Integrand size = 26, antiderivative size = 277 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {143 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} a^{5/2} d}+\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
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Time = 0.21 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {143 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} a^{5/2} d}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}} \]
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Rule 44
Rule 53
Rule 65
Rule 212
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {\left (13 i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d} \\ & = -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac {\left (143 i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d} \\ & = \frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac {\left (143 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{64 d} \\ & = \frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}-\frac {(143 i a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d} \\ & = \frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {(143 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d} \\ & = \frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(143 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{512 a d} \\ & = \frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(143 i) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{1024 a^2 d} \\ & = \frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(143 i) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{512 a^2 d} \\ & = -\frac {143 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} a^{5/2} d}+\frac {143 i a^2}{288 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{9/2}}-\frac {13 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {143 i a}{448 d (a+i a \tan (c+d x))^{7/2}}+\frac {143 i}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i}{768 a d (a+i a \tan (c+d x))^{3/2}}+\frac {143 i}{512 a^2 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.19 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i a^2 \operatorname {Hypergeometric2F1}\left (-\frac {9}{2},3,-\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{36 d (a+i a \tan (c+d x))^{9/2}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 913 vs. \(2 (220 ) = 440\).
Time = 9.16 (sec) , antiderivative size = 914, normalized size of antiderivative = 3.30
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Time = 0.26 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-630 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 8505 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 42709 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 69392 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 26752 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10144 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2480 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 280 i\right )}\right )} e^{\left (-9 i \, d x - 9 i \, c\right )}}{322560 \, a^{3} d} \]
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\[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.44 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \, {\left (\frac {4 \, {\left (45045 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} - 150150 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a + 96096 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{2} + 27456 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{3} + 18304 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} + 16640 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} + 17920 \, a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2} + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{3}} + \frac {45045 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}}\right )}}{645120 \, a d} \]
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\[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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