Integrand size = 15, antiderivative size = 39 \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=-\frac {\sqrt {16-x^4}}{64 x^4}-\frac {1}{256} \text {arctanh}\left (\frac {\sqrt {16-x^4}}{4}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 44, 65, 212} \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=-\frac {1}{256} \text {arctanh}\left (\frac {\sqrt {16-x^4}}{4}\right )-\frac {\sqrt {16-x^4}}{64 x^4} \]
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Rule 44
Rule 65
Rule 212
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {16-x} x^2} \, dx,x,x^4\right ) \\ & = -\frac {\sqrt {16-x^4}}{64 x^4}+\frac {1}{128} \text {Subst}\left (\int \frac {1}{\sqrt {16-x} x} \, dx,x,x^4\right ) \\ & = -\frac {\sqrt {16-x^4}}{64 x^4}-\frac {1}{64} \text {Subst}\left (\int \frac {1}{16-x^2} \, dx,x,\sqrt {16-x^4}\right ) \\ & = -\frac {\sqrt {16-x^4}}{64 x^4}-\frac {1}{256} \tanh ^{-1}\left (\frac {\sqrt {16-x^4}}{4}\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=-\frac {\sqrt {16-x^4}}{64 x^4}-\frac {1}{256} \text {arctanh}\left (\frac {\sqrt {16-x^4}}{4}\right ) \]
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Time = 4.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {\sqrt {-x^{4}+16}}{64 x^{4}}-\frac {\operatorname {arctanh}\left (\frac {4}{\sqrt {-x^{4}+16}}\right )}{256}\) | \(30\) |
elliptic | \(-\frac {\sqrt {-x^{4}+16}}{64 x^{4}}-\frac {\operatorname {arctanh}\left (\frac {4}{\sqrt {-x^{4}+16}}\right )}{256}\) | \(30\) |
trager | \(-\frac {\sqrt {-x^{4}+16}}{64 x^{4}}-\frac {\ln \left (\frac {\sqrt {-x^{4}+16}+4}{x^{2}}\right )}{256}\) | \(34\) |
risch | \(\frac {x^{4}-16}{64 x^{4} \sqrt {-x^{4}+16}}-\frac {\operatorname {arctanh}\left (\frac {4}{\sqrt {-x^{4}+16}}\right )}{256}\) | \(35\) |
pseudoelliptic | \(\frac {-\operatorname {arctanh}\left (\frac {4}{\sqrt {-x^{4}+16}}\right ) x^{4}-4 \sqrt {-x^{4}+16}}{256 x^{4}}\) | \(35\) |
meijerg | \(-\frac {\frac {16 \sqrt {\pi }}{x^{4}}-\frac {\left (1-6 \ln \left (2\right )+4 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{2}-\frac {2 \sqrt {\pi }\, \left (-\frac {x^{4}}{4}+8\right )}{x^{4}}+\frac {16 \sqrt {\pi }\, \sqrt {1-\frac {x^{4}}{16}}}{x^{4}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {x^{4}}{16}}}{2}\right )}{256 \sqrt {\pi }}\) | \(84\) |
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Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=-\frac {x^{4} \log \left (\sqrt {-x^{4} + 16} + 4\right ) - x^{4} \log \left (\sqrt {-x^{4} + 16} - 4\right ) + 8 \, \sqrt {-x^{4} + 16}}{512 \, x^{4}} \]
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Result contains complex when optimal does not.
Time = 1.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.92 \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=\begin {cases} - \frac {\operatorname {acosh}{\left (\frac {4}{x^{2}} \right )}}{256} + \frac {1}{64 x^{2} \sqrt {-1 + \frac {16}{x^{4}}}} - \frac {1}{4 x^{6} \sqrt {-1 + \frac {16}{x^{4}}}} & \text {for}\: \frac {1}{\left |{x^{4}}\right |} > \frac {1}{16} \\\frac {i \operatorname {asin}{\left (\frac {4}{x^{2}} \right )}}{256} - \frac {i \sqrt {1 - \frac {16}{x^{4}}}}{64 x^{2}} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=-\frac {\sqrt {-x^{4} + 16}}{64 \, x^{4}} - \frac {1}{512} \, \log \left (\sqrt {-x^{4} + 16} + 4\right ) + \frac {1}{512} \, \log \left (\sqrt {-x^{4} + 16} - 4\right ) \]
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Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=-\frac {\sqrt {-x^{4} + 16}}{64 \, x^{4}} - \frac {1}{512} \, \log \left (\sqrt {-x^{4} + 16} + 4\right ) + \frac {1}{512} \, \log \left (-\sqrt {-x^{4} + 16} + 4\right ) \]
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Time = 5.49 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^5 \sqrt {16-x^4}} \, dx=-\frac {\mathrm {atanh}\left (\frac {\sqrt {16-x^4}}{4}\right )}{256}-\frac {\sqrt {16-x^4}}{64\,x^4} \]
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