Integrand size = 14, antiderivative size = 183 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}} \]
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Time = 0.07 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}} \]
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Rule 8
Rule 3554
Rule 3739
Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^2(e+f x) \int \cot ^{10}(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}} \\ & = -\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan ^2(e+f x) \int \cot ^8(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}} \\ & = \frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\tan ^2(e+f x) \int \cot ^6(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}} \\ & = -\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan ^2(e+f x) \int \cot ^4(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}} \\ & = \frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\tan ^2(e+f x) \int \cot ^2(e+f x) \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}} \\ & = \frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan ^2(e+f x) \int 1 \, dx}{b^2 \sqrt {b \tan ^4(e+f x)}} \\ & = \frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.25 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {9}{2},1,-\frac {7}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{9 f \left (b \tan ^4(e+f x)\right )^{5/2}} \]
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Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45
method | result | size |
derivativedivides | \(-\frac {\tan \left (f x +e \right ) \left (315 \arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{9}+315 \tan \left (f x +e \right )^{8}-105 \tan \left (f x +e \right )^{6}+63 \tan \left (f x +e \right )^{4}-45 \tan \left (f x +e \right )^{2}+35\right )}{315 f \left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}}}\) | \(83\) |
default | \(-\frac {\tan \left (f x +e \right ) \left (315 \arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{9}+315 \tan \left (f x +e \right )^{8}-105 \tan \left (f x +e \right )^{6}+63 \tan \left (f x +e \right )^{4}-45 \tan \left (f x +e \right )^{2}+35\right )}{315 f \left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}}}\) | \(83\) |
risch | \(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} x}{b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}}+\frac {2 i \left (1575 \,{\mathrm e}^{16 i \left (f x +e \right )}-6300 \,{\mathrm e}^{14 i \left (f x +e \right )}+21000 \,{\mathrm e}^{12 i \left (f x +e \right )}-31500 \,{\mathrm e}^{10 i \left (f x +e \right )}+39438 \,{\mathrm e}^{8 i \left (f x +e \right )}-26292 \,{\mathrm e}^{6 i \left (f x +e \right )}+13968 \,{\mathrm e}^{4 i \left (f x +e \right )}-3492 \,{\mathrm e}^{2 i \left (f x +e \right )}+563\right )}{315 b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{7} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, f}\) | \(218\) |
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Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (315 \, f x \tan \left (f x + e\right )^{9} + 315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{315 \, b^{3} f \tan \left (f x + e\right )^{11}} \]
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\[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {315 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (f x + e\right )^{9}}}{315 \, f} \]
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Time = 1.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {161280 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {121590 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 18480 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3528 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 495 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} - \frac {35 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 495 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 3528 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18480 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 121590 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{b^{\frac {45}{2}}}}{161280 \, f} \]
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Timed out. \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{5/2}} \,d x \]
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