Integrand size = 33, antiderivative size = 190 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 (5 b d-a e) \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Time = 0.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1264, 467, 464, 211} \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {x (b d-a e)}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (5 b d-a e)}{8 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x (7 b d-3 a e)}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
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Rule 211
Rule 464
Rule 467
Rule 1264
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {d+e x^2}{x^2 \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {-\frac {4 d}{a b}+\frac {3 (b d-a e) x^2}{a^2 b}}{x^2 \left (a b+b^2 x^2\right )^2} \, dx}{4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {\frac {8 d}{a^2 b^2}-\frac {(7 b d-3 a e) x^2}{a^3 b^2}}{x^2 \left (a b+b^2 x^2\right )} \, dx}{8 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (3 (5 b d-a e) \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ & = -\frac {(7 b d-3 a e) x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 (5 b d-a e) \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}} \\ \end{align*}
Time = 1.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.65 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {a} \sqrt {b} \left (-15 b^2 d x^4+a^2 \left (-8 d+5 e x^2\right )+a b \left (-25 d x^2+3 e x^4\right )\right )+3 (-5 b d+a e) x \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b} x \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.22 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {3 b \left (a e -5 b d \right ) x^{4}}{8 a^{3}}+\frac {5 \left (a e -5 b d \right ) x^{2}}{8 a^{2}}-\frac {d}{a}\right )}{\left (b \,x^{2}+a \right )^{3} x}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2} b +e^{2} a^{2}-10 a b d e +25 b^{2} d^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{7} b +2 e^{2} a^{2}-20 a b d e +50 b^{2} d^{2}\right ) x +\left (-e \,a^{5}+5 a^{4} b d \right ) \textit {\_R} \right )\right )}{16 \left (b \,x^{2}+a \right )}\) | \(176\) |
default | \(-\frac {\left (-3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} e \,x^{5}+15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{3} d \,x^{5}-3 \sqrt {a b}\, a b e \,x^{4}+15 \sqrt {a b}\, b^{2} d \,x^{4}-6 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b e \,x^{3}+30 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} d \,x^{3}-5 \sqrt {a b}\, a^{2} e \,x^{2}+25 \sqrt {a b}\, a b d \,x^{2}-3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3} e x +15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b d x +8 \sqrt {a b}\, a^{2} d \right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, x \,a^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(206\) |
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Time = 0.27 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.76 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [-\frac {16 \, a^{3} b d + 6 \, {\left (5 \, a b^{3} d - a^{2} b^{2} e\right )} x^{4} + 10 \, {\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x^{2} - 3 \, {\left ({\left (5 \, b^{3} d - a b^{2} e\right )} x^{5} + 2 \, {\left (5 \, a b^{2} d - a^{2} b e\right )} x^{3} + {\left (5 \, a^{2} b d - a^{3} e\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac {8 \, a^{3} b d + 3 \, {\left (5 \, a b^{3} d - a^{2} b^{2} e\right )} x^{4} + 5 \, {\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x^{2} + 3 \, {\left ({\left (5 \, b^{3} d - a b^{2} e\right )} x^{5} + 2 \, {\left (5 \, a b^{2} d - a^{2} b e\right )} x^{3} + {\left (5 \, a^{2} b d - a^{3} e\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \]
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\[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {d + e x^{2}}{x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.71 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {1}{8} \, d {\left (\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 8 \, a^{2}}{a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x} + \frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}}\right )} + \frac {1}{8} \, e {\left (\frac {3 \, b x^{3} + 5 \, a x}{a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.59 \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (5 \, b d - a e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {d}{a^{3} x \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {7 \, b^{2} d x^{3} - 3 \, a b e x^{3} + 9 \, a b d x - 5 \, a^{2} e x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \]
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Timed out. \[ \int \frac {d+e x^2}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {e\,x^2+d}{x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]
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