Integrand size = 25, antiderivative size = 97 \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \]
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Time = 0.14 (sec) , antiderivative size = 97, 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 = {2688, 2696, 2721, 2719} \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}} \]
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Rule 2688
Rule 2696
Rule 2719
Rule 2721
Rubi steps \begin{align*} \text {integral}& = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {d^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx}{b^2} \\ & = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\left (d^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {b \sin (e+f x)} \, dx}{b^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \\ & = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\left (d^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {\sin (e+f x)} \, dx}{b^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \\ & = -\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (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 = 0.82 \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 d^2 \left (3+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} \tan ^2(e+f x)\right )}{3 b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \]
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Result contains complex when optimal does not.
Time = 1.98 (sec) , antiderivative size = 365, normalized size of antiderivative = 3.76
method | result | size |
default | \(\frac {\csc \left (f x +e \right ) \left (2 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1\right ) {\left (-\frac {d \left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}\right )}^{\frac {3}{2}} \left (1-\cos \left (f x +e \right )\right ) \sqrt {2}}{f {\left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1\right )}^{2} {\left (-\frac {b \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}\right )}^{\frac {3}{2}}}\) | \(365\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=-\frac {2 \, d \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + i \, \sqrt {-2 i \, b d} d \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2 i \, b d} d \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{b^{2} f \sin \left (f x + e\right )} \]
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\[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]
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