Integrand size = 16, antiderivative size = 216 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=-\frac {a^{10} A}{7 x^7}-\frac {a^9 (10 A b+a B)}{6 x^6}-\frac {a^8 b (9 A b+2 a B)}{x^5}-\frac {15 a^7 b^2 (8 A b+3 a B)}{4 x^4}-\frac {10 a^6 b^3 (7 A b+4 a B)}{x^3}-\frac {21 a^5 b^4 (6 A b+5 a B)}{x^2}-\frac {42 a^4 b^5 (5 A b+6 a B)}{x}+15 a^2 b^7 (3 A b+8 a B) x+\frac {5}{2} a b^8 (2 A b+9 a B) x^2+\frac {1}{3} b^9 (A b+10 a B) x^3+\frac {1}{4} b^{10} B x^4+30 a^3 b^6 (4 A b+7 a B) \log (x) \]
-1/7*a^10*A/x^7-1/6*a^9*(10*A*b+B*a)/x^6-a^8*b*(9*A*b+2*B*a)/x^5-15/4*a^7* b^2*(8*A*b+3*B*a)/x^4-10*a^6*b^3*(7*A*b+4*B*a)/x^3-21*a^5*b^4*(6*A*b+5*B*a )/x^2-42*a^4*b^5*(5*A*b+6*B*a)/x+15*a^2*b^7*(3*A*b+8*B*a)*x+5/2*a*b^8*(2*A *b+9*B*a)*x^2+1/3*b^9*(A*b+10*B*a)*x^3+1/4*b^10*B*x^4+30*a^3*b^6*(4*A*b+7* B*a)*ln(x)
Time = 0.07 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=-\frac {210 a^4 A b^6}{x}+120 a^3 b^7 B x+\frac {45}{2} a^2 b^8 x (2 A+B x)-\frac {126 a^5 b^5 (A+2 B x)}{x^2}+\frac {5}{3} a b^9 x^2 (3 A+2 B x)-\frac {35 a^6 b^4 (2 A+3 B x)}{x^3}+\frac {1}{12} b^{10} x^3 (4 A+3 B x)-\frac {10 a^7 b^3 (3 A+4 B x)}{x^4}-\frac {9 a^8 b^2 (4 A+5 B x)}{4 x^5}-\frac {a^9 b (5 A+6 B x)}{3 x^6}-\frac {a^{10} (6 A+7 B x)}{42 x^7}+30 a^3 b^6 (4 A b+7 a B) \log (x) \]
(-210*a^4*A*b^6)/x + 120*a^3*b^7*B*x + (45*a^2*b^8*x*(2*A + B*x))/2 - (126 *a^5*b^5*(A + 2*B*x))/x^2 + (5*a*b^9*x^2*(3*A + 2*B*x))/3 - (35*a^6*b^4*(2 *A + 3*B*x))/x^3 + (b^10*x^3*(4*A + 3*B*x))/12 - (10*a^7*b^3*(3*A + 4*B*x) )/x^4 - (9*a^8*b^2*(4*A + 5*B*x))/(4*x^5) - (a^9*b*(5*A + 6*B*x))/(3*x^6) - (a^10*(6*A + 7*B*x))/(42*x^7) + 30*a^3*b^6*(4*A*b + 7*a*B)*Log[x]
Time = 0.38 (sec) , antiderivative size = 216, 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.
