Integrand size = 24, antiderivative size = 165 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {15 x}{64 a^2}-\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )} \]
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Time = 0.15 (sec) , antiderivative size = 165, 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 = {3568, 46, 212} \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {15 x}{64 a^2}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}-\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2} \]
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Rule 46
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)^5} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \left (\frac {1}{32 a^5 (a-x)^3}+\frac {5}{64 a^6 (a-x)^2}+\frac {1}{8 a^3 (a+x)^5}+\frac {3}{16 a^4 (a+x)^4}+\frac {3}{16 a^5 (a+x)^3}+\frac {5}{32 a^6 (a+x)^2}+\frac {15}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {(15 i) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,i a \tan (c+d x)\right )}{64 a d} \\ & = \frac {15 x}{64 a^2}-\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i \sec ^6(c+d x) (-80-65 \cos (2 (c+d x))+16 \cos (4 (c+d x))+\cos (6 (c+d x))+120 i \arctan (\tan (c+d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-5 i \sin (2 (c+d x))+32 i \sin (4 (c+d x))+3 i \sin (6 (c+d x)))}{512 a^2 d (-i+\tan (c+d x))^4 (i+\tan (c+d x))^2} \]
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Time = 0.75 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.69
method | result | size |
risch | \(\frac {15 x}{64 a^{2}}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{64 a^{2} d}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{512 a^{2} d}+\frac {7 i \cos \left (4 d x +4 c \right )}{128 a^{2} d}+\frac {\sin \left (4 d x +4 c \right )}{16 a^{2} d}+\frac {7 i \cos \left (2 d x +2 c \right )}{64 a^{2} d}+\frac {13 \sin \left (2 d x +2 c \right )}{64 a^{2} d}\) | \(114\) |
derivativedivides | \(\frac {-\frac {15 i \ln \left (\tan \left (d x +c \right )-i\right )}{128}+\frac {i}{32 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{16 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5}{32 \left (\tan \left (d x +c \right )-i\right )}+\frac {i}{64 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {15 i \ln \left (\tan \left (d x +c \right )+i\right )}{128}+\frac {5}{64 \left (\tan \left (d x +c \right )+i\right )}}{d \,a^{2}}\) | \(116\) |
default | \(\frac {-\frac {15 i \ln \left (\tan \left (d x +c \right )-i\right )}{128}+\frac {i}{32 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{16 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5}{32 \left (\tan \left (d x +c \right )-i\right )}+\frac {i}{64 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {15 i \ln \left (\tan \left (d x +c \right )+i\right )}{128}+\frac {5}{64 \left (\tan \left (d x +c \right )+i\right )}}{d \,a^{2}}\) | \(116\) |
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Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (120 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 2 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 24 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 80 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 30 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{512 \, a^{2} d} \]
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Time = 0.30 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.56 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 17179869184 i a^{10} d^{5} e^{24 i c} e^{4 i d x} - 206158430208 i a^{10} d^{5} e^{22 i c} e^{2 i d x} + 687194767360 i a^{10} d^{5} e^{18 i c} e^{- 2 i d x} + 257698037760 i a^{10} d^{5} e^{16 i c} e^{- 4 i d x} + 68719476736 i a^{10} d^{5} e^{14 i c} e^{- 6 i d x} + 8589934592 i a^{10} d^{5} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{4398046511104 a^{12} d^{6}} & \text {for}\: a^{12} d^{6} e^{20 i c} \neq 0 \\x \left (\frac {\left (e^{12 i c} + 6 e^{10 i c} + 15 e^{8 i c} + 20 e^{6 i c} + 15 e^{4 i c} + 6 e^{2 i c} + 1\right ) e^{- 8 i c}}{64 a^{2}} - \frac {15}{64 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {15 x}{64 a^{2}} \]
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Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.51 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {-\frac {60 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {60 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {2 \, {\left (45 i \, \tan \left (d x + c\right )^{2} - 110 \, \tan \left (d x + c\right ) - 69 i\right )}}{a^{2} {\left (\tan \left (d x + c\right ) + i\right )}^{2}} + \frac {-125 i \, \tan \left (d x + c\right )^{4} - 580 \, \tan \left (d x + c\right )^{3} + 1038 i \, \tan \left (d x + c\right )^{2} + 868 \, \tan \left (d x + c\right ) - 301 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{512 \, d} \]
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Time = 5.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {15\,x}{64\,a^2}+\frac {\frac {1}{4\,a^2}-\frac {\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{64\,a^2}+\frac {25\,{\mathrm {tan}\left (c+d\,x\right )}^2}{32\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,5{}\mathrm {i}}{32\,a^2}+\frac {15\,{\mathrm {tan}\left (c+d\,x\right )}^4}{32\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,15{}\mathrm {i}}{64\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^5+{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}+4\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]
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