Integrand size = 10, antiderivative size = 13 \[ \int x \sec \left (5-x^2\right ) \, dx=-\frac {1}{2} \text {arctanh}\left (\sin \left (5-x^2\right )\right ) \]
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Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4289, 3855} \[ \int x \sec \left (5-x^2\right ) \, dx=-\frac {1}{2} \text {arctanh}\left (\sin \left (5-x^2\right )\right ) \]
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Rule 3855
Rule 4289
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \sec (5-x) \, dx,x,x^2\right ) \\ & = -\frac {1}{2} \text {arctanh}\left (\sin \left (5-x^2\right )\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int x \sec \left (5-x^2\right ) \, dx=-\frac {1}{2} \text {arctanh}\left (\sin \left (5-x^2\right )\right ) \]
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Time = 0.34 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31
method | result | size |
derivativedivides | \(\frac {\ln \left (\sec \left (x^{2}-5\right )+\tan \left (x^{2}-5\right )\right )}{2}\) | \(17\) |
default | \(\frac {\ln \left (\sec \left (x^{2}-5\right )+\tan \left (x^{2}-5\right )\right )}{2}\) | \(17\) |
norman | \(-\frac {\ln \left (\tan \left (-\frac {5}{2}+\frac {x^{2}}{2}\right )-1\right )}{2}+\frac {\ln \left (\tan \left (-\frac {5}{2}+\frac {x^{2}}{2}\right )+1\right )}{2}\) | \(28\) |
parallelrisch | \(\ln \left (\frac {1}{\sqrt {\tan \left (-\frac {5}{2}+\frac {x^{2}}{2}\right )-1}}\right )+\ln \left (\sqrt {\tan \left (-\frac {5}{2}+\frac {x^{2}}{2}\right )+1}\right )\) | \(28\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{i \left (x^{2}-5\right )}-i\right )}{2}+\frac {\ln \left (i+{\mathrm e}^{i \left (x^{2}-5\right )}\right )}{2}\) | \(32\) |
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Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (9) = 18\).
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.92 \[ \int x \sec \left (5-x^2\right ) \, dx=\frac {1}{4} \, \log \left (\sin \left (x^{2} - 5\right ) + 1\right ) - \frac {1}{4} \, \log \left (-\sin \left (x^{2} - 5\right ) + 1\right ) \]
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Time = 7.64 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int x \sec \left (5-x^2\right ) \, dx=\frac {\log {\left (\tan {\left (x^{2} - 5 \right )} + \sec {\left (x^{2} - 5 \right )} \right )}}{2} \]
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none
Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int x \sec \left (5-x^2\right ) \, dx=\frac {1}{2} \, \log \left (\sec \left (x^{2} - 5\right ) + \tan \left (x^{2} - 5\right )\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (9) = 18\).
Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 3.15 \[ \int x \sec \left (5-x^2\right ) \, dx=\frac {1}{8} \, \log \left ({\left | \frac {1}{\sin \left (x^{2} - 5\right )} + \sin \left (x^{2} - 5\right ) + 2 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | \frac {1}{\sin \left (x^{2} - 5\right )} + \sin \left (x^{2} - 5\right ) - 2 \right |}\right ) \]
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Time = 26.88 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int x \sec \left (5-x^2\right ) \, dx=-\mathrm {atan}\left ({\mathrm {e}}^{-5{}\mathrm {i}}\,{\mathrm {e}}^{x^2\,1{}\mathrm {i}}\right )\,1{}\mathrm {i} \]
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