Integrand size = 17, antiderivative size = 92 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=\frac {(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac {(3 A b+a B) x}{8 a^2 b \left (a+b x^2\right )}+\frac {(3 A b+a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}} \]
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Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {393, 205, 211} \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=\frac {(a B+3 A b) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}}+\frac {x (a B+3 A b)}{8 a^2 b \left (a+b x^2\right )}+\frac {x (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]
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Rule 205
Rule 211
Rule 393
Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac {(3 A b+a B) \int \frac {1}{\left (a+b x^2\right )^2} \, dx}{4 a b} \\ & = \frac {(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac {(3 A b+a B) x}{8 a^2 b \left (a+b x^2\right )}+\frac {(3 A b+a B) \int \frac {1}{a+b x^2} \, dx}{8 a^2 b} \\ & = \frac {(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac {(3 A b+a B) x}{8 a^2 b \left (a+b x^2\right )}+\frac {(3 A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (-a^2 B+3 A b^2 x^2+a b \left (5 A+B x^2\right )\right )}{8 a^2 b \left (a+b x^2\right )^2}+\frac {(3 A b+a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}} \]
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Time = 2.54 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\frac {\left (3 A b +B a \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A b -B a \right ) x}{8 a b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (3 A b +B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} b \sqrt {a b}}\) | \(77\) |
risch | \(\frac {\frac {\left (3 A b +B a \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A b -B a \right ) x}{8 a b}}{\left (b \,x^{2}+a \right )^{2}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) A}{16 \sqrt {-a b}\, a^{2}}-\frac {\ln \left (b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b a}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) A}{16 \sqrt {-a b}\, a^{2}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b a}\) | \(147\) |
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Time = 0.28 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.26 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=\left [\frac {2 \, {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} - {\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{4} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 2 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{16 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, \frac {{\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} + {\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{4} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{8 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \]
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Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.63 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \cdot \left (3 A b + B a\right ) \log {\left (- a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \cdot \left (3 A b + B a\right ) \log {\left (a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 A b^{2} + B a b\right ) + x \left (5 A a b - B a^{2}\right )}{8 a^{4} b + 16 a^{3} b^{2} x^{2} + 8 a^{2} b^{3} x^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (B a b + 3 \, A b^{2}\right )} x^{3} - {\left (B a^{2} - 5 \, A a b\right )} x}{8 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} + \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} \]
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Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} + \frac {B a b x^{3} + 3 \, A b^{2} x^{3} - B a^{2} x + 5 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b} \]
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Time = 5.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {x^3\,\left (3\,A\,b+B\,a\right )}{8\,a^2}+\frac {x\,\left (5\,A\,b-B\,a\right )}{8\,a\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,A\,b+B\,a\right )}{8\,a^{5/2}\,b^{3/2}} \]
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