Integrand size = 17, antiderivative size = 58 \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {1}{5 b x^5}+\frac {c}{3 b^2 x^3}-\frac {c^2}{b^3 x}-\frac {c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}} \]
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Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1598, 331, 211} \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}}-\frac {c^2}{b^3 x}+\frac {c}{3 b^2 x^3}-\frac {1}{5 b x^5} \]
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Rule 211
Rule 331
Rule 1598
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^6 \left (b+c x^2\right )} \, dx \\ & = -\frac {1}{5 b x^5}-\frac {c \int \frac {1}{x^4 \left (b+c x^2\right )} \, dx}{b} \\ & = -\frac {1}{5 b x^5}+\frac {c}{3 b^2 x^3}+\frac {c^2 \int \frac {1}{x^2 \left (b+c x^2\right )} \, dx}{b^2} \\ & = -\frac {1}{5 b x^5}+\frac {c}{3 b^2 x^3}-\frac {c^2}{b^3 x}-\frac {c^3 \int \frac {1}{b+c x^2} \, dx}{b^3} \\ & = -\frac {1}{5 b x^5}+\frac {c}{3 b^2 x^3}-\frac {c^2}{b^3 x}-\frac {c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {1}{5 b x^5}+\frac {c}{3 b^2 x^3}-\frac {c^2}{b^3 x}-\frac {c^{5/2} \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2}} \]
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Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {1}{5 b \,x^{5}}-\frac {c^{2}}{b^{3} x}+\frac {c}{3 b^{2} x^{3}}-\frac {c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{b^{3} \sqrt {b c}}\) | \(52\) |
risch | \(\frac {-\frac {c^{2} x^{4}}{b^{3}}+\frac {c \,x^{2}}{3 b^{2}}-\frac {1}{5 b}}{x^{5}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{7} \textit {\_Z}^{2}+c^{5}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{7}+2 c^{5}\right ) x +b^{4} c^{2} \textit {\_R} \right )\right )}{2}\) | \(77\) |
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Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.28 \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=\left [\frac {15 \, c^{2} x^{5} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) - 30 \, c^{2} x^{4} + 10 \, b c x^{2} - 6 \, b^{2}}{30 \, b^{3} x^{5}}, -\frac {15 \, c^{2} x^{5} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 15 \, c^{2} x^{4} - 5 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{3} x^{5}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.72 \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=\frac {\sqrt {- \frac {c^{5}}{b^{7}}} \log {\left (- \frac {b^{4} \sqrt {- \frac {c^{5}}{b^{7}}}}{c^{3}} + x \right )}}{2} - \frac {\sqrt {- \frac {c^{5}}{b^{7}}} \log {\left (\frac {b^{4} \sqrt {- \frac {c^{5}}{b^{7}}}}{c^{3}} + x \right )}}{2} + \frac {- 3 b^{2} + 5 b c x^{2} - 15 c^{2} x^{4}}{15 b^{3} x^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} - \frac {15 \, c^{2} x^{4} - 5 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{3} x^{5}} \]
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Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} b^{3}} - \frac {15 \, c^{2} x^{4} - 5 \, b c x^{2} + 3 \, b^{2}}{15 \, b^{3} x^{5}} \]
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Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^4 \left (b x^2+c x^4\right )} \, dx=-\frac {\frac {1}{5\,b}-\frac {c\,x^2}{3\,b^2}+\frac {c^2\,x^4}{b^3}}{x^5}-\frac {c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{b^{7/2}} \]
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