Integrand size = 22, antiderivative size = 85 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{-\frac {1}{2}+n} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-n,\frac {1}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^n}{d} \]
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Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3586, 3604, 72, 71} \[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{n-\frac {1}{2}} \cos (c+d x) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-n,\frac {1}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d} \]
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \left (\cos (c+d x) \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{-\frac {1}{2}+n}}{\sqrt {a-i a \tan (c+d x)}} \, dx \\ & = \frac {\left (a^2 \cos (c+d x) \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-\frac {3}{2}+n}}{(a-i a x)^{3/2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{-\frac {3}{2}+n} a \cos (c+d x) \sqrt {a-i a \tan (c+d x)} (a+i a \tan (c+d x))^n \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {1}{2}-n}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {3}{2}+n}}{(a-i a x)^{3/2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {i 2^{-\frac {1}{2}+n} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-n,\frac {1}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^n}{d} \\ \end{align*}
Time = 13.43 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.72 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{-1+n} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-1+n} \left (1+e^{2 i (c+d x)}\right )^{-1+n} \operatorname {Hypergeometric2F1}\left (-1+n,-\frac {1}{2}+n,\frac {1}{2}+n,-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (-1+2 n)} \]
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\[\int \cos \left (d x +c \right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
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\[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right ) \,d x } \]
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\[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \cos {\left (c + d x \right )}\, dx \]
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\[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right ) \,d x } \]
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\[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right ) \,d x } \]
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Timed out. \[ \int \cos (c+d x) (a+i a \tan (c+d x))^n \, dx=\int \cos \left (c+d\,x\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]
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