Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}} \]
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Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2689, 2696, 2721, 2720} \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {4 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}}-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}} \]
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Rule 2689
Rule 2696
Rule 2720
Rule 2721
Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{3 b^2} \\ & = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{3 b^2 \sqrt {b \tan (e+f x)}} \\ & = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{3 b^2 \sqrt {b \tan (e+f x)}} \\ & = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.82 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {2 d^2 \left (\csc ^2(e+f x)+2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}} \]
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Result contains complex when optimal does not.
Time = 2.07 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.52
method | result | size |
default | \(-\frac {\sqrt {d \sec \left (f x +e \right )}\, \left (2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \cot \left (f x +e \right )\right ) \sqrt {2}}{3 f \sqrt {b \tan \left (f x +e \right )}\, b^{2}}\) | \(239\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - \sqrt {-2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - \sqrt {2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{3 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}} \]
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\[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {d \sec {\left (e + f x \right )}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {d \sec \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {d \sec \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]
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