Integrand size = 14, antiderivative size = 51 \[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=-\frac {\tan (e+f x)}{f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{\sqrt {b \tan ^4(e+f x)}} \]
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Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=-\frac {\tan (e+f x)}{f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{\sqrt {b \tan ^4(e+f x)}} \]
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Rule 8
Rule 3554
Rule 3739
Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^2(e+f x) \int \cot ^2(e+f x) \, dx}{\sqrt {b \tan ^4(e+f x)}} \\ & = -\frac {\tan (e+f x)}{f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan ^2(e+f x) \int 1 \, dx}{\sqrt {b \tan ^4(e+f x)}} \\ & = -\frac {\tan (e+f x)}{f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{\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 = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{f \sqrt {b \tan ^4(e+f x)}} \]
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Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(-\frac {\tan \left (f x +e \right ) \left (\arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )+1\right )}{f \sqrt {b \tan \left (f x +e \right )^{4}}}\) | \(40\) |
default | \(-\frac {\tan \left (f x +e \right ) \left (\arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )+1\right )}{f \sqrt {b \tan \left (f x +e \right )^{4}}}\) | \(40\) |
risch | \(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} x}{\sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {2 i \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{\sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} f}\) | \(120\) |
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=-\frac {\sqrt {b \tan \left (f x + e\right )^{4}} {\left (f x \tan \left (f x + e\right ) + 1\right )}}{b f \tan \left (f x + e\right )^{3}} \]
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\[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=\int \frac {1}{\sqrt {b \tan ^{4}{\left (e + f x \right )}}}\, dx \]
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Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=-\frac {\frac {f x + e}{\sqrt {b}} + \frac {1}{\sqrt {b} \tan \left (f x + e\right )}}{f} \]
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Time = 0.38 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=-\frac {\frac {2 \, {\left (f x + e\right )}}{\sqrt {b}} - \frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {b}} + \frac {1}{\sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}}{2 \, f} \]
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Timed out. \[ \int \frac {1}{\sqrt {b \tan ^4(e+f x)}} \, dx=\int \frac {1}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^4}} \,d x \]
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