| \(\displaystyle \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \int \left (\frac {a^{10} A}{x^8}+\frac {a^9 (a B+10 A b)}{x^7}+\frac {5 a^8 b (2 a B+9 A b)}{x^6}+\frac {15 a^7 b^2 (3 a B+8 A b)}{x^5}+\frac {30 a^6 b^3 (4 a B+7 A b)}{x^4}+\frac {42 a^5 b^4 (5 a B+6 A b)}{x^3}+\frac {42 a^4 b^5 (6 a B+5 A b)}{x^2}+\frac {30 a^3 b^6 (7 a B+4 A b)}{x}+15 a^2 b^7 (8 a B+3 A b)+b^9 x^2 (10 a B+A b)+5 a b^8 x (9 a B+2 A b)+b^{10} B x^3\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^{10} A}{7 x^7}-\frac {a^9 (a B+10 A b)}{6 x^6}-\frac {a^8 b (2 a B+9 A b)}{x^5}-\frac {15 a^7 b^2 (3 a B+8 A b)}{4 x^4}-\frac {10 a^6 b^3 (4 a B+7 A b)}{x^3}-\frac {21 a^5 b^4 (5 a B+6 A b)}{x^2}-\frac {42 a^4 b^5 (6 a B+5 A b)}{x}+30 a^3 b^6 \log (x) (7 a B+4 A b)+15 a^2 b^7 x (8 a B+3 A b)+\frac {1}{3} b^9 x^3 (10 a B+A b)+\frac {5}{2} a b^8 x^2 (9 a B+2 A b)+\frac {1}{4} b^{10} B x^4\) |
-1/7*(a^10*A)/x^7 - (a^9*(10*A*b + a*B))/(6*x^6) - (a^8*b*(9*A*b + 2*a*B)) /x^5 - (15*a^7*b^2*(8*A*b + 3*a*B))/(4*x^4) - (10*a^6*b^3*(7*A*b + 4*a*B)) /x^3 - (21*a^5*b^4*(6*A*b + 5*a*B))/x^2 - (42*a^4*b^5*(5*A*b + 6*a*B))/x + 15*a^2*b^7*(3*A*b + 8*a*B)*x + (5*a*b^8*(2*A*b + 9*a*B)*x^2)/2 + (b^9*(A* b + 10*a*B)*x^3)/3 + (b^10*B*x^4)/4 + 30*a^3*b^6*(4*A*b + 7*a*B)*Log[x]
3.2.55.3.1 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 = 0.41 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.99
| method | result | size |
| default | \(\frac {b^{10} B \,x^{4}}{4}+\frac {A \,b^{10} x^{3}}{3}+\frac {10 B a \,b^{9} x^{3}}{3}+5 A a \,b^{9} x^{2}+\frac {45 B \,a^{2} b^{8} x^{2}}{2}+45 A \,a^{2} b^{8} x +120 B \,a^{3} b^{7} x +30 a^{3} b^{6} \left (4 A b +7 B a \right ) \ln \left (x \right )-\frac {a^{9} \left (10 A b +B a \right )}{6 x^{6}}-\frac {a^{10} A}{7 x^{7}}-\frac {10 a^{6} b^{3} \left (7 A b +4 B a \right )}{x^{3}}-\frac {42 a^{4} b^{5} \left (5 A b +6 B a \right )}{x}-\frac {21 a^{5} b^{4} \left (6 A b +5 B a \right )}{x^{2}}-\frac {15 a^{7} b^{2} \left (8 A b +3 B a \right )}{4 x^{4}}-\frac {a^{8} b \left (9 A b +2 B a \right )}{x^{5}}\) | \(214\) |
| risch | \(\frac {b^{10} B \,x^{4}}{4}+\frac {A \,b^{10} x^{3}}{3}+\frac {10 B a \,b^{9} x^{3}}{3}+5 A a \,b^{9} x^{2}+\frac {45 B \,a^{2} b^{8} x^{2}}{2}+45 A \,a^{2} b^{8} x +120 B \,a^{3} b^{7} x +\frac {\left (-210 a^{4} b^{6} A -252 a^{5} b^{5} B \right ) x^{6}+\left (-126 a^{5} b^{5} A -105 a^{6} b^{4} B \right ) x^{5}+\left (-70 a^{6} b^{4} A -40 a^{7} b^{3} B \right ) x^{4}+\left (-30 a^{7} b^{3} A -\frac {45}{4} a^{8} b^{2} B \right ) x^{3}+\left (-9 a^{8} b^{2} A -2 a^{9} b B \right ) x^{2}+\left (-\frac {5}{3} a^{9} b A -\frac {1}{6} a^{10} B \right ) x -\frac {a^{10} A}{7}}{x^{7}}+120 A \ln \left (x \right ) a^{3} b^{7}+210 B \ln \left (x \right ) a^{4} b^{6}\) | \(234\) |
| norman | \(\frac {\left (\frac {1}{3} b^{10} A +\frac {10}{3} a \,b^{9} B \right ) x^{10}+\left (5 a \,b^{9} A +\frac {45}{2} a^{2} b^{8} B \right ) x^{9}+\left (-30 a^{7} b^{3} A -\frac {45}{4} a^{8} b^{2} B \right ) x^{3}+\left (-\frac {5}{3} a^{9} b A -\frac {1}{6} a^{10} B \right ) x +\left (45 a^{2} b^{8} A +120 a^{3} b^{7} B \right ) x^{8}+\left (-210 a^{4} b^{6} A -252 a^{5} b^{5} B \right ) x^{6}+\left (-126 a^{5} b^{5} A -105 a^{6} b^{4} B \right ) x^{5}+\left (-70 a^{6} b^{4} A -40 a^{7} b^{3} B \right ) x^{4}+\left (-9 a^{8} b^{2} A -2 a^{9} b B \right ) x^{2}-\frac {a^{10} A}{7}+\frac {b^{10} B \,x^{11}}{4}}{x^{7}}+\left (120 a^{3} b^{7} A +210 a^{4} b^{6} B \right ) \ln \left (x \right )\) | \(235\) |
