Integrand size = 16, antiderivative size = 218 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx=-\frac {a^{10} A}{6 x^6}-\frac {a^9 (10 A b+a B)}{5 x^5}-\frac {5 a^8 b (9 A b+2 a B)}{4 x^4}-\frac {5 a^7 b^2 (8 A b+3 a B)}{x^3}-\frac {15 a^6 b^3 (7 A b+4 a B)}{x^2}-\frac {42 a^5 b^4 (6 A b+5 a B)}{x}+30 a^3 b^6 (4 A b+7 a B) x+\frac {15}{2} a^2 b^7 (3 A b+8 a B) x^2+\frac {5}{3} a b^8 (2 A b+9 a B) x^3+\frac {1}{4} b^9 (A b+10 a B) x^4+\frac {1}{5} b^{10} B x^5+42 a^4 b^5 (5 A b+6 a B) \log (x) \]
-1/6*a^10*A/x^6-1/5*a^9*(10*A*b+B*a)/x^5-5/4*a^8*b*(9*A*b+2*B*a)/x^4-5*a^7 *b^2*(8*A*b+3*B*a)/x^3-15*a^6*b^3*(7*A*b+4*B*a)/x^2-42*a^5*b^4*(6*A*b+5*B* a)/x+30*a^3*b^6*(4*A*b+7*B*a)*x+15/2*a^2*b^7*(3*A*b+8*B*a)*x^2+5/3*a*b^8*( 2*A*b+9*B*a)*x^3+1/4*b^9*(A*b+10*B*a)*x^4+1/5*b^10*B*x^5+42*a^4*b^5*(5*A*b +6*B*a)*ln(x)
Time = 0.06 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx=-\frac {252 a^5 A b^5}{x}+210 a^4 b^6 B x+60 a^3 b^7 x (2 A+B x)-\frac {105 a^6 b^4 (A+2 B x)}{x^2}+\frac {15}{2} a^2 b^8 x^2 (3 A+2 B x)-\frac {20 a^7 b^3 (2 A+3 B x)}{x^3}+\frac {5}{6} a b^9 x^3 (4 A+3 B x)-\frac {15 a^8 b^2 (3 A+4 B x)}{4 x^4}+\frac {1}{20} b^{10} x^4 (5 A+4 B x)-\frac {a^9 b (4 A+5 B x)}{2 x^5}-\frac {a^{10} (5 A+6 B x)}{30 x^6}+42 a^4 b^5 (5 A b+6 a B) \log (x) \]
(-252*a^5*A*b^5)/x + 210*a^4*b^6*B*x + 60*a^3*b^7*x*(2*A + B*x) - (105*a^6 *b^4*(A + 2*B*x))/x^2 + (15*a^2*b^8*x^2*(3*A + 2*B*x))/2 - (20*a^7*b^3*(2* A + 3*B*x))/x^3 + (5*a*b^9*x^3*(4*A + 3*B*x))/6 - (15*a^8*b^2*(3*A + 4*B*x ))/(4*x^4) + (b^10*x^4*(5*A + 4*B*x))/20 - (a^9*b*(4*A + 5*B*x))/(2*x^5) - (a^10*(5*A + 6*B*x))/(30*x^6) + 42*a^4*b^5*(5*A*b + 6*a*B)*Log[x]
Time = 0.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {85, 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.
