Integrand size = 26, antiderivative size = 147 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {256 i a^4 \sec ^{13}(c+d x)}{20995 d (a+i a \tan (c+d x))^{13/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}} \]
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Time = 0.34 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {256 i a^4 \sec ^{13}(c+d x)}{20995 d (a+i a \tan (c+d x))^{13/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}} \]
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Rule 3574
Rule 3575
Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}+\frac {1}{19} (12 a) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx \\ & = \frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}+\frac {1}{323} \left (96 a^2\right ) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{9/2}} \, dx \\ & = \frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}+\frac {\left (128 a^3\right ) \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{11/2}} \, dx}{1615} \\ & = \frac {256 i a^4 \sec ^{13}(c+d x)}{20995 d (a+i a \tan (c+d x))^{13/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}} \\ \end{align*}
Time = 1.99 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^{12}(c+d x) (798 \cos (c+d x)+1631 \cos (3 (c+d x))+13 i (38 \sin (c+d x)+123 \sin (3 (c+d x)))) (-2 i \cos (4 (c+d x))-2 \sin (4 (c+d x)))}{20995 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
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Timed out.
\[\int \frac {\sec ^{13}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]
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none
Time = 0.36 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {1024 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1615 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 646 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 152 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{20995 \, {\left (a^{3} d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]
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Timed out. \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 902 vs. \(2 (115) = 230\).
Time = 1.05 (sec) , antiderivative size = 902, normalized size of antiderivative = 6.14 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
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\[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{13}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Time = 12.75 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{13\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{5\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,3072{}\mathrm {i}}{17\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{19\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^9} \]
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