Integrand size = 16, antiderivative size = 216 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=-\frac {a^{10} A}{7 x^7}-\frac {a^9 (10 A b+a B)}{6 x^6}-\frac {a^8 b (9 A b+2 a B)}{x^5}-\frac {15 a^7 b^2 (8 A b+3 a B)}{4 x^4}-\frac {10 a^6 b^3 (7 A b+4 a B)}{x^3}-\frac {21 a^5 b^4 (6 A b+5 a B)}{x^2}-\frac {42 a^4 b^5 (5 A b+6 a B)}{x}+15 a^2 b^7 (3 A b+8 a B) x+\frac {5}{2} a b^8 (2 A b+9 a B) x^2+\frac {1}{3} b^9 (A b+10 a B) x^3+\frac {1}{4} b^{10} B x^4+30 a^3 b^6 (4 A b+7 a B) \log (x) \]
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Time = 0.10 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {77} \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=-\frac {a^{10} A}{7 x^7}-\frac {a^9 (a B+10 A b)}{6 x^6}-\frac {a^8 b (2 a B+9 A b)}{x^5}-\frac {15 a^7 b^2 (3 a B+8 A b)}{4 x^4}-\frac {10 a^6 b^3 (4 a B+7 A b)}{x^3}-\frac {21 a^5 b^4 (5 a B+6 A b)}{x^2}-\frac {42 a^4 b^5 (6 a B+5 A b)}{x}+30 a^3 b^6 \log (x) (7 a B+4 A b)+15 a^2 b^7 x (8 a B+3 A b)+\frac {1}{3} b^9 x^3 (10 a B+A b)+\frac {5}{2} a b^8 x^2 (9 a B+2 A b)+\frac {1}{4} b^{10} B x^4 \]
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Rule 77
Rubi steps \begin{align*} \text {integral}& = \int \left (15 a^2 b^7 (3 A b+8 a B)+\frac {a^{10} A}{x^8}+\frac {a^9 (10 A b+a B)}{x^7}+\frac {5 a^8 b (9 A b+2 a B)}{x^6}+\frac {15 a^7 b^2 (8 A b+3 a B)}{x^5}+\frac {30 a^6 b^3 (7 A b+4 a B)}{x^4}+\frac {42 a^5 b^4 (6 A b+5 a B)}{x^3}+\frac {42 a^4 b^5 (5 A b+6 a B)}{x^2}+\frac {30 a^3 b^6 (4 A b+7 a B)}{x}+5 a b^8 (2 A b+9 a B) x+b^9 (A b+10 a B) x^2+b^{10} B x^3\right ) \, 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) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=-\frac {210 a^4 A b^6}{x}+120 a^3 b^7 B x+\frac {45}{2} a^2 b^8 x (2 A+B x)-\frac {126 a^5 b^5 (A+2 B x)}{x^2}+\frac {5}{3} a b^9 x^2 (3 A+2 B x)-\frac {35 a^6 b^4 (2 A+3 B x)}{x^3}+\frac {1}{12} b^{10} x^3 (4 A+3 B x)-\frac {10 a^7 b^3 (3 A+4 B x)}{x^4}-\frac {9 a^8 b^2 (4 A+5 B x)}{4 x^5}-\frac {a^9 b (5 A+6 B x)}{3 x^6}-\frac {a^{10} (6 A+7 B x)}{42 x^7}+30 a^3 b^6 (4 A b+7 a B) \log (x) \]
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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 | \(\frac {21 b^{10} B \,x^{11}+28 A \,b^{10} x^{10}+280 B a \,b^{9} x^{10}+420 a A \,b^{9} x^{9}+1890 B \,a^{2} b^{8} x^{9}+10080 A \ln \left (x \right ) x^{7} a^{3} b^{7}+3780 a^{2} A \,b^{8} x^{8}+17640 B \ln \left (x \right ) x^{7} a^{4} b^{6}+10080 B \,a^{3} b^{7} x^{8}-17640 a^{4} A \,b^{6} x^{6}-21168 B \,a^{5} b^{5} x^{6}-10584 a^{5} A \,b^{5} x^{5}-8820 B \,a^{6} b^{4} x^{5}-5880 a^{6} A \,b^{4} x^{4}-3360 B \,a^{7} b^{3} x^{4}-2520 a^{7} A \,b^{3} x^{3}-945 B \,a^{8} b^{2} x^{3}-756 a^{8} A \,b^{2} x^{2}-168 B \,a^{9} b \,x^{2}-140 a^{9} A b x -14 a^{10} B x -12 a^{10} A}{84 x^{7}}\) | \(248\) |
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Time = 0.23 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {21 \, B b^{10} x^{11} - 12 \, A a^{10} + 28 \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{10} + 210 \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{9} + 1260 \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{8} + 2520 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x^{7} \log \left (x\right ) - 3528 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} - 1764 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} - 840 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} - 315 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} - 84 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} - 14 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{84 \, x^{7}} \]
