Integrand size = 24, antiderivative size = 94 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {5}{2}+n} a^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-\frac {3}{2}-n,\frac {7}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^5(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-2+n}}{5 d} \]
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Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3586, 3604, 72, 71} \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i a^2 2^{n+\frac {5}{2}} \sec ^5(c+d x) (1+i \tan (c+d x))^{-n-\frac {1}{2}} (a+i a \tan (c+d x))^{n-2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-n-\frac {3}{2},\frac {7}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d} \]
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^5(c+d x) \int (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{\frac {5}{2}+n} \, dx}{(a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}} \\ & = \frac {\left (a^2 \sec ^5(c+d x)\right ) \text {Subst}\left (\int (a-i a x)^{3/2} (a+i a x)^{\frac {3}{2}+n} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}} \\ & = \frac {\left (2^{\frac {3}{2}+n} a^3 \sec ^5(c+d x) (a+i a \tan (c+d x))^{-2+n} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-n}\right ) \text {Subst}\left (\int \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 (a-i a \tan (c+d x))^{5/2}} \\ & = \frac {i 2^{\frac {5}{2}+n} a^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-\frac {3}{2}-n,\frac {7}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^5(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-2+n}}{5 d} \\ \end{align*}
Time = 14.72 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.59 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{5+n} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,\frac {7}{2}+n,-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d \left (1+e^{2 i (c+d x)}\right )^4 (5+2 n)} \]
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\[\int \left (\sec ^{5}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
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\[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{5} \,d x } \]
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\[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \sec ^{5}{\left (c + d x \right )}\, dx \]
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\[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{5} \,d x } \]
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\[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{5} \,d x } \]
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Timed out. \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\cos \left (c+d\,x\right )}^5} \,d x \]
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