Integrand size = 20, antiderivative size = 142 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=-\frac {A}{5 a^3 x^5}+\frac {3 A b-a B}{3 a^4 x^3}-\frac {3 b (2 A b-a B)}{a^5 x}-\frac {b^2 (A b-a B) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (15 A b-11 a B) x}{8 a^5 \left (a+b x^2\right )}-\frac {7 b^{3/2} (9 A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2}} \]
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Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {467, 1819, 1816, 211} \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=-\frac {7 b^{3/2} (9 A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2}}-\frac {b^2 x (15 A b-11 a B)}{8 a^5 \left (a+b x^2\right )}-\frac {3 b (2 A b-a B)}{a^5 x}-\frac {b^2 x (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}+\frac {3 A b-a B}{3 a^4 x^3}-\frac {A}{5 a^3 x^5} \]
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Rule 211
Rule 467
Rule 1816
Rule 1819
Rubi steps \begin{align*} \text {integral}& = -\frac {b^2 (A b-a B) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {1}{4} b^2 \int \frac {-\frac {4 A}{a b^2}+\frac {4 (A b-a B) x^2}{a^2 b^2}-\frac {4 (A b-a B) x^4}{a^3 b}+\frac {3 (A b-a B) x^6}{a^4}}{x^6 \left (a+b x^2\right )^2} \, dx \\ & = -\frac {b^2 (A b-a B) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (15 A b-11 a B) x}{8 a^5 \left (a+b x^2\right )}+\frac {b^2 \int \frac {\frac {8 A}{a b^2}-\frac {8 (2 A b-a B) x^2}{a^2 b^2}+\frac {8 (3 A b-2 a B) x^4}{a^3 b}-\frac {(15 A b-11 a B) x^6}{a^4}}{x^6 \left (a+b x^2\right )} \, dx}{8 a} \\ & = -\frac {b^2 (A b-a B) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (15 A b-11 a B) x}{8 a^5 \left (a+b x^2\right )}+\frac {b^2 \int \left (\frac {8 A}{a^2 b^2 x^6}+\frac {8 (-3 A b+a B)}{a^3 b^2 x^4}-\frac {24 (-2 A b+a B)}{a^4 b x^2}+\frac {7 (-9 A b+5 a B)}{a^4 \left (a+b x^2\right )}\right ) \, dx}{8 a} \\ & = -\frac {A}{5 a^3 x^5}+\frac {3 A b-a B}{3 a^4 x^3}-\frac {3 b (2 A b-a B)}{a^5 x}-\frac {b^2 (A b-a B) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (15 A b-11 a B) x}{8 a^5 \left (a+b x^2\right )}-\frac {\left (7 b^2 (9 A b-5 a B)\right ) \int \frac {1}{a+b x^2} \, dx}{8 a^5} \\ & = -\frac {A}{5 a^3 x^5}+\frac {3 A b-a B}{3 a^4 x^3}-\frac {3 b (2 A b-a B)}{a^5 x}-\frac {b^2 (A b-a B) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (15 A b-11 a B) x}{8 a^5 \left (a+b x^2\right )}-\frac {7 b^{3/2} (9 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2}} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {-945 A b^4 x^8+525 a b^3 x^6 \left (-3 A+B x^2\right )-8 a^4 \left (3 A+5 B x^2\right )+8 a^3 b x^2 \left (9 A+35 B x^2\right )+7 a^2 b^2 x^4 \left (-72 A+125 B x^2\right )}{120 a^5 x^5 \left (a+b x^2\right )^2}+\frac {7 b^{3/2} (-9 A b+5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2}} \]
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Time = 2.66 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {A}{5 a^{3} x^{5}}-\frac {-3 A b +B a}{3 a^{4} x^{3}}-\frac {3 b \left (2 A b -B a \right )}{a^{5} x}-\frac {b^{2} \left (\frac {\left (\frac {15}{8} b^{2} A -\frac {11}{8} a b B \right ) x^{3}+\frac {a \left (17 A b -13 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {7 \left (9 A b -5 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}\) | \(119\) |
risch | \(\frac {-\frac {7 b^{3} \left (9 A b -5 B a \right ) x^{8}}{8 a^{5}}-\frac {35 b^{2} \left (9 A b -5 B a \right ) x^{6}}{24 a^{4}}-\frac {7 b \left (9 A b -5 B a \right ) x^{4}}{15 a^{3}}+\frac {\left (9 A b -5 B a \right ) x^{2}}{15 a^{2}}-\frac {A}{5 a}}{x^{5} \left (b \,x^{2}+a \right )^{2}}+\frac {63 \sqrt {-a b}\, b^{2} \ln \left (-b x +\sqrt {-a b}\right ) A}{16 a^{6}}-\frac {35 \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right ) B}{16 a^{5}}-\frac {63 \sqrt {-a b}\, b^{2} \ln \left (-b x -\sqrt {-a b}\right ) A}{16 a^{6}}+\frac {35 \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right ) B}{16 a^{5}}\) | \(205\) |
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Time = 0.28 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.00 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\left [\frac {210 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{8} + 350 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{6} - 48 \, A a^{4} + 112 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{4} - 16 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x^{2} - 105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{9} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{7} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{5}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{240 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}}, \frac {105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{8} + 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{6} - 24 \, A a^{4} + 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x^{2} + 105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{9} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{7} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{5}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{120 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}}\right ] \]
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Time = 0.47 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.83 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=- \frac {7 \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right ) \log {\left (- \frac {7 a^{6} \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right )}{- 63 A b^{3} + 35 B a b^{2}} + x \right )}}{16} + \frac {7 \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right ) \log {\left (\frac {7 a^{6} \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right )}{- 63 A b^{3} + 35 B a b^{2}} + x \right )}}{16} + \frac {- 24 A a^{4} + x^{8} \left (- 945 A b^{4} + 525 B a b^{3}\right ) + x^{6} \left (- 1575 A a b^{3} + 875 B a^{2} b^{2}\right ) + x^{4} \left (- 504 A a^{2} b^{2} + 280 B a^{3} b\right ) + x^{2} \cdot \left (72 A a^{3} b - 40 B a^{4}\right )}{120 a^{7} x^{5} + 240 a^{6} b x^{7} + 120 a^{5} b^{2} x^{9}} \]
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Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{8} + 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{6} - 24 \, A a^{4} + 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x^{2}}{120 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}} + \frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} \]
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Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} + \frac {11 \, B a b^{3} x^{3} - 15 \, A b^{4} x^{3} + 13 \, B a^{2} b^{2} x - 17 \, A a b^{3} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{5}} + \frac {45 \, B a b x^{4} - 90 \, A b^{2} x^{4} - 5 \, B a^{2} x^{2} + 15 \, A a b x^{2} - 3 \, A a^{2}}{15 \, a^{5} x^{5}} \]
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Time = 5.03 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=-\frac {\frac {A}{5\,a}-\frac {x^2\,\left (9\,A\,b-5\,B\,a\right )}{15\,a^2}+\frac {35\,b^2\,x^6\,\left (9\,A\,b-5\,B\,a\right )}{24\,a^4}+\frac {7\,b^3\,x^8\,\left (9\,A\,b-5\,B\,a\right )}{8\,a^5}+\frac {7\,b\,x^4\,\left (9\,A\,b-5\,B\,a\right )}{15\,a^3}}{a^2\,x^5+2\,a\,b\,x^7+b^2\,x^9}-\frac {7\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (9\,A\,b-5\,B\,a\right )}{8\,a^{11/2}} \]
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