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Time = 2.69 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {B b^{10} x^{4}}{4} + 30 a^{3} b^{6} \cdot \left (4 A b + 7 B a\right ) \log {\left (x \right )} + x^{3} \left (\frac {A b^{10}}{3} + \frac {10 B a b^{9}}{3}\right ) + x^{2} \cdot \left (5 A a b^{9} + \frac {45 B a^{2} b^{8}}{2}\right ) + x \left (45 A a^{2} b^{8} + 120 B a^{3} b^{7}\right ) + \frac {- 12 A a^{10} + x^{6} \left (- 17640 A a^{4} b^{6} - 21168 B a^{5} b^{5}\right ) + x^{5} \left (- 10584 A a^{5} b^{5} - 8820 B a^{6} b^{4}\right ) + x^{4} \left (- 5880 A a^{6} b^{4} - 3360 B a^{7} b^{3}\right ) + x^{3} \left (- 2520 A a^{7} b^{3} - 945 B a^{8} b^{2}\right ) + x^{2} \left (- 756 A a^{8} b^{2} - 168 B a^{9} b\right ) + x \left (- 140 A a^{9} b - 14 B a^{10}\right )}{84 x^{7}} \]
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Time = 0.21 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {1}{4} \, B b^{10} x^{4} + \frac {1}{3} \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{3} + \frac {5}{2} \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{2} + 15 \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x + 30 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} \log \left (x\right ) - \frac {12 \, A a^{10} + 3528 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} + 1764 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + 840 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 315 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 84 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 14 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{84 \, x^{7}} \]
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Time = 0.27 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=\frac {1}{4} \, B b^{10} x^{4} + \frac {10}{3} \, B a b^{9} x^{3} + \frac {1}{3} \, A b^{10} x^{3} + \frac {45}{2} \, B a^{2} b^{8} x^{2} + 5 \, A a b^{9} x^{2} + 120 \, B a^{3} b^{7} x + 45 \, A a^{2} b^{8} x + 30 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, A a^{10} + 3528 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} + 1764 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + 840 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 315 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 84 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 14 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{84 \, x^{7}} \]
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Time = 0.09 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^8} \, dx=x^3\,\left (\frac {A\,b^{10}}{3}+\frac {10\,B\,a\,b^9}{3}\right )-\frac {x\,\left (\frac {B\,a^{10}}{6}+\frac {5\,A\,b\,a^9}{3}\right )+\frac {A\,a^{10}}{7}+x^2\,\left (2\,B\,a^9\,b+9\,A\,a^8\,b^2\right )+x^3\,\left (\frac {45\,B\,a^8\,b^2}{4}+30\,A\,a^7\,b^3\right )+x^4\,\left (40\,B\,a^7\,b^3+70\,A\,a^6\,b^4\right )+x^5\,\left (105\,B\,a^6\,b^4+126\,A\,a^5\,b^5\right )+x^6\,\left (252\,B\,a^5\,b^5+210\,A\,a^4\,b^6\right )}{x^7}+\ln \left (x\right )\,\left (210\,B\,a^4\,b^6+120\,A\,a^3\,b^7\right )+\frac {B\,b^{10}\,x^4}{4}+15\,a^2\,b^7\,x\,\left (3\,A\,b+8\,B\,a\right )+\frac {5\,a\,b^8\,x^2\,\left (2\,A\,b+9\,B\,a\right )}{2} \]
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