Integrand size = 11, antiderivative size = 74 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=-x+\frac {7 x}{4 \sqrt {2}}+\frac {7 \arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{4 \sqrt {2}}+\frac {\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac {\tan (x)}{4 \left (1+2 \tan ^2(x)\right )} \]
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Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {425, 541, 536, 209} \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\frac {7 \arctan \left (\frac {\sin (x) \cos (x)}{\sin ^2(x)+\sqrt {2}+1}\right )}{4 \sqrt {2}}+\frac {7 x}{4 \sqrt {2}}-x-\frac {\tan (x)}{4 \left (2 \tan ^2(x)+1\right )}+\frac {\tan (x)}{2 \left (2 \tan ^2(x)+1\right )^2} \]
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Rule 209
Rule 425
Rule 536
Rule 541
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \left (1+2 x^2\right )^3} \, dx,x,\tan (x)\right ) \\ & = \frac {\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac {1}{4} \text {Subst}\left (\int \frac {-2-6 x^2}{\left (1+x^2\right ) \left (1+2 x^2\right )^2} \, dx,x,\tan (x)\right ) \\ & = \frac {\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac {\tan (x)}{4 \left (1+2 \tan ^2(x)\right )}+\frac {1}{8} \text {Subst}\left (\int \frac {6-2 x^2}{\left (1+x^2\right ) \left (1+2 x^2\right )} \, dx,x,\tan (x)\right ) \\ & = \frac {\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac {\tan (x)}{4 \left (1+2 \tan ^2(x)\right )}+\frac {7}{4} \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\tan (x)\right )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right ) \\ & = -x+\frac {7 x}{4 \sqrt {2}}+\frac {7 \arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{4 \sqrt {2}}+\frac {\tan (x)}{2 \left (1+2 \tan ^2(x)\right )^2}-\frac {\tan (x)}{4 \left (1+2 \tan ^2(x)\right )} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=-\frac {(-3+\cos (2 x)) \sec ^6(x) \left (-76 x+7 \sqrt {2} \arctan \left (\sqrt {2} \tan (x)\right ) (-3+\cos (2 x))^2+48 x \cos (2 x)-4 x \cos (4 x)-2 \sin (2 x)+3 \sin (4 x)\right )}{64 \left (\sec ^2(x)+\tan ^2(x)\right )^3} \]
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Time = 14.44 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.03
method | result | size |
parallelrisch | \(0\) | \(2\) |
default | \(-\arctan \left (\tan \left (x \right )\right )+\frac {-\frac {\tan \left (x \right )^{3}}{2}+\frac {\tan \left (x \right )}{4}}{\left (1+2 \tan \left (x \right )^{2}\right )^{2}}+\frac {7 \sqrt {2}\, \arctan \left (\sqrt {2}\, \tan \left (x \right )\right )}{8}\) | \(42\) |
risch | \(-x -\frac {i \left (17 \,{\mathrm e}^{6 i x}-57 \,{\mathrm e}^{4 i x}+19 \,{\mathrm e}^{2 i x}-3\right )}{2 \left ({\mathrm e}^{4 i x}-6 \,{\mathrm e}^{2 i x}+1\right )^{2}}+\frac {7 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}-3\right )}{16}-\frac {7 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}-3\right )}{16}\) | \(85\) |
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Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=-\frac {16 \, x \cos \left (x\right )^{4} - 64 \, x \cos \left (x\right )^{2} + 7 \, {\left (\sqrt {2} \cos \left (x\right )^{4} - 4 \, \sqrt {2} \cos \left (x\right )^{2} + 4 \, \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - 2 \, \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) - 4 \, {\left (3 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )\right )} \sin \left (x\right ) + 64 \, x}{16 \, {\left (\cos \left (x\right )^{4} - 4 \, \cos \left (x\right )^{2} + 4\right )}} \]
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\[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\int \frac {1}{\left (\tan ^{2}{\left (x \right )} + \sec ^{2}{\left (x \right )}\right )^{3}}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\frac {7}{8} \, \sqrt {2} \arctan \left (\sqrt {2} \tan \left (x\right )\right ) - x - \frac {2 \, \tan \left (x\right )^{3} - \tan \left (x\right )}{4 \, {\left (4 \, \tan \left (x\right )^{4} + 4 \, \tan \left (x\right )^{2} + 1\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\frac {7}{8} \, \sqrt {2} \arctan \left (\sqrt {2} \tan \left (x\right )\right ) - x - \frac {2 \, \tan \left (x\right )^{3} - \tan \left (x\right )}{4 \, {\left (2 \, \tan \left (x\right )^{2} + 1\right )}^{2}} \]
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Time = 28.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (\sec ^2(x)+\tan ^2(x)\right )^3} \, dx=\frac {\frac {\mathrm {tan}\left (x\right )}{16}-\frac {{\mathrm {tan}\left (x\right )}^3}{8}}{{\mathrm {tan}\left (x\right )}^4+{\mathrm {tan}\left (x\right )}^2+\frac {1}{4}}-x+\frac {7\,\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\right )}{8} \]
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