Integrand size = 23, antiderivative size = 114 \[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\frac {2^{\frac {5}{2}+n} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2}+n,1,\frac {7}{4},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {3}{2}+n} (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]
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Time = 0.07 (sec) , antiderivative size = 114, 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} \[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\frac {2^{n+\frac {5}{2}} \tan ^{\frac {3}{2}}(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {3}{2}} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (\frac {3}{4},n+\frac {1}{2},1,\frac {7}{4},-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \]
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Rule 3974
Rubi steps \begin{align*} \text {integral}& = \frac {2^{\frac {5}{2}+n} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2}+n,1,\frac {7}{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 {3}{2}+n} (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x)}{3 d} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(114)=228\).
Time = 4.88 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.09 \[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\frac {56 \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2}+n,1,\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (a (1+\sec (c+d x)))^n \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\tan (c+d x)}}{d \left (6 \left (2 \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2}+n,2,\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-(1+2 n) \operatorname {AppellF1}\left (\frac {7}{4},\frac {3}{2}+n,1,\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+21 \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2}+n,1,\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \]
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\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \sqrt {\tan \left (d x +c \right )}d x\]
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\[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sqrt {\tan \left (d x + c\right )} \,d x } \]
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\[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \sqrt {\tan {\left (c + d x \right )}}\, dx \]
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\[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sqrt {\tan \left (d x + c\right )} \,d x } \]
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\[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sqrt {\tan \left (d x + c\right )} \,d x } \]
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Timed out. \[ \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx=\int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]
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