| parallelrisch | \(\frac {21 b^{10} B \,x^{11}+28 A \,b^{10} x^{10}+280 B a \,b^{9} x^{10}+420 a A \,b^{9} x^{9}+1890 B \,a^{2} b^{8} x^{9}+10080 A \ln \left (x \right ) x^{7} a^{3} b^{7}+3780 a^{2} A \,b^{8} x^{8}+17640 B \ln \left (x \right ) x^{7} a^{4} b^{6}+10080 B \,a^{3} b^{7} x^{8}-17640 a^{4} A \,b^{6} x^{6}-21168 B \,a^{5} b^{5} x^{6}-10584 a^{5} A \,b^{5} x^{5}-8820 B \,a^{6} b^{4} x^{5}-5880 a^{6} A \,b^{4} x^{4}-3360 B \,a^{7} b^{3} x^{4}-2520 a^{7} A \,b^{3} x^{3}-945 B \,a^{8} b^{2} x^{3}-756 a^{8} A \,b^{2} x^{2}-168 B \,a^{9} b \,x^{2}-140 a^{9} A b x -14 a^{10} B x -12 a^{10} A}{84 x^{7}}\) | \(248\) |
1/4*b^10*B*x^4+1/3*A*b^10*x^3+10/3*B*a*b^9*x^3+5*A*a*b^9*x^2+45/2*B*a^2*b^ 8*x^2+45*A*a^2*b^8*x+120*B*a^3*b^7*x+30*a^3*b^6*(4*A*b+7*B*a)*ln(x)-1/6*a^ 9*(10*A*b+B*a)/x^6-1/7*a^10*A/x^7-10*a^6*b^3*(7*A*b+4*B*a)/x^3-42*a^4*b^5* (5*A*b+6*B*a)/x-21*a^5*b^4*(6*A*b+5*B*a)/x^2-15/4*a^7*b^2*(8*A*b+3*B*a)/x^ 4-a^8*b*(9*A*b+2*B*a)/x^5
Time = 0.23 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {21 \, B b^{10} x^{11} - 12 \, A a^{10} + 28 \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{10} + 210 \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{9} + 1260 \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{8} + 2520 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x^{7} \log \left (x\right ) - 3528 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} - 1764 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} - 840 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} - 315 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} - 84 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} - 14 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{84 \, x^{7}} \]
1/84*(21*B*b^10*x^11 - 12*A*a^10 + 28*(10*B*a*b^9 + A*b^10)*x^10 + 210*(9* B*a^2*b^8 + 2*A*a*b^9)*x^9 + 1260*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x^8 + 2520*( 7*B*a^4*b^6 + 4*A*a^3*b^7)*x^7*log(x) - 3528*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x ^6 - 1764*(5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 - 840*(4*B*a^7*b^3 + 7*A*a^6*b^4 )*x^4 - 315*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 - 84*(2*B*a^9*b + 9*A*a^8*b^2) *x^2 - 14*(B*a^10 + 10*A*a^9*b)*x)/x^7
Time = 2.69 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {B b^{10} x^{4}}{4} + 30 a^{3} b^{6} \cdot \left (4 A b + 7 B a\right ) \log {\left (x \right )} + x^{3} \left (\frac {A b^{10}}{3} + \frac {10 B a b^{9}}{3}\right ) + x^{2} \cdot \left (5 A a b^{9} + \frac {45 B a^{2} b^{8}}{2}\right ) + x \left (45 A a^{2} b^{8} + 120 B a^{3} b^{7}\right ) + \frac {- 12 A a^{10} + x^{6} \left (- 17640 A a^{4} b^{6} - 21168 B a^{5} b^{5}\right ) + x^{5} \left (- 10584 A a^{5} b^{5} - 8820 B a^{6} b^{4}\right ) + x^{4} \left (- 5880 A a^{6} b^{4} - 3360 B a^{7} b^{3}\right ) + x^{3} \left (- 2520 A a^{7} b^{3} - 945 B a^{8} b^{2}\right ) + x^{2} \left (- 756 A a^{8} b^{2} - 168 B a^{9} b\right ) + x \left (- 140 A a^{9} b - 14 B a^{10}\right )}{84 x^{7}} \]
B*b**10*x**4/4 + 30*a**3*b**6*(4*A*b + 7*B*a)*log(x) + x**3*(A*b**10/3 + 1 0*B*a*b**9/3) + x**2*(5*A*a*b**9 + 45*B*a**2*b**8/2) + x*(45*A*a**2*b**8 + 120*B*a**3*b**7) + (-12*A*a**10 + x**6*(-17640*A*a**4*b**6 - 21168*B*a**5 *b**5) + x**5*(-10584*A*a**5*b**5 - 8820*B*a**6*b**4) + x**4*(-5880*A*a**6 *b**4 - 3360*B*a**7*b**3) + x**3*(-2520*A*a**7*b**3 - 945*B*a**8*b**2) + x **2*(-756*A*a**8*b**2 - 168*B*a**9*b) + x*(-140*A*a**9*b - 14*B*a**10))/(8 4*x**7)
