Integrand size = 20, antiderivative size = 94 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x}{b^3}+\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}} \]
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Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {466, 1171, 396, 211} \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}-\frac {x (5 A b-9 a B)}{8 b^3 \left (a+b x^2\right )}+\frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}+\frac {B x}{b^3} \]
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Rule 211
Rule 396
Rule 466
Rule 1171
Rubi steps \begin{align*} \text {integral}& = \frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {\int \frac {a (A b-a B)-4 b (A b-a B) x^2-4 b^2 B x^4}{\left (a+b x^2\right )^2} \, dx}{4 b^3} \\ & = \frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {\int \frac {a (3 A b-7 a B)+8 a b B x^2}{a+b x^2} \, dx}{8 a b^3} \\ & = \frac {B x}{b^3}+\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {(3 (A b-5 a B)) \int \frac {1}{a+b x^2} \, dx}{8 b^3} \\ & = \frac {B x}{b^3}+\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (15 a^2 B+b^2 x^2 \left (-5 A+8 B x^2\right )+a \left (-3 A b+25 b B x^2\right )\right )}{8 b^3 \left (a+b x^2\right )^2}+\frac {3 (A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}} \]
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Time = 2.54 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {B x}{b^{3}}+\frac {\frac {\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{3}-\frac {a \left (3 A b -7 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {3 \left (A b -5 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{3}}\) | \(77\) |
risch | \(\frac {B x}{b^{3}}+\frac {\left (-\frac {5}{8} b^{2} A +\frac {9}{8} a b B \right ) x^{3}-\frac {a \left (3 A b -7 B a \right ) x}{8}}{b^{3} \left (b \,x^{2}+a \right )^{2}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) A}{16 b^{2} \sqrt {-a b}}+\frac {15 \ln \left (b x +\sqrt {-a b}\right ) B a}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) A}{16 b^{2} \sqrt {-a b}}-\frac {15 \ln \left (-b x +\sqrt {-a b}\right ) B a}{16 b^{3} \sqrt {-a b}}\) | \(147\) |
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Time = 0.32 (sec) , antiderivative size = 328, normalized size of antiderivative = 3.49 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, B a b^{3} x^{5} + 10 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} + 3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{4} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {8 \, B a b^{3} x^{5} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{4} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (92) = 184\).
Time = 0.58 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.06 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x}{b^{3}} + \frac {3 \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right ) \log {\left (- \frac {3 a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right )}{- 3 A b + 15 B a} + x \right )}}{16} - \frac {3 \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right ) \log {\left (\frac {3 a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right )}{- 3 A b + 15 B a} + x \right )}}{16} + \frac {x^{3} \left (- 5 A b^{2} + 9 B a b\right ) + x \left (- 3 A a b + 7 B a^{2}\right )}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (9 \, B a b - 5 \, A b^{2}\right )} x^{3} + {\left (7 \, B a^{2} - 3 \, A a b\right )} x}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {B x}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {9 \, B a b x^{3} - 5 \, A b^{2} x^{3} + 7 \, B a^{2} x - 3 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3}} \]
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Time = 5.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B\,x}{b^3}-\frac {x^3\,\left (\frac {5\,A\,b^2}{8}-\frac {9\,B\,a\,b}{8}\right )-x\,\left (\frac {7\,B\,a^2}{8}-\frac {3\,A\,a\,b}{8}\right )}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b-5\,B\,a\right )}{8\,\sqrt {a}\,b^{7/2}} \]
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