Integrand size = 25, antiderivative size = 131 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\frac {4 b^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{21 d^4 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}} \]
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Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2690, 2692, 2696, 2721, 2720} \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\frac {4 b^2 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{21 d^4 f \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}} \]
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Rule 2690
Rule 2692
Rule 2696
Rule 2720
Rule 2721
Rubi steps \begin{align*} \text {integral}& = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {b^2 \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx}{7 d^2} \\ & = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2\right ) \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{21 d^4} \\ & = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{21 d^4 \sqrt {b \tan (e+f x)}} \\ & = -\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac {\left (2 b^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{21 d^4 \sqrt {b \tan (e+f x)}} \\ & = \frac {4 b^2 \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)}}{21 d^4 f \sqrt {b \tan (e+f x)}}-\frac {2 b \sqrt {b \tan (e+f x)}}{7 f (d \sec (e+f x))^{7/2}}+\frac {2 b \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.93 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.62 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=-\frac {b \left (1+3 \cos (2 (e+f x))-4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{21 d^2 f (d \sec (e+f x))^{3/2}} \]
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Result contains complex when optimal does not.
Time = 3.12 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.23
method | result | size |
default | \(\frac {\sin \left (f x +e \right ) \left (-2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+3 \sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {2}\, \cos \left (f x +e \right ) \sin \left (f x +e \right )\right ) \sqrt {b \tan \left (f x +e \right )}\, b \sqrt {2}}{21 f \left (\cos \left (f x +e \right )-1\right ) d^{3} \sqrt {d \sec \left (f x +e \right )}\, \left (\cos \left (f x +e \right )+1\right )}\) | \(292\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.89 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\frac {2 \, {\left (\sqrt {-2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2 i \, b d} b {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - {\left (3 \, b \cos \left (f x + e\right )^{4} - b \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}\right )}}{21 \, d^{4} f} \]
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Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \]
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\[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \]
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