Integrand size = 30, antiderivative size = 163 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\frac {486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f} \]
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Time = 0.38 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3575, 3574} \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\frac {486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {81 i a^3 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f} \]
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Rule 3574
Rule 3575
Rubi steps \begin{align*} \text {integral}& = \frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac {1}{5} (9 a) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx \\ & = \frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac {1}{35} \left (108 a^2\right ) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx \\ & = \frac {81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac {1}{35} \left (162 a^3\right ) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3} \, dx \\ & = \frac {486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f} \\ \end{align*}
Time = 2.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\frac {3 a^3 (d \sec (e+f x))^{5/3} (i \cos (e-2 f x)+\sin (e-2 f x)) (364+442 \cos (2 (e+f x))+59 i \sec (e+f x) \sin (3 (e+f x))+45 i \tan (e+f x)) (a+i a \tan (e+f x))^{2/3}}{140 d f (\cos (f x)+i \sin (f x))^3} \]
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\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{\frac {11}{3}}d x\]
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none
Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.82 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=-\frac {6 \cdot 2^{\frac {1}{3}} {\left (-140 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - 315 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 270 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 81 i \, a^{3}\right )} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}}{35 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]
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Timed out. \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\text {Timed out} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 983 vs. \(2 (123) = 246\).
Time = 0.40 (sec) , antiderivative size = 983, normalized size of antiderivative = 6.03 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\text {Too large to display} \]
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\[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {11}{3}} \,d x } \]
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Time = 8.95 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.86 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\frac {{\left (-\frac {d}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,243{}\mathrm {i}}{35\,f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+1\right )\,162{}\mathrm {i}}{7\,f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,27{}\mathrm {i}}{f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (3\,e+3\,f\,x\right )}^2+\sin \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}+1\right )\,12{}\mathrm {i}}{f}\right )}{4\,\left ({\sin \left (e+f\,x\right )}^2-1\right )} \]
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