3*B*b)*log(x) + (-10*A*a**3 + x**5*(-180*A*b*c**2 - 180*B*a*c**2 - 180*B*b**2*c) + x**4*(-90*A*a*c**2 - 90*A*b**2*c - 180*B*a *b*c - 30*B*b**3) + x**3*(-120*A*a*b*c - 20*A*b**3 - 60*B*a**2*c - 60*B*a* b**2) + x**2*(-45*A*a**2*c - 45*A*a*b**2 - 45*B*a**2*b) + x*(-36*A*a**2*b - 12*B*a**3))/(60*x**6)
Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx=B c^{3} x + {\left (3 \, B b c^{2} + A c^{3}\right )} \log \left (x\right ) - \frac {180 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 30 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + 10 \, A a^{3} + 20 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 45 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 12 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{60 \, x^{6}} \]
B*c^3*x + (3*B*b*c^2 + A*c^3)*log(x) - 1/60*(180*(B*b^2*c + (B*a + A*b)*c^ 2)*x^5 + 30*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 + 10*A*a^3 + 2 0*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 45*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 12*(B*a^3 + 3*A*a^2*b)*x)/x^6
Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx=B c^{3} x + {\left (3 \, B b c^{2} + A c^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac {180 \, {\left (B b^{2} c + B a c^{2} + A b c^{2}\right )} x^{5} + 30 \, {\left (B b^{3} + 6 \, B a b c + 3 \, A b^{2} c + 3 \, A a c^{2}\right )} x^{4} + 10 \, A a^{3} + 20 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, B a^{2} c + 6 \, A a b c\right )} x^{3} + 45 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 12 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{60 \, x^{6}} \]
B*c^3*x + (3*B*b*c^2 + A*c^3)*log(abs(x)) - 1/60*(180*(B*b^2*c + B*a*c^2 + A*b*c^2)*x^5 + 30*(B*b^3 + 6*B*a*b*c + 3*A*b^2*c + 3*A*a*c^2)*x^4 + 10*A* a^3 + 20*(3*B*a*b^2 + A*b^3 + 3*B*a^2*c + 6*A*a*b*c)*x^3 + 45*(B*a^2*b + A *a*b^2 + A*a^2*c)*x^2 + 12*(B*a^3 + 3*A*a^2*b)*x)/x^6
Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^7} \, dx=\ln \left (x\right )\,\left (A\,c^3+3\,B\,b\,c^2\right )-\frac {x^3\,\left (B\,c\,a^2+B\,a\,b^2+2\,A\,c\,a\,b+\frac {A\,b^3}{3}\right )+x^4\,\left (\frac {B\,b^3}{2}+\frac {3\,A\,b^2\,c}{2}+3\,B\,a\,b\,c+\frac {3\,A\,a\,c^2}{2}\right )+x\,\left (\frac {B\,a^3}{5}+\frac {3\,A\,b\,a^2}{5}\right )+\frac {A\,a^3}{6}+x^2\,\left (\frac {3\,B\,a^2\,b}{4}+\frac {3\,A\,c\,a^2}{4}+\frac {3\,A\,a\,b^2}{4}\right )+x^5\,\left (3\,B\,b^2\,c+3\,A\,b\,c^2+3\,B\,a\,c^2\right )}{x^6}+B\,c^3\,x \]
log(x)*(A*c^3 + 3*B*b*c^2) - (x^3*((A*b^3)/3 + B*a*b^2 + B*a^2*c + 2*A*a*b *c) + x^4*((B*b^3)/2 + (3*A*a*c^2)/2 + (3*A*b^2*c)/2 + 3*B*a*b*c) + x*((B* a^3)/5 + (3*A*a^2*b)/5) + (A*a^3)/6 + x^2*((3*A*a*b^2)/4 + (3*A*a^2*c)/4 + (3*B*a^2*b)/4) + x^5*(3*A*b*c^2 + 3*B*a*c^2 + 3*B*b^2*c))/x^6 + B*c^3*x