| \(\displaystyle \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \int \left (\frac {a^{10} A}{x^7}+\frac {a^9 (a B+10 A b)}{x^6}+\frac {5 a^8 b (2 a B+9 A b)}{x^5}+\frac {15 a^7 b^2 (3 a B+8 A b)}{x^4}+\frac {30 a^6 b^3 (4 a B+7 A b)}{x^3}+\frac {42 a^5 b^4 (5 a B+6 A b)}{x^2}+\frac {42 a^4 b^5 (6 a B+5 A b)}{x}+30 a^3 b^6 (7 a B+4 A b)+15 a^2 b^7 x (8 a B+3 A b)+b^9 x^3 (10 a B+A b)+5 a b^8 x^2 (9 a B+2 A b)+b^{10} B x^4\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^{10} A}{6 x^6}-\frac {a^9 (a B+10 A b)}{5 x^5}-\frac {5 a^8 b (2 a B+9 A b)}{4 x^4}-\frac {5 a^7 b^2 (3 a B+8 A b)}{x^3}-\frac {15 a^6 b^3 (4 a B+7 A b)}{x^2}-\frac {42 a^5 b^4 (5 a B+6 A b)}{x}+42 a^4 b^5 \log (x) (6 a B+5 A b)+30 a^3 b^6 x (7 a B+4 A b)+\frac {15}{2} a^2 b^7 x^2 (8 a B+3 A b)+\frac {1}{4} b^9 x^4 (10 a B+A b)+\frac {5}{3} a b^8 x^3 (9 a B+2 A b)+\frac {1}{5} b^{10} B x^5\) |
-1/6*(a^10*A)/x^6 - (a^9*(10*A*b + a*B))/(5*x^5) - (5*a^8*b*(9*A*b + 2*a*B ))/(4*x^4) - (5*a^7*b^2*(8*A*b + 3*a*B))/x^3 - (15*a^6*b^3*(7*A*b + 4*a*B) )/x^2 - (42*a^5*b^4*(6*A*b + 5*a*B))/x + 30*a^3*b^6*(4*A*b + 7*a*B)*x + (1 5*a^2*b^7*(3*A*b + 8*a*B)*x^2)/2 + (5*a*b^8*(2*A*b + 9*a*B)*x^3)/3 + (b^9* (A*b + 10*a*B)*x^4)/4 + (b^10*B*x^5)/5 + 42*a^4*b^5*(5*A*b + 6*a*B)*Log[x]
3.2.54.3.1 Defintions of rubi rules used
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Time = 0.40 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00
| method | result | size |
| default | \(\frac {b^{10} B \,x^{5}}{5}+\frac {A \,b^{10} x^{4}}{4}+\frac {5 B a \,b^{9} x^{4}}{2}+\frac {10 A a \,b^{9} x^{3}}{3}+15 B \,a^{2} b^{8} x^{3}+\frac {45 A \,a^{2} b^{8} x^{2}}{2}+60 B \,a^{3} b^{7} x^{2}+120 A \,a^{3} b^{7} x +210 B \,a^{4} b^{6} x +42 a^{4} b^{5} \left (5 A b +6 B a \right ) \ln \left (x \right )-\frac {a^{10} A}{6 x^{6}}-\frac {5 a^{7} b^{2} \left (8 A b +3 B a \right )}{x^{3}}-\frac {42 a^{5} b^{4} \left (6 A b +5 B a \right )}{x}-\frac {15 a^{6} b^{3} \left (7 A b +4 B a \right )}{x^{2}}-\frac {5 a^{8} b \left (9 A b +2 B a \right )}{4 x^{4}}-\frac {a^{9} \left (10 A b +B a \right )}{5 x^{5}}\) | \(218\) |
| norman | \(\frac {\left (\frac {1}{4} b^{10} A +\frac {5}{2} a \,b^{9} B \right ) x^{10}+\left (\frac {10}{3} a \,b^{9} A +15 a^{2} b^{8} B \right ) x^{9}+\left (\frac {45}{2} a^{2} b^{8} A +60 a^{3} b^{7} B \right ) x^{8}+\left (-\frac {45}{4} a^{8} b^{2} A -\frac {5}{2} a^{9} b B \right ) x^{2}+\left (-2 a^{9} b A -\frac {1}{5} a^{10} B \right ) x +\left (120 a^{3} b^{7} A +210 a^{4} b^{6} B \right ) x^{7}+\left (-252 a^{5} b^{5} A -210 a^{6} b^{4} B \right ) x^{5}+\left (-105 a^{6} b^{4} A -60 a^{7} b^{3} B \right ) x^{4}+\left (-40 a^{7} b^{3} A -15 a^{8} b^{2} B \right ) x^{3}-\frac {a^{10} A}{6}+\frac {b^{10} B \,x^{11}}{5}}{x^{6}}+\left (210 a^{4} b^{6} A +252 a^{5} b^{5} B \right ) \ln \left (x \right )\) | \(235\) |
