Integrand size = 26, antiderivative size = 146 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {5 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}+\frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {5 i}{8 d \sqrt {a+i a \tan (c+d x)}} \]
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Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac {5 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}+\frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i}{8 d \sqrt {a+i a \tan (c+d x)}} \]
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Rule 44
Rule 53
Rule 65
Rule 212
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac {\left (5 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d} \\ & = \frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac {(5 i a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d} \\ & = \frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {5 i}{8 d \sqrt {a+i a \tan (c+d x)}}-\frac {(5 i) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{16 d} \\ & = \frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {5 i}{8 d \sqrt {a+i a \tan (c+d x)}}-\frac {(5 i) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{8 d} \\ & = -\frac {5 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}+\frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {5 i}{8 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 = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.35 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{6 d (a+i a \tan (c+d x))^{3/2}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (114 ) = 228\).
Time = 10.67 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.78
method | result | size |
default | \(-\frac {4 i \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-20 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-15 i \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-30 i \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-20 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-15 i \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-30 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+15 \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )}{48 d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(406\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (105) = 210\).
Time = 0.25 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 11 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{48 \, a d} \]
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\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
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Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (15 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 20 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}\right )}}{96 \, a d} \]
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\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]
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