Integrand size = 30, antiderivative size = 97 \[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{\frac {5}{2}-n} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (-3+2 n),\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (e \sec (c+d x))^{5-2 n} (1-i \tan (c+d x))^{-\frac {5}{2}+n} (a+i a \tan (c+d x))^n}{5 d} \]
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Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3586, 3604, 7, 72, 71} \[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{\frac {5}{2}-n} (1-i \tan (c+d x))^{n-\frac {5}{2}} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{5-2 n} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (2 n-3),\frac {7}{2},\frac {1}{2} (i \tan (c+d x)+1)\right )}{5 d} \]
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Rule 7
Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \left ((e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \int (a-i a \tan (c+d x))^{\frac {1}{2} (5-2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (5-2 n)+n} \, dx \\ & = \frac {\left (a^2 (e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {1}{2} (5-2 n)} (a+i a x)^{-1+\frac {1}{2} (5-2 n)+n} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (a^2 (e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {1}{2} (5-2 n)} (a+i a x)^{3/2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{\frac {3}{2}-n} a^3 (e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2}-n+\frac {1}{2} (-5+2 n)} \left (\frac {a-i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}+n} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {i x}{2}\right )^{-1+\frac {1}{2} (5-2 n)} (a+i a x)^{3/2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {i 2^{\frac {5}{2}-n} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (-3+2 n),\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (e \sec (c+d x))^{5-2 n} (1-i \tan (c+d x))^{-\frac {5}{2}+n} (a+i a \tan (c+d x))^n}{5 d} \\ \end{align*}
Time = 15.72 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.71 \[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{5-n} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \left (1+e^{2 i (c+d x)}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},5-n,\frac {7}{2},-e^{2 i (c+d x)}\right ) \sec ^{-5+n}(c+d x) (e \sec (c+d x))^{5-2 n} (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{5 d} \]
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\[\int \left (e \sec \left (d x +c \right )\right )^{5-2 n} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
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\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{5 - 2 n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]
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\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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\[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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Timed out. \[ \int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5-2\,n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]
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