Integrand size = 9, antiderivative size = 26 \[ \int \cos (4 x) \sec ^5(x) \, dx=\frac {35}{8} \text {arctanh}(\sin (x))-\frac {29}{8} \sec (x) \tan (x)+\frac {1}{4} \sec ^3(x) \tan (x) \]
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Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4449, 1171, 393, 212} \[ \int \cos (4 x) \sec ^5(x) \, dx=\frac {35}{8} \text {arctanh}(\sin (x))+\frac {1}{4} \tan (x) \sec ^3(x)-\frac {29}{8} \tan (x) \sec (x) \]
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Rule 212
Rule 393
Rule 1171
Rule 4449
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1-8 x^2+8 x^4}{\left (1-x^2\right )^3} \, dx,x,\sin (x)\right ) \\ & = \frac {1}{4} \sec ^3(x) \tan (x)-\frac {1}{4} \text {Subst}\left (\int \frac {-3+32 x^2}{\left (1-x^2\right )^2} \, dx,x,\sin (x)\right ) \\ & = -\frac {29}{8} \sec (x) \tan (x)+\frac {1}{4} \sec ^3(x) \tan (x)+\frac {35}{8} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (x)\right ) \\ & = \frac {35}{8} \text {arctanh}(\sin (x))-\frac {29}{8} \sec (x) \tan (x)+\frac {1}{4} \sec ^3(x) \tan (x) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \cos (4 x) \sec ^5(x) \, dx=\frac {1}{8} \left (35 \text {arctanh}(\sin (x))-27 \sec ^3(x) \tan (x)+29 \sec (x) \tan ^3(x)\right ) \]
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Time = 38.58 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19
method | result | size |
default | \(-\left (-\frac {\left (\sec ^{3}\left (x \right )\right )}{4}-\frac {3 \sec \left (x \right )}{8}\right ) \tan \left (x \right )+\frac {35 \ln \left (\sec \left (x \right )+\tan \left (x \right )\right )}{8}-4 \sec \left (x \right ) \tan \left (x \right )\) | \(31\) |
risch | \(\frac {i \left (29 \,{\mathrm e}^{7 i x}+21 \,{\mathrm e}^{5 i x}-21 \,{\mathrm e}^{3 i x}-29 \,{\mathrm e}^{i x}\right )}{4 \left ({\mathrm e}^{2 i x}+1\right )^{4}}+\frac {35 \ln \left (i+{\mathrm e}^{i x}\right )}{8}-\frac {35 \ln \left ({\mathrm e}^{i x}-i\right )}{8}\) | \(65\) |
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (20) = 40\).
Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \cos (4 x) \sec ^5(x) \, dx=\frac {35 \, \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - 35 \, \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, {\left (29 \, \cos \left (x\right )^{2} - 2\right )} \sin \left (x\right )}{16 \, \cos \left (x\right )^{4}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (27) = 54\).
Time = 6.98 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \cos (4 x) \sec ^5(x) \, dx=- \frac {35 \log {\left (\sin {\left (x \right )} - 1 \right )}}{16} + \frac {35 \log {\left (\sin {\left (x \right )} + 1 \right )}}{16} - \frac {3 \sin ^{3}{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} + \frac {5 \sin {\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} + \frac {8 \sin {\left (x \right )}}{2 \sin ^{2}{\left (x \right )} - 2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (20) = 40\).
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \cos (4 x) \sec ^5(x) \, dx=\frac {5 \, \sin \left (x\right )^{3} - 3 \, \sin \left (x\right )}{8 \, {\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac {3 \, \sin \left (x\right )}{\sin \left (x\right )^{2} - 1} + \frac {35}{16} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac {35}{16} \, \log \left (\sin \left (x\right ) - 1\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \cos (4 x) \sec ^5(x) \, dx=\frac {29 \, \sin \left (x\right )^{3} - 27 \, \sin \left (x\right )}{8 \, {\left (\sin \left (x\right )^{2} - 1\right )}^{2}} + \frac {35}{16} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac {35}{16} \, \log \left (-\sin \left (x\right ) + 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \cos (4 x) \sec ^5(x) \, dx=\frac {35\,\mathrm {atanh}\left (\sin \left (x\right )\right )}{8}-\frac {\frac {27\,\sin \left (x\right )}{8}-\frac {29\,{\sin \left (x\right )}^3}{8}}{{\sin \left (x\right )}^4-2\,{\sin \left (x\right )}^2+1} \]
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