Time = 0.21 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {1}{4} \, B b^{10} x^{4} + \frac {1}{3} \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{3} + \frac {5}{2} \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{2} + 15 \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x + 30 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} \log \left (x\right ) - \frac {12 \, A a^{10} + 3528 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} + 1764 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + 840 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 315 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 84 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 14 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{84 \, x^{7}} \]
1/4*B*b^10*x^4 + 1/3*(10*B*a*b^9 + A*b^10)*x^3 + 5/2*(9*B*a^2*b^8 + 2*A*a* b^9)*x^2 + 15*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x + 30*(7*B*a^4*b^6 + 4*A*a^3*b^ 7)*log(x) - 1/84*(12*A*a^10 + 3528*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x^6 + 1764* (5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 + 840*(4*B*a^7*b^3 + 7*A*a^6*b^4)*x^4 + 31 5*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 84*(2*B*a^9*b + 9*A*a^8*b^2)*x^2 + 14* (B*a^10 + 10*A*a^9*b)*x)/x^7
Time = 0.27 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {1}{4} \, B b^{10} x^{4} + \frac {10}{3} \, B a b^{9} x^{3} + \frac {1}{3} \, A b^{10} x^{3} + \frac {45}{2} \, B a^{2} b^{8} x^{2} + 5 \, A a b^{9} x^{2} + 120 \, B a^{3} b^{7} x + 45 \, A a^{2} b^{8} x + 30 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, A a^{10} + 3528 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} + 1764 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + 840 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 315 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 84 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 14 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{84 \, x^{7}} \]
1/4*B*b^10*x^4 + 10/3*B*a*b^9*x^3 + 1/3*A*b^10*x^3 + 45/2*B*a^2*b^8*x^2 + 5*A*a*b^9*x^2 + 120*B*a^3*b^7*x + 45*A*a^2*b^8*x + 30*(7*B*a^4*b^6 + 4*A*a ^3*b^7)*log(abs(x)) - 1/84*(12*A*a^10 + 3528*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x ^6 + 1764*(5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 + 840*(4*B*a^7*b^3 + 7*A*a^6*b^4 )*x^4 + 315*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 84*(2*B*a^9*b + 9*A*a^8*b^2) *x^2 + 14*(B*a^10 + 10*A*a^9*b)*x)/x^7
Time = 0.09 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=x^3\,\left (\frac {A\,b^{10}}{3}+\frac {10\,B\,a\,b^9}{3}\right )-\frac {x\,\left (\frac {B\,a^{10}}{6}+\frac {5\,A\,b\,a^9}{3}\right )+\frac {A\,a^{10}}{7}+x^2\,\left (2\,B\,a^9\,b+9\,A\,a^8\,b^2\right )+x^3\,\left (\frac {45\,B\,a^8\,b^2}{4}+30\,A\,a^7\,b^3\right )+x^4\,\left (40\,B\,a^7\,b^3+70\,A\,a^6\,b^4\right )+x^5\,\left (105\,B\,a^6\,b^4+126\,A\,a^5\,b^5\right )+x^6\,\left (252\,B\,a^5\,b^5+210\,A\,a^4\,b^6\right )}{x^7}+\ln \left (x\right )\,\left (210\,B\,a^4\,b^6+120\,A\,a^3\,b^7\right )+\frac {B\,b^{10}\,x^4}{4}+15\,a^2\,b^7\,x\,\left (3\,A\,b+8\,B\,a\right )+\frac {5\,a\,b^8\,x^2\,\left (2\,A\,b+9\,B\,a\right )}{2} \]
x^3*((A*b^10)/3 + (10*B*a*b^9)/3) - (x*((B*a^10)/6 + (5*A*a^9*b)/3) + (A*a ^10)/7 + x^2*(9*A*a^8*b^2 + 2*B*a^9*b) + x^3*(30*A*a^7*b^3 + (45*B*a^8*b^2 )/4) + x^4*(70*A*a^6*b^4 + 40*B*a^7*b^3) + x^5*(126*A*a^5*b^5 + 105*B*a^6* b^4) + x^6*(210*A*a^4*b^6 + 252*B*a^5*b^5))/x^7 + log(x)*(120*A*a^3*b^7 + 210*B*a^4*b^6) + (B*b^10*x^4)/4 + 15*a^2*b^7*x*(3*A*b + 8*B*a) + (5*a*b^8* x^2*(2*A*b + 9*B*a))/2