| risch | \(\frac {b^{10} B \,x^{5}}{5}+\frac {A \,b^{10} x^{4}}{4}+\frac {5 B a \,b^{9} x^{4}}{2}+\frac {10 A a \,b^{9} x^{3}}{3}+15 B \,a^{2} b^{8} x^{3}+\frac {45 A \,a^{2} b^{8} x^{2}}{2}+60 B \,a^{3} b^{7} x^{2}+120 A \,a^{3} b^{7} x +210 B \,a^{4} b^{6} x +\frac {\left (-252 a^{5} b^{5} A -210 a^{6} b^{4} B \right ) x^{5}+\left (-105 a^{6} b^{4} A -60 a^{7} b^{3} B \right ) x^{4}+\left (-40 a^{7} b^{3} A -15 a^{8} b^{2} B \right ) x^{3}+\left (-\frac {45}{4} a^{8} b^{2} A -\frac {5}{2} a^{9} b B \right ) x^{2}+\left (-2 a^{9} b A -\frac {1}{5} a^{10} B \right ) x -\frac {a^{10} A}{6}}{x^{6}}+210 A \ln \left (x \right ) a^{4} b^{6}+252 B \ln \left (x \right ) a^{5} b^{5}\) | \(235\) |
| parallelrisch | \(\frac {12 b^{10} B \,x^{11}+15 A \,b^{10} x^{10}+150 B a \,b^{9} x^{10}+200 a A \,b^{9} x^{9}+900 B \,a^{2} b^{8} x^{9}+1350 a^{2} A \,b^{8} x^{8}+3600 B \,a^{3} b^{7} x^{8}+12600 A \ln \left (x \right ) x^{6} a^{4} b^{6}+7200 a^{3} A \,b^{7} x^{7}+15120 B \ln \left (x \right ) x^{6} a^{5} b^{5}+12600 B \,a^{4} b^{6} x^{7}-15120 a^{5} A \,b^{5} x^{5}-12600 B \,a^{6} b^{4} x^{5}-6300 a^{6} A \,b^{4} x^{4}-3600 B \,a^{7} b^{3} x^{4}-2400 a^{7} A \,b^{3} x^{3}-900 B \,a^{8} b^{2} x^{3}-675 a^{8} A \,b^{2} x^{2}-150 B \,a^{9} b \,x^{2}-120 a^{9} A b x -12 a^{10} B x -10 a^{10} A}{60 x^{6}}\) | \(248\) |
1/5*b^10*B*x^5+1/4*A*b^10*x^4+5/2*B*a*b^9*x^4+10/3*A*a*b^9*x^3+15*B*a^2*b^ 8*x^3+45/2*A*a^2*b^8*x^2+60*B*a^3*b^7*x^2+120*A*a^3*b^7*x+210*B*a^4*b^6*x+ 42*a^4*b^5*(5*A*b+6*B*a)*ln(x)-1/6*a^10*A/x^6-5*a^7*b^2*(8*A*b+3*B*a)/x^3- 42*a^5*b^4*(6*A*b+5*B*a)/x-15*a^6*b^3*(7*A*b+4*B*a)/x^2-5/4*a^8*b*(9*A*b+2 *B*a)/x^4-1/5*a^9*(10*A*b+B*a)/x^5
Time = 0.21 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx=\frac {12 \, B b^{10} x^{11} - 10 \, A a^{10} + 15 \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{10} + 100 \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{9} + 450 \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{8} + 1800 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x^{7} + 2520 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} \log \left (x\right ) - 2520 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} - 900 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} - 300 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} - 75 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} - 12 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{60 \, x^{6}} \]
