Integrand size = 28, antiderivative size = 91 \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{\frac {3}{4}+n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n}{d \sqrt {e \sec (c+d x)}} \]
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Time = 0.23 (sec) , antiderivative size = 91, 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} \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{n+\frac {3}{4}} (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{d \sqrt {e \sec (c+d x)}} \]
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{-\frac {1}{4}+n}}{\sqrt [4]{a-i a \tan (c+d x)}} \, dx}{\sqrt {e \sec (c+d x)}} \\ & = \frac {\left (a^2 \sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-\frac {5}{4}+n}}{(a-i a x)^{5/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}} \\ & = \frac {\left (2^{-\frac {5}{4}+n} a \sqrt [4]{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}{4}-n}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {5}{4}+n}}{(a-i a x)^{5/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}} \\ & = -\frac {i 2^{\frac {3}{4}+n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4}-n,\frac {3}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{4}-n} (a+i a \tan (c+d x))^n}{d \sqrt {e \sec (c+d x)}} \\ \end{align*}
Time = 11.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.57 \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {i 2^{\frac {1}{2}+n} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{-\frac {1}{2}+n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}+n,-\frac {1}{4}+n,\frac {3}{4}+n,-e^{2 i (c+d x)}\right ) \sec ^{\frac {1}{2}-n}(c+d x) (a+i a \tan (c+d x))^n}{d (-1+4 n) \sqrt {e \sec (c+d x)}} \]
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\[\int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{n}}{\sqrt {e \sec \left (d x +c \right )}}d x\]
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\[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {e \sec \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {e \sec \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {e \sec \left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {e \sec (c+d x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}} \,d x \]
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