Integrand size = 20, antiderivative size = 138 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^3}{3 b^4}+\frac {B x^5}{5 b^3}+\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {7 a^{3/2} (5 A b-9 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \]
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Time = 0.15 (sec) , antiderivative size = 138, 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 = {466, 1828, 1824, 211} \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {7 a^{3/2} (5 A b-9 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}}+\frac {a^3 x (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 x (13 A b-17 a B)}{8 b^5 \left (a+b x^2\right )}-\frac {3 a x (A b-2 a B)}{b^5}+\frac {x^3 (A b-3 a B)}{3 b^4}+\frac {B x^5}{5 b^3} \]
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Rule 211
Rule 466
Rule 1824
Rule 1828
Rubi steps \begin{align*} \text {integral}& = \frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {\int \frac {a^3 (A b-a B)-4 a^2 b (A b-a B) x^2+4 a b^2 (A b-a B) x^4-4 b^3 (A b-a B) x^6-4 b^4 B x^8}{\left (a+b x^2\right )^2} \, dx}{4 b^5} \\ & = \frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {\int \frac {a^3 (11 A b-15 a B)-8 a^2 b (2 A b-3 a B) x^2+8 a b^2 (A b-2 a B) x^4+8 a b^3 B x^6}{a+b x^2} \, dx}{8 a b^5} \\ & = \frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {\int \left (-24 a^2 (A b-2 a B)+8 a b (A b-3 a B) x^2+8 a b^2 B x^4-\frac {7 \left (-5 a^3 A b+9 a^4 B\right )}{a+b x^2}\right ) \, dx}{8 a b^5} \\ & = -\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^3}{3 b^4}+\frac {B x^5}{5 b^3}+\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {\left (7 a^2 (5 A b-9 a B)\right ) \int \frac {1}{a+b x^2} \, dx}{8 b^5} \\ & = -\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^3}{3 b^4}+\frac {B x^5}{5 b^3}+\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (945 a^4 B-525 a^3 b \left (A-3 B x^2\right )+8 b^4 x^6 \left (5 A+3 B x^2\right )-8 a b^3 x^4 \left (35 A+9 B x^2\right )+7 a^2 b^2 x^2 \left (-125 A+72 B x^2\right )\right )}{120 b^5 \left (a+b x^2\right )^2}-\frac {7 a^{3/2} (-5 A b+9 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \]
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Time = 2.63 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {-\frac {1}{5} b^{2} B \,x^{5}-\frac {1}{3} A \,b^{2} x^{3}+B a b \,x^{3}+3 a A b x -6 a^{2} B x}{b^{5}}+\frac {a^{2} \left (\frac {\left (-\frac {13}{8} b^{2} A +\frac {17}{8} a b B \right ) x^{3}-\frac {a \left (11 A b -15 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {7 \left (5 A b -9 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}\) | \(119\) |
risch | \(\frac {B \,x^{5}}{5 b^{3}}+\frac {A \,x^{3}}{3 b^{3}}-\frac {B a \,x^{3}}{b^{4}}-\frac {3 a A x}{b^{4}}+\frac {6 a^{2} B x}{b^{5}}+\frac {\left (-\frac {13}{8} A \,a^{2} b^{2}+\frac {17}{8} B \,a^{3} b \right ) x^{3}-\frac {a^{3} \left (11 A b -15 B a \right ) x}{8}}{b^{5} \left (b \,x^{2}+a \right )^{2}}+\frac {35 \sqrt {-a b}\, a \ln \left (-\sqrt {-a b}\, x +a \right ) A}{16 b^{5}}-\frac {63 \sqrt {-a b}\, a^{2} \ln \left (-\sqrt {-a b}\, x +a \right ) B}{16 b^{6}}-\frac {35 \sqrt {-a b}\, a \ln \left (\sqrt {-a b}\, x +a \right ) A}{16 b^{5}}+\frac {63 \sqrt {-a b}\, a^{2} \ln \left (\sqrt {-a b}\, x +a \right ) B}{16 b^{6}}\) | \(200\) |
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Time = 0.31 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.01 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {48 \, B b^{4} x^{9} - 16 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{7} + 112 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + 350 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} - 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 210 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b\right )} x}{240 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}, \frac {24 \, B b^{4} x^{9} - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{7} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} - 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b\right )} x}{120 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \]
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Time = 0.81 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.83 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x^{5}}{5 b^{3}} + x^{3} \left (\frac {A}{3 b^{3}} - \frac {B a}{b^{4}}\right ) + x \left (- \frac {3 A a}{b^{4}} + \frac {6 B a^{2}}{b^{5}}\right ) + \frac {7 \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right ) \log {\left (- \frac {7 b^{5} \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right )}{- 35 A a b + 63 B a^{2}} + x \right )}}{16} - \frac {7 \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right ) \log {\left (\frac {7 b^{5} \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right )}{- 35 A a b + 63 B a^{2}} + x \right )}}{16} + \frac {x^{3} \left (- 13 A a^{2} b^{2} + 17 B a^{3} b\right ) + x \left (- 11 A a^{3} b + 15 B a^{4}\right )}{8 a^{2} b^{5} + 16 a b^{6} x^{2} + 8 b^{7} x^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (17 \, B a^{3} b - 13 \, A a^{2} b^{2}\right )} x^{3} + {\left (15 \, B a^{4} - 11 \, A a^{3} b\right )} x}{8 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} - \frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {3 \, B b^{2} x^{5} - 5 \, {\left (3 \, B a b - A b^{2}\right )} x^{3} + 45 \, {\left (2 \, B a^{2} - A a b\right )} x}{15 \, b^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {17 \, B a^{3} b x^{3} - 13 \, A a^{2} b^{2} x^{3} + 15 \, B a^{4} x - 11 \, A a^{3} b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{5}} + \frac {3 \, B b^{12} x^{5} - 15 \, B a b^{11} x^{3} + 5 \, A b^{12} x^{3} + 90 \, B a^{2} b^{10} x - 45 \, A a b^{11} x}{15 \, b^{15}} \]
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Time = 0.07 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.28 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x\,\left (\frac {15\,B\,a^4}{8}-\frac {11\,A\,a^3\,b}{8}\right )-x^3\,\left (\frac {13\,A\,a^2\,b^2}{8}-\frac {17\,B\,a^3\,b}{8}\right )}{a^2\,b^5+2\,a\,b^6\,x^2+b^7\,x^4}-x\,\left (\frac {3\,a\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{b}+\frac {3\,B\,a^2}{b^5}\right )+x^3\,\left (\frac {A}{3\,b^3}-\frac {B\,a}{b^4}\right )+\frac {B\,x^5}{5\,b^3}-\frac {7\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,x\,\left (5\,A\,b-9\,B\,a\right )}{9\,B\,a^3-5\,A\,a^2\,b}\right )\,\left (5\,A\,b-9\,B\,a\right )}{8\,b^{11/2}} \]
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