Integrand size = 24, antiderivative size = 122 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}} \]
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Time = 0.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3583, 3570, 212} \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}} \]
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Rule 212
Rule 3570
Rule 3583
Rubi steps \begin{align*} \text {integral}& = \frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{8 a} \\ & = \frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac {3 \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{32 a^2} \\ & = \frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{2-a x^2} \, dx,x,\frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}}\right )}{16 a^2 d} \\ & = \frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}} \\ \end{align*}
Time = 1.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \sec ^3(c+d x) \left (7+3 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )+7 \cos (2 (c+d x))+3 i \sin (2 (c+d x))\right )}{32 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 605 vs. \(2 (97 ) = 194\).
Time = 9.23 (sec) , antiderivative size = 606, normalized size of antiderivative = 4.97
method | result | size |
default | \(-\frac {i \left (12 i \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )+6 i \tan \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+6 i \tan \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+12 \cos \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-3 i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+6 i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+6 \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+14 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-9 \sec \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+14 \sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-3 \left (\sec ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right )}{32 d \left (\tan \left (d x +c \right )-i\right )^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{2} \left (\cos \left (d x +c \right )+1\right )}\) | \(606\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (91) = 182\).
Time = 0.26 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.19 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (-3 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, a^{2} d}\right ) + 3 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, a^{2} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{3} d} \]
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\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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