Integrand size = 20, antiderivative size = 97 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {A}{a^3 x}-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac {3 (5 A b-a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}} \]
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Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {467, 464, 211} \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {3 (5 A b-a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}-\frac {x (7 A b-3 a B)}{8 a^3 \left (a+b x^2\right )}-\frac {A}{a^3 x}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2} \]
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Rule 211
Rule 464
Rule 467
Rubi steps \begin{align*} \text {integral}& = -\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {1}{4} \int \frac {-\frac {4 A}{a}+\frac {3 (A b-a B) x^2}{a^2}}{x^2 \left (a+b x^2\right )^2} \, dx \\ & = -\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}+\frac {1}{8} \int \frac {\frac {8 A}{a^2}-\frac {(7 A b-3 a B) x^2}{a^3}}{x^2 \left (a+b x^2\right )} \, dx \\ & = -\frac {A}{a^3 x}-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac {(3 (5 A b-a B)) \int \frac {1}{a+b x^2} \, dx}{8 a^3} \\ & = -\frac {A}{a^3 x}-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac {3 (5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {A}{a^3 x}+\frac {(-A b+a B) x}{4 a^2 \left (a+b x^2\right )^2}+\frac {(-7 A b+3 a B) x}{8 a^3 \left (a+b x^2\right )}+\frac {3 (-5 A b+a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}} \]
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Time = 2.50 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {A}{a^{3} x}-\frac {\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{3}+\frac {a \left (9 A b -5 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{a^{3}}\) | \(82\) |
risch | \(\frac {-\frac {3 b \left (5 A b -B a \right ) x^{4}}{8 a^{3}}-\frac {5 \left (5 A b -B a \right ) x^{2}}{8 a^{2}}-\frac {A}{a}}{x \left (b \,x^{2}+a \right )^{2}}-\frac {15 \ln \left (-\sqrt {-a b}\, x -a \right ) A b}{16 \sqrt {-a b}\, a^{3}}+\frac {3 \ln \left (-\sqrt {-a b}\, x -a \right ) B}{16 \sqrt {-a b}\, a^{2}}+\frac {15 \ln \left (-\sqrt {-a b}\, x +a \right ) A b}{16 \sqrt {-a b}\, a^{3}}-\frac {3 \ln \left (-\sqrt {-a b}\, x +a \right ) B}{16 \sqrt {-a b}\, a^{2}}\) | \(159\) |
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Time = 0.27 (sec) , antiderivative size = 324, normalized size of antiderivative = 3.34 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=\left [-\frac {16 \, A a^{3} b - 6 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 10 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac {8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (94) = 188\).
Time = 0.38 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.00 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=- \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (- \frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (\frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {- 8 A a^{2} + x^{4} \left (- 15 A b^{2} + 3 B a b\right ) + x^{2} \left (- 25 A a b + 5 B a^{2}\right )}{8 a^{5} x + 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} \]
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Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {3 \, {\left (B a b - 5 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} + 5 \, {\left (B a^{2} - 5 \, A a b\right )} x^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} + \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {A}{a^{3} x} + \frac {3 \, B a b x^{3} - 7 \, A b^{2} x^{3} + 5 \, B a^{2} x - 9 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3}} \]
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Time = 5.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x^2}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {\frac {A}{a}+\frac {5\,x^2\,\left (5\,A\,b-B\,a\right )}{8\,a^2}+\frac {3\,b\,x^4\,\left (5\,A\,b-B\,a\right )}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,x\,\left (5\,A\,b-B\,a\right )}{\sqrt {a}\,\left (15\,A\,b-3\,B\,a\right )}\right )\,\left (5\,A\,b-B\,a\right )}{8\,a^{7/2}\,\sqrt {b}} \]
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