Integrand size = 23, antiderivative size = 171 \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=\frac {(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac {i 2^{-2-m} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{a^2 f}+\frac {i 4^{-2-m} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i f (c+d x)}{d}\right )}{a^2 f} \]
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Time = 0.23 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3810, 2212} \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=\frac {i 2^{-m-2} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{a^2 f}+\frac {i 4^{-m-2} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i f (c+d x)}{d}\right )}{a^2 f}+\frac {(c+d x)^{m+1}}{4 a^2 d (m+1)} \]
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Rule 2212
Rule 3810
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(c+d x)^m}{4 a^2}+\frac {e^{-2 i e-2 i f x} (c+d x)^m}{2 a^2}+\frac {e^{-4 i e-4 i f x} (c+d x)^m}{4 a^2}\right ) \, dx \\ & = \frac {(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac {\int e^{-4 i e-4 i f x} (c+d x)^m \, dx}{4 a^2}+\frac {\int e^{-2 i e-2 i f x} (c+d x)^m \, dx}{2 a^2} \\ & = \frac {(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac {i 2^{-2-m} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{a^2 f}+\frac {i 4^{-2-m} e^{-4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i f (c+d x)}{d}\right )}{a^2 f} \\ \end{align*}
Time = 4.49 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=\frac {(c+d x)^m \left (\frac {4 e^{2 i e} f (c+d x)}{d (1+m)}+i 2^{2-m} e^{\frac {2 i c f}{d}} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )+i 4^{-m} e^{-2 i e+\frac {4 i c f}{d}} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i f (c+d x)}{d}\right )\right ) \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2}{16 f (a+i a \tan (e+f x))^2} \]
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\[\int \frac {\left (d x +c \right )^{m}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}d x\]
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Time = 0.09 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.84 \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=-\frac {4 \, {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - {\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {4 i \, f}{d}\right ) + 4 i \, d e - 4 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - 4 \, {\left (d f x + c f\right )} {\left (d x + c\right )}^{m}}{16 \, {\left (a^{2} d f m + a^{2} d f\right )}} \]
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\[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {\left (c + d x\right )^{m}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
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\[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx=\int \frac {{\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]
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