Integrand size = 13, antiderivative size = 44 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{2} \sin (5 x)+\frac {1}{6} \sin ^3(5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x) \]
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Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2672, 294, 308, 212} \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{6} \sin ^3(5 x)+\frac {1}{2} \sin (5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x) \]
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Rule 212
Rule 294
Rule 308
Rule 2672
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \text {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (5 x)\right ) \\ & = \frac {1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac {1}{2} \text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (5 x)\right ) \\ & = \frac {1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac {1}{2} \text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (5 x)\right ) \\ & = \frac {1}{2} \sin (5 x)+\frac {1}{6} \sin ^3(5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x)-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (5 x)\right ) \\ & = -\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{2} \sin (5 x)+\frac {1}{6} \sin ^3(5 x)+\frac {1}{10} \sin ^3(5 x) \tan ^2(5 x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.18 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {1}{2} \text {arctanh}(\sin (5 x))+\frac {1}{2} \sec (5 x) \tan (5 x)-\frac {1}{3} \sin (5 x) \tan ^2(5 x)-\frac {1}{15} \sin ^3(5 x) \tan ^2(5 x) \]
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Time = 3.52 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {\sin \left (5 x \right )^{7}}{10 \cos \left (5 x \right )^{2}}+\frac {\sin \left (5 x \right )^{5}}{10}+\frac {\sin \left (5 x \right )^{3}}{6}+\frac {\sin \left (5 x \right )}{2}-\frac {\ln \left (\sec \left (5 x \right )+\tan \left (5 x \right )\right )}{2}\) | \(50\) |
default | \(\frac {\sin \left (5 x \right )^{7}}{10 \cos \left (5 x \right )^{2}}+\frac {\sin \left (5 x \right )^{5}}{10}+\frac {\sin \left (5 x \right )^{3}}{6}+\frac {\sin \left (5 x \right )}{2}-\frac {\ln \left (\sec \left (5 x \right )+\tan \left (5 x \right )\right )}{2}\) | \(50\) |
risch | \(\frac {i {\mathrm e}^{15 i x}}{120}-\frac {9 i {\mathrm e}^{5 i x}}{40}+\frac {9 i {\mathrm e}^{-5 i x}}{40}-\frac {i {\mathrm e}^{-15 i x}}{120}-\frac {i \left ({\mathrm e}^{15 i x}-{\mathrm e}^{5 i x}\right )}{5 \left ({\mathrm e}^{10 i x}+1\right )^{2}}-\frac {\ln \left (i+{\mathrm e}^{5 i x}\right )}{2}+\frac {\ln \left ({\mathrm e}^{5 i x}-i\right )}{2}\) | \(81\) |
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Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=-\frac {15 \, \cos \left (5 \, x\right )^{2} \log \left (\sin \left (5 \, x\right ) + 1\right ) - 15 \, \cos \left (5 \, x\right )^{2} \log \left (-\sin \left (5 \, x\right ) + 1\right ) + 2 \, {\left (2 \, \cos \left (5 \, x\right )^{4} - 14 \, \cos \left (5 \, x\right )^{2} - 3\right )} \sin \left (5 \, x\right )}{60 \, \cos \left (5 \, x\right )^{2}} \]
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Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {\log {\left (\sin {\left (5 x \right )} - 1 \right )}}{4} - \frac {\log {\left (\sin {\left (5 x \right )} + 1 \right )}}{4} + \frac {\sin ^{3}{\left (5 x \right )}}{15} + \frac {2 \sin {\left (5 x \right )}}{5} - \frac {\sin {\left (5 x \right )}}{5 \cdot \left (2 \sin ^{2}{\left (5 x \right )} - 2\right )} \]
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Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {1}{15} \, \sin \left (5 \, x\right )^{3} - \frac {\sin \left (5 \, x\right )}{10 \, {\left (\sin \left (5 \, x\right )^{2} - 1\right )}} - \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) + 1\right ) + \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) - 1\right ) + \frac {2}{5} \, \sin \left (5 \, x\right ) \]
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Time = 0.37 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {1}{15} \, \sin \left (5 \, x\right )^{3} - \frac {\sin \left (5 \, x\right )}{10 \, {\left (\sin \left (5 \, x\right )^{2} - 1\right )}} - \frac {1}{4} \, \log \left (\sin \left (5 \, x\right ) + 1\right ) + \frac {1}{4} \, \log \left (-\sin \left (5 \, x\right ) + 1\right ) + \frac {2}{5} \, \sin \left (5 \, x\right ) \]
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Time = 26.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.57 \[ \int \sin ^3(5 x) \tan ^3(5 x) \, dx=\frac {5\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^9+\frac {20\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^7}{3}-\frac {22\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^5}{3}+\frac {20\,{\mathrm {tan}\left (\frac {5\,x}{2}\right )}^3}{3}+5\,\mathrm {tan}\left (\frac {5\,x}{2}\right )}{5\,{\left ({\mathrm {tan}\left (\frac {5\,x}{2}\right )}^2-1\right )}^2\,{\left ({\mathrm {tan}\left (\frac {5\,x}{2}\right )}^2+1\right )}^3}-\mathrm {atanh}\left (\mathrm {tan}\left (\frac {5\,x}{2}\right )\right ) \]
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