Integrand size = 28, antiderivative size = 94 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {5}{4}+n} a \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4}-n,\frac {5}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) \sqrt {e \sec (c+d x)} (1+i \tan (c+d x))^{\frac {3}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{d} \]
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Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3586, 3604, 72, 71} \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=\frac {i a 2^{n+\frac {5}{4}} \sqrt {e \sec (c+d x)} (1+i \tan (c+d x))^{\frac {3}{4}-n} (a+i a \tan (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4}-n,\frac {5}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{d} \]
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {e \sec (c+d x)} \int \sqrt [4]{a-i a \tan (c+d x)} (a+i a \tan (c+d x))^{\frac {1}{4}+n} \, dx}{\sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}} \\ & = \frac {\left (a^2 \sqrt {e \sec (c+d x)}\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-\frac {3}{4}+n}}{(a-i a x)^{3/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}} \\ & = \frac {\left (2^{-\frac {3}{4}+n} a^2 \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{-1+n} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {3}{4}-n}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {3}{4}+n}}{(a-i a x)^{3/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt [4]{a-i a \tan (c+d x)}} \\ & = \frac {i 2^{\frac {5}{4}+n} a \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4}-n,\frac {5}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) \sqrt {e \sec (c+d x)} (1+i \tan (c+d x))^{\frac {3}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{d} \\ \end{align*}
Time = 10.43 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.81 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{\frac {3}{2}+n} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{\frac {1}{2}+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{4}+n,\frac {1}{2}+n,\frac {5}{4}+n,-e^{2 i (c+d x)}\right ) \sec ^{-\frac {1}{2}-n}(c+d x) \sqrt {e \sec (c+d x)} (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (1+4 n)} \]
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\[\int \sqrt {e \sec \left (d x +c \right )}\, \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
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\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=\int { \sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=\int \sqrt {e \sec {\left (c + d x \right )}} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]
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\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=\int { \sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=\int { \sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
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Timed out. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx=\int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]
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