Integrand size = 20, antiderivative size = 116 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {(A b-3 a B) x}{b^4}+\frac {B x^3}{3 b^3}-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}-\frac {5 \sqrt {a} (3 A b-7 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}} \]
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Time = 0.11 (sec) , antiderivative size = 116, 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, 1167, 211} \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {a^2 x (A b-a B)}{4 b^4 \left (a+b x^2\right )^2}-\frac {5 \sqrt {a} (3 A b-7 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}}+\frac {a x (9 A b-13 a B)}{8 b^4 \left (a+b x^2\right )}+\frac {x (A b-3 a B)}{b^4}+\frac {B x^3}{3 b^3} \]
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Rule 211
Rule 466
Rule 1167
Rule 1828
Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}-\frac {\int \frac {-a^2 (A b-a B)+4 a b (A b-a B) x^2-4 b^2 (A b-a B) x^4-4 b^3 B x^6}{\left (a+b x^2\right )^2} \, dx}{4 b^4} \\ & = -\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}+\frac {\int \frac {-a^2 (7 A b-11 a B)+8 a b (A b-2 a B) x^2+8 a b^2 B x^4}{a+b x^2} \, dx}{8 a b^4} \\ & = -\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}+\frac {\int \left (8 a (A b-3 a B)+8 a b B x^2+\frac {5 \left (-3 a^2 A b+7 a^3 B\right )}{a+b x^2}\right ) \, dx}{8 a b^4} \\ & = \frac {(A b-3 a B) x}{b^4}+\frac {B x^3}{3 b^3}-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}-\frac {(5 a (3 A b-7 a B)) \int \frac {1}{a+b x^2} \, dx}{8 b^4} \\ & = \frac {(A b-3 a B) x}{b^4}+\frac {B x^3}{3 b^3}-\frac {a^2 (A b-a B) x}{4 b^4 \left (a+b x^2\right )^2}+\frac {a (9 A b-13 a B) x}{8 b^4 \left (a+b x^2\right )}-\frac {5 \sqrt {a} (3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {-105 a^3 B x+a b^2 x^3 \left (75 A-56 B x^2\right )+5 a^2 b x \left (9 A-35 B x^2\right )+8 b^3 x^5 \left (3 A+B x^2\right )}{24 b^4 \left (a+b x^2\right )^2}+\frac {5 \sqrt {a} (-3 A b+7 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{9/2}} \]
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Time = 2.64 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {\frac {1}{3} b B \,x^{3}+A b x -3 B a x}{b^{4}}-\frac {a \left (\frac {\left (-\frac {9}{8} b^{2} A +\frac {13}{8} a b B \right ) x^{3}-\frac {a \left (7 A b -11 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {5 \left (3 A b -7 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\) | \(95\) |
risch | \(\frac {B \,x^{3}}{3 b^{3}}+\frac {A x}{b^{3}}-\frac {3 B a x}{b^{4}}+\frac {\left (\frac {9}{8} a \,b^{2} A -\frac {13}{8} a^{2} b B \right ) x^{3}+\frac {a^{2} \left (7 A b -11 B a \right ) x}{8}}{b^{4} \left (b \,x^{2}+a \right )^{2}}+\frac {15 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right ) A}{16 b^{4}}-\frac {35 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right ) B a}{16 b^{5}}-\frac {15 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right ) A}{16 b^{4}}+\frac {35 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right ) B a}{16 b^{5}}\) | \(177\) |
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Time = 0.30 (sec) , antiderivative size = 358, normalized size of antiderivative = 3.09 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, B b^{3} x^{7} - 16 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{5} - 50 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{3} - 15 \, {\left ({\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{4} + 7 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 30 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} x}{48 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {8 \, B b^{3} x^{7} - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{5} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{3} + 15 \, {\left ({\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{4} + 7 \, B a^{3} - 3 \, A a^{2} b + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} x}{24 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \]
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Time = 0.71 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.84 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x^{3}}{3 b^{3}} + x \left (\frac {A}{b^{3}} - \frac {3 B a}{b^{4}}\right ) - \frac {5 \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right ) \log {\left (- \frac {5 b^{4} \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right )}{- 15 A b + 35 B a} + x \right )}}{16} + \frac {5 \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right ) \log {\left (\frac {5 b^{4} \sqrt {- \frac {a}{b^{9}}} \left (- 3 A b + 7 B a\right )}{- 15 A b + 35 B a} + x \right )}}{16} + \frac {x^{3} \cdot \left (9 A a b^{2} - 13 B a^{2} b\right ) + x \left (7 A a^{2} b - 11 B a^{3}\right )}{8 a^{2} b^{4} + 16 a b^{5} x^{2} + 8 b^{6} x^{4}} \]
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Time = 0.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {{\left (13 \, B a^{2} b - 9 \, A a b^{2}\right )} x^{3} + {\left (11 \, B a^{3} - 7 \, A a^{2} b\right )} x}{8 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} + \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} + \frac {B b x^{3} - 3 \, {\left (3 \, B a - A b\right )} x}{3 \, b^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} - \frac {13 \, B a^{2} b x^{3} - 9 \, A a b^{2} x^{3} + 11 \, B a^{3} x - 7 \, A a^{2} b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{4}} + \frac {B b^{6} x^{3} - 9 \, B a b^{5} x + 3 \, A b^{6} x}{3 \, b^{9}} \]
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Time = 4.97 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.19 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x^3\,\left (\frac {9\,A\,a\,b^2}{8}-\frac {13\,B\,a^2\,b}{8}\right )-x\,\left (\frac {11\,B\,a^3}{8}-\frac {7\,A\,a^2\,b}{8}\right )}{a^2\,b^4+2\,a\,b^5\,x^2+b^6\,x^4}+x\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )+\frac {B\,x^3}{3\,b^3}+\frac {5\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (3\,A\,b-7\,B\,a\right )}{7\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-7\,B\,a\right )}{8\,b^{9/2}} \]
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