Integrand size = 23, antiderivative size = 111 \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\frac {2^{\frac {3}{2}+n} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2}+n,1,\frac {5}{4},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {1}{2}+n} (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)}}{d} \]
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Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {3974} \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\frac {2^{n+\frac {3}{2}} \sqrt {\tan (c+d x)} \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {1}{2}} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (\frac {1}{4},n-\frac {1}{2},1,\frac {5}{4},-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d} \]
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Rule 3974
Rubi steps \begin{align*} \text {integral}& = \frac {2^{\frac {3}{2}+n} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2}+n,1,\frac {5}{4},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {1}{2}+n} (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)}}{d} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(229\) vs. \(2(111)=222\).
Time = 9.02 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.06 \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\frac {10 \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2}+n,1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n \sqrt {\tan (c+d x)}}{d \left (2 \left (2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2}+n,2,\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+(1-2 n) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2}+n,1,\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+5 \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2}+n,1,\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \]
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\[\int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n}}{\sqrt {\tan \left (d x +c \right )}}d x\]
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\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
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\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \,d x \]
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