1/60*(12*B*b^10*x^11 - 10*A*a^10 + 15*(10*B*a*b^9 + A*b^10)*x^10 + 100*(9* B*a^2*b^8 + 2*A*a*b^9)*x^9 + 450*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x^8 + 1800*(7 *B*a^4*b^6 + 4*A*a^3*b^7)*x^7 + 2520*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x^6*log(x ) - 2520*(5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 - 900*(4*B*a^7*b^3 + 7*A*a^6*b^4) *x^4 - 300*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 - 75*(2*B*a^9*b + 9*A*a^8*b^2)* x^2 - 12*(B*a^10 + 10*A*a^9*b)*x)/x^6
Time = 1.85 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx=\frac {B b^{10} x^{5}}{5} + 42 a^{4} b^{5} \cdot \left (5 A b + 6 B a\right ) \log {\left (x \right )} + x^{4} \left (\frac {A b^{10}}{4} + \frac {5 B a b^{9}}{2}\right ) + x^{3} \cdot \left (\frac {10 A a b^{9}}{3} + 15 B a^{2} b^{8}\right ) + x^{2} \cdot \left (\frac {45 A a^{2} b^{8}}{2} + 60 B a^{3} b^{7}\right ) + x \left (120 A a^{3} b^{7} + 210 B a^{4} b^{6}\right ) + \frac {- 10 A a^{10} + x^{5} \left (- 15120 A a^{5} b^{5} - 12600 B a^{6} b^{4}\right ) + x^{4} \left (- 6300 A a^{6} b^{4} - 3600 B a^{7} b^{3}\right ) + x^{3} \left (- 2400 A a^{7} b^{3} - 900 B a^{8} b^{2}\right ) + x^{2} \left (- 675 A a^{8} b^{2} - 150 B a^{9} b\right ) + x \left (- 120 A a^{9} b - 12 B a^{10}\right )}{60 x^{6}} \]
B*b**10*x**5/5 + 42*a**4*b**5*(5*A*b + 6*B*a)*log(x) + x**4*(A*b**10/4 + 5 *B*a*b**9/2) + x**3*(10*A*a*b**9/3 + 15*B*a**2*b**8) + x**2*(45*A*a**2*b** 8/2 + 60*B*a**3*b**7) + x*(120*A*a**3*b**7 + 210*B*a**4*b**6) + (-10*A*a** 10 + x**5*(-15120*A*a**5*b**5 - 12600*B*a**6*b**4) + x**4*(-6300*A*a**6*b* *4 - 3600*B*a**7*b**3) + x**3*(-2400*A*a**7*b**3 - 900*B*a**8*b**2) + x**2 *(-675*A*a**8*b**2 - 150*B*a**9*b) + x*(-120*A*a**9*b - 12*B*a**10))/(60*x **6)
Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx=\frac {1}{5} \, B b^{10} x^{5} + \frac {1}{4} \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{4} + \frac {5}{3} \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{3} + \frac {15}{2} \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{2} + 30 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x + 42 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} \log \left (x\right ) - \frac {10 \, A a^{10} + 2520 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + 900 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 300 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 75 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 12 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{60 \, x^{6}} \]
