Integrand size = 20, antiderivative size = 108 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=-\frac {a^5 A}{3 x^3}-\frac {a^4 (5 A b+a B)}{x}+5 a^3 b (2 A b+a B) x+\frac {10}{3} a^2 b^2 (A b+a B) x^3+a b^3 (A b+2 a B) x^5+\frac {1}{7} b^4 (A b+5 a B) x^7+\frac {1}{9} b^5 B x^9 \]
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Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=-\frac {a^5 A}{3 x^3}-\frac {a^4 (a B+5 A b)}{x}+5 a^3 b x (a B+2 A b)+\frac {10}{3} a^2 b^2 x^3 (a B+A b)+\frac {1}{7} b^4 x^7 (5 a B+A b)+a b^3 x^5 (2 a B+A b)+\frac {1}{9} b^5 B x^9 \]
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Rule 459
Rubi steps \begin{align*} \text {integral}& = \int \left (5 a^3 b (2 A b+a B)+\frac {a^5 A}{x^4}+\frac {a^4 (5 A b+a B)}{x^2}+10 a^2 b^2 (A b+a B) x^2+5 a b^3 (A b+2 a B) x^4+b^4 (A b+5 a B) x^6+b^5 B x^8\right ) \, dx \\ & = -\frac {a^5 A}{3 x^3}-\frac {a^4 (5 A b+a B)}{x}+5 a^3 b (2 A b+a B) x+\frac {10}{3} a^2 b^2 (A b+a B) x^3+a b^3 (A b+2 a B) x^5+\frac {1}{7} b^4 (A b+5 a B) x^7+\frac {1}{9} b^5 B x^9 \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=-\frac {a^5 A}{3 x^3}+\frac {-5 a^4 A b-a^5 B}{x}+5 a^3 b (2 A b+a B) x+\frac {10}{3} a^2 b^2 (A b+a B) x^3+a b^3 (A b+2 a B) x^5+\frac {1}{7} b^4 (A b+5 a B) x^7+\frac {1}{9} b^5 B x^9 \]
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Time = 2.55 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {b^{5} B \,x^{9}}{9}+\frac {A \,b^{5} x^{7}}{7}+\frac {5 B a \,b^{4} x^{7}}{7}+A a \,b^{4} x^{5}+2 B \,a^{2} b^{3} x^{5}+\frac {10 a^{2} A \,b^{3} x^{3}}{3}+\frac {10 B \,a^{3} b^{2} x^{3}}{3}+10 a^{3} b^{2} A x +5 a^{4} b B x -\frac {a^{5} A}{3 x^{3}}-\frac {a^{4} \left (5 A b +B a \right )}{x}\) | \(118\) |
norman | \(\frac {\frac {b^{5} B \,x^{12}}{9}+\left (\frac {1}{7} b^{5} A +\frac {5}{7} a \,b^{4} B \right ) x^{10}+\left (a \,b^{4} A +2 a^{2} b^{3} B \right ) x^{8}+\left (\frac {10}{3} a^{2} b^{3} A +\frac {10}{3} a^{3} b^{2} B \right ) x^{6}+\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) x^{4}+\left (-5 a^{4} b A -a^{5} B \right ) x^{2}-\frac {a^{5} A}{3}}{x^{3}}\) | \(121\) |
risch | \(\frac {b^{5} B \,x^{9}}{9}+\frac {A \,b^{5} x^{7}}{7}+\frac {5 B a \,b^{4} x^{7}}{7}+A a \,b^{4} x^{5}+2 B \,a^{2} b^{3} x^{5}+\frac {10 a^{2} A \,b^{3} x^{3}}{3}+\frac {10 B \,a^{3} b^{2} x^{3}}{3}+10 a^{3} b^{2} A x +5 a^{4} b B x +\frac {\left (-5 a^{4} b A -a^{5} B \right ) x^{2}-\frac {a^{5} A}{3}}{x^{3}}\) | \(122\) |
gosper | \(-\frac {-7 b^{5} B \,x^{12}-9 A \,b^{5} x^{10}-45 B a \,b^{4} x^{10}-63 a A \,b^{4} x^{8}-126 B \,a^{2} b^{3} x^{8}-210 a^{2} A \,b^{3} x^{6}-210 B \,a^{3} b^{2} x^{6}-630 a^{3} A \,b^{2} x^{4}-315 B \,a^{4} b \,x^{4}+315 a^{4} A b \,x^{2}+63 a^{5} B \,x^{2}+21 a^{5} A}{63 x^{3}}\) | \(128\) |
parallelrisch | \(\frac {7 b^{5} B \,x^{12}+9 A \,b^{5} x^{10}+45 B a \,b^{4} x^{10}+63 a A \,b^{4} x^{8}+126 B \,a^{2} b^{3} x^{8}+210 a^{2} A \,b^{3} x^{6}+210 B \,a^{3} b^{2} x^{6}+630 a^{3} A \,b^{2} x^{4}+315 B \,a^{4} b \,x^{4}-315 a^{4} A b \,x^{2}-63 a^{5} B \,x^{2}-21 a^{5} A}{63 x^{3}}\) | \(128\) |
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Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=\frac {7 \, B b^{5} x^{12} + 9 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} + 63 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} + 210 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} - 21 \, A a^{5} + 315 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} - 63 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{63 \, x^{3}} \]
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Time = 0.17 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=\frac {B b^{5} x^{9}}{9} + x^{7} \left (\frac {A b^{5}}{7} + \frac {5 B a b^{4}}{7}\right ) + x^{5} \left (A a b^{4} + 2 B a^{2} b^{3}\right ) + x^{3} \cdot \left (\frac {10 A a^{2} b^{3}}{3} + \frac {10 B a^{3} b^{2}}{3}\right ) + x \left (10 A a^{3} b^{2} + 5 B a^{4} b\right ) + \frac {- A a^{5} + x^{2} \left (- 15 A a^{4} b - 3 B a^{5}\right )}{3 x^{3}} \]
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Time = 0.18 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=\frac {1}{9} \, B b^{5} x^{9} + \frac {1}{7} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{7} + {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{5} + \frac {10}{3} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 5 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x - \frac {A a^{5} + 3 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{3 \, x^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=\frac {1}{9} \, B b^{5} x^{9} + \frac {5}{7} \, B a b^{4} x^{7} + \frac {1}{7} \, A b^{5} x^{7} + 2 \, B a^{2} b^{3} x^{5} + A a b^{4} x^{5} + \frac {10}{3} \, B a^{3} b^{2} x^{3} + \frac {10}{3} \, A a^{2} b^{3} x^{3} + 5 \, B a^{4} b x + 10 \, A a^{3} b^{2} x - \frac {3 \, B a^{5} x^{2} + 15 \, A a^{4} b x^{2} + A a^{5}}{3 \, x^{3}} \]
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Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^4} \, dx=x^7\,\left (\frac {A\,b^5}{7}+\frac {5\,B\,a\,b^4}{7}\right )-\frac {\frac {A\,a^5}{3}+x^2\,\left (B\,a^5+5\,A\,b\,a^4\right )}{x^3}+\frac {B\,b^5\,x^9}{9}+\frac {10\,a^2\,b^2\,x^3\,\left (A\,b+B\,a\right )}{3}+5\,a^3\,b\,x\,\left (2\,A\,b+B\,a\right )+a\,b^3\,x^5\,\left (A\,b+2\,B\,a\right ) \]
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