Integrand size = 26, antiderivative size = 120 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\frac {15 \text {arctanh}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{8 f}-\frac {15 \sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{8 f}-\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f} \]
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Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3255, 3286, 2672, 294, 327, 212} \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\frac {15 \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \text {arctanh}(\sin (e+f x))}{8 f}+\frac {\tan ^5(e+f x) \sqrt {a \cos ^2(e+f x)}}{4 f}-\frac {5 \tan ^3(e+f x) \sqrt {a \cos ^2(e+f x)}}{8 f}-\frac {15 \tan (e+f x) \sqrt {a \cos ^2(e+f x)}}{8 f} \]
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Rule 212
Rule 294
Rule 327
Rule 2672
Rule 3255
Rule 3286
Rubi steps \begin{align*} \text {integral}& = \int \sqrt {a \cos ^2(e+f x)} \tan ^6(e+f x) \, dx \\ & = \left (\sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \int \sin (e+f x) \tan ^5(e+f x) \, dx \\ & = \frac {\left (\sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f}-\frac {\left (5 \sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 f} \\ & = -\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f}+\frac {\left (15 \sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (e+f x)\right )}{8 f} \\ & = -\frac {15 \sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{8 f}-\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f}+\frac {\left (15 \sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{8 f} \\ & = \frac {15 \text {arctanh}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{8 f}-\frac {15 \sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{8 f}-\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f} \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.62 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\frac {\sqrt {a \cos ^2(e+f x)} \sec ^5(e+f x) \left (60 \text {arctanh}(\sin (e+f x)) \cos ^4(e+f x)-5 \sin (e+f x)-15 \sin (3 (e+f x))-2 \sin (5 (e+f x))\right )}{32 f} \]
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Time = 1.05 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.31
method | result | size |
default | \(-\frac {\sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (8 \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {a}-15 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}+2 a}{\cos \left (f x +e \right )}\right ) a \left (\cos ^{4}\left (f x +e \right )\right )+9 \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {a}-2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{8 \cos \left (f x +e \right )^{3} \sqrt {a}\, \sin \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\) | \(157\) |
risch | \(\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}+\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (9 \,{\mathrm e}^{8 i \left (f x +e \right )}+{\mathrm e}^{6 i \left (f x +e \right )}-{\mathrm e}^{4 i \left (f x +e \right )}-9 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{4 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}-\frac {15 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{8 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {15 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{8 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(327\) |
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Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=-\frac {{\left (15 \, \cos \left (f x + e\right )^{4} \log \left (-\frac {\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \, {\left (8 \, \cos \left (f x + e\right )^{4} + 9 \, \cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{16 \, f \cos \left (f x + e\right )^{5}} \]
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\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{6}{\left (e + f x \right )}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 1955 vs. \(2 (104) = 208\).
Time = 1.13 (sec) , antiderivative size = 1955, normalized size of antiderivative = 16.29 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (104) = 208\).
Time = 2.31 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.95 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=-\frac {{\left (15 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 15 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \frac {32 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - \frac {4 \, {\left (7 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 36 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )\right )}}{{\left ({\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} - 4\right )}^{2}}\right )} \sqrt {a}}{16 \, f} \]
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Timed out. \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^6\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \]
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