1/5*B*b^10*x^5 + 1/4*(10*B*a*b^9 + A*b^10)*x^4 + 5/3*(9*B*a^2*b^8 + 2*A*a* b^9)*x^3 + 15/2*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x^2 + 30*(7*B*a^4*b^6 + 4*A*a^ 3*b^7)*x + 42*(6*B*a^5*b^5 + 5*A*a^4*b^6)*log(x) - 1/60*(10*A*a^10 + 2520* (5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 + 900*(4*B*a^7*b^3 + 7*A*a^6*b^4)*x^4 + 30 0*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 75*(2*B*a^9*b + 9*A*a^8*b^2)*x^2 + 12* (B*a^10 + 10*A*a^9*b)*x)/x^6
Time = 0.28 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx=\frac {1}{5} \, B b^{10} x^{5} + \frac {5}{2} \, B a b^{9} x^{4} + \frac {1}{4} \, A b^{10} x^{4} + 15 \, B a^{2} b^{8} x^{3} + \frac {10}{3} \, A a b^{9} x^{3} + 60 \, B a^{3} b^{7} x^{2} + \frac {45}{2} \, A a^{2} b^{8} x^{2} + 210 \, B a^{4} b^{6} x + 120 \, A a^{3} b^{7} x + 42 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} \log \left ({\left | x \right |}\right ) - \frac {10 \, A a^{10} + 2520 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + 900 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 300 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 75 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 12 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{60 \, x^{6}} \]
1/5*B*b^10*x^5 + 5/2*B*a*b^9*x^4 + 1/4*A*b^10*x^4 + 15*B*a^2*b^8*x^3 + 10/ 3*A*a*b^9*x^3 + 60*B*a^3*b^7*x^2 + 45/2*A*a^2*b^8*x^2 + 210*B*a^4*b^6*x + 120*A*a^3*b^7*x + 42*(6*B*a^5*b^5 + 5*A*a^4*b^6)*log(abs(x)) - 1/60*(10*A* a^10 + 2520*(5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 + 900*(4*B*a^7*b^3 + 7*A*a^6*b ^4)*x^4 + 300*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 75*(2*B*a^9*b + 9*A*a^8*b^ 2)*x^2 + 12*(B*a^10 + 10*A*a^9*b)*x)/x^6
Time = 0.13 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^7} \, dx=x^4\,\left (\frac {A\,b^{10}}{4}+\frac {5\,B\,a\,b^9}{2}\right )-\frac {x\,\left (\frac {B\,a^{10}}{5}+2\,A\,b\,a^9\right )+\frac {A\,a^{10}}{6}+x^2\,\left (\frac {5\,B\,a^9\,b}{2}+\frac {45\,A\,a^8\,b^2}{4}\right )+x^3\,\left (15\,B\,a^8\,b^2+40\,A\,a^7\,b^3\right )+x^4\,\left (60\,B\,a^7\,b^3+105\,A\,a^6\,b^4\right )+x^5\,\left (210\,B\,a^6\,b^4+252\,A\,a^5\,b^5\right )}{x^6}+\ln \left (x\right )\,\left (252\,B\,a^5\,b^5+210\,A\,a^4\,b^6\right )+\frac {B\,b^{10}\,x^5}{5}+\frac {15\,a^2\,b^7\,x^2\,\left (3\,A\,b+8\,B\,a\right )}{2}+30\,a^3\,b^6\,x\,\left (4\,A\,b+7\,B\,a\right )+\frac {5\,a\,b^8\,x^3\,\left (2\,A\,b+9\,B\,a\right )}{3} \]
x^4*((A*b^10)/4 + (5*B*a*b^9)/2) - (x*((B*a^10)/5 + 2*A*a^9*b) + (A*a^10)/ 6 + x^2*((45*A*a^8*b^2)/4 + (5*B*a^9*b)/2) + x^3*(40*A*a^7*b^3 + 15*B*a^8* b^2) + x^4*(105*A*a^6*b^4 + 60*B*a^7*b^3) + x^5*(252*A*a^5*b^5 + 210*B*a^6 *b^4))/x^6 + log(x)*(210*A*a^4*b^6 + 252*B*a^5*b^5) + (B*b^10*x^5)/5 + (15 *a^2*b^7*x^2*(3*A*b + 8*B*a))/2 + 30*a^3*b^6*x*(4*A*b + 7*B*a) + (5*a*b^8* x^3*(2*A*b + 9*B*a))/3