Integrand size = 26, antiderivative size = 105 \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{-\frac {m}{2}+n} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {1}{2} (2+m-2 n),1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{2} (m-2 n)} (a+i a \tan (c+d x))^n}{d m} \]
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Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3596, 3586, 3604, 72, 71} \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{n-\frac {m}{2}} (a+i a \tan (c+d x))^n (e \cos (c+d x))^m (1+i \tan (c+d x))^{\frac {1}{2} (m-2 n)} \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {1}{2} (m-2 n+2),1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m} \]
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Rule 71
Rule 72
Rule 3586
Rule 3596
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \left ((e \cos (c+d x))^m (e \sec (c+d x))^m\right ) \int (e \sec (c+d x))^{-m} (a+i a \tan (c+d x))^n \, dx \\ & = \left ((e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \int (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-\frac {m}{2}+n} \, dx \\ & = \frac {\left (a^2 (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \text {Subst}\left (\int (a-i a x)^{-1-\frac {m}{2}} (a+i a x)^{-1-\frac {m}{2}+n} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{-1-\frac {m}{2}+n} a (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^n \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {m}{2}-n}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-1-\frac {m}{2}+n} (a-i a x)^{-1-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {i 2^{-\frac {m}{2}+n} (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {1}{2} (2+m-2 n),1-\frac {m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{2} (m-2 n)} (a+i a \tan (c+d x))^n}{d m} \\ \end{align*}
Time = 14.50 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.92 \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{-m+n} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^{-m+n} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \cos ^{-m}(c+d x) (e \cos (c+d x))^m \operatorname {Hypergeometric2F1}\left (-m+n,-\frac {m}{2}+n,1-\frac {m}{2}+n,-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (m-2 n)} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
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\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{m} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{m} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]
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\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{m} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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\[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{m} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^n \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]
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