Integrand size = 26, antiderivative size = 91 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}}-\frac {3 \tan (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}}+\frac {\tan ^3(e+f x)}{4 f \sqrt {a \cos ^2(e+f x)}} \]
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Time = 0.16 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3255, 3286, 2691, 3855} \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {3 \cos (e+f x) \text {arctanh}(\sin (e+f x))}{8 f \sqrt {a \cos ^2(e+f x)}}+\frac {\tan ^3(e+f x)}{4 f \sqrt {a \cos ^2(e+f x)}}-\frac {3 \tan (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}} \]
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Rule 2691
Rule 3255
Rule 3286
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \int \frac {\tan ^4(e+f x)}{\sqrt {a \cos ^2(e+f x)}} \, dx \\ & = \frac {\cos (e+f x) \int \sec (e+f x) \tan ^4(e+f x) \, dx}{\sqrt {a \cos ^2(e+f x)}} \\ & = \frac {\tan ^3(e+f x)}{4 f \sqrt {a \cos ^2(e+f x)}}-\frac {(3 \cos (e+f x)) \int \sec (e+f x) \tan ^2(e+f x) \, dx}{4 \sqrt {a \cos ^2(e+f x)}} \\ & = -\frac {3 \tan (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}}+\frac {\tan ^3(e+f x)}{4 f \sqrt {a \cos ^2(e+f x)}}+\frac {(3 \cos (e+f x)) \int \sec (e+f x) \, dx}{8 \sqrt {a \cos ^2(e+f x)}} \\ & = \frac {3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}}-\frac {3 \tan (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}}+\frac {\tan ^3(e+f x)}{4 f \sqrt {a \cos ^2(e+f x)}} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)+\tan (e+f x) \left (3-6 \sec ^2(e+f x)+8 \tan ^2(e+f x)\right )}{8 f \sqrt {a \cos ^2(e+f x)}} \]
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Time = 1.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.45
method | result | size |
default | \(-\frac {\sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (-3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}+2 a}{\cos \left (f x +e \right )}\right ) a \left (\cos ^{4}\left (f x +e \right )\right )+5 \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {a}-2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{8 \cos \left (f x +e \right )^{3} a^{\frac {3}{2}} \sin \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\) | \(132\) |
risch | \(\frac {i \left (5 \,{\mathrm e}^{6 i \left (f x +e \right )}-3 \,{\mathrm e}^{4 i \left (f x +e \right )}+3 \,{\mathrm e}^{2 i \left (f x +e \right )}-5\right )}{4 \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} f}-\frac {3 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}+\frac {3 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}\) | \(188\) |
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Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {{\left (3 \, \cos \left (f x + e\right )^{4} \log \left (-\frac {\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \, {\left (5 \, \cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{16 \, a f \cos \left (f x + e\right )^{5}} \]
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\[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 1518 vs. \(2 (79) = 158\).
Time = 0.55 (sec) , antiderivative size = 1518, normalized size of antiderivative = 16.68 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (79) = 158\).
Time = 1.46 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.07 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\frac {3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} - \frac {3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} - \frac {4 \, {\left (3 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} - \frac {20}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - 20 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left ({\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} - 4\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}}{16 \, \sqrt {a} f} \]
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Timed out. \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2}} \,d x \]
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