Integrand size = 23, antiderivative size = 70 \[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},1-n,\frac {1}{2}-m,\frac {3}{2},1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {1-\sec (e+f x)}} \]
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Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3910, 138} \[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\frac {2^{m+\frac {1}{2}} \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1-n,\frac {1}{2}-m,\frac {3}{2},\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f \sqrt {1-\sec (e+f x)}} \]
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Rule 138
Rule 3910
Rubi steps \begin{align*} \text {integral}& = \frac {\tan (e+f x) \text {Subst}\left (\int \frac {(1-x)^{-1+n} (2-x)^{-\frac {1}{2}+m}}{\sqrt {x}} \, dx,x,1+\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \\ & = \frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},1-n,\frac {1}{2}-m,\frac {3}{2},1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {1-\sec (e+f x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(257\) vs. \(2(70)=140\).
Time = 0.34 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.67 \[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\frac {(3+2 m) \operatorname {AppellF1}\left (\frac {1}{2}+m,m+n,1-n,\frac {3}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1-\sec (e+f x))^m (-\sec (e+f x))^n \sin (e+f x)}{f (1+2 m) \left ((3+2 m) \operatorname {AppellF1}\left (\frac {1}{2}+m,m+n,1-n,\frac {3}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+2 \left ((-1+n) \operatorname {AppellF1}\left (\frac {3}{2}+m,m+n,2-n,\frac {5}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(m+n) \operatorname {AppellF1}\left (\frac {3}{2}+m,1+m+n,1-n,\frac {5}{2}+m,\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]
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\[\int \left (1-\sec \left (f x +e \right )\right )^{m} \left (-\sec \left (f x +e \right )\right )^{n}d x\]
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\[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\int { \left (-\sec \left (f x + e\right )\right )^{n} {\left (-\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
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\[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\int \left (- \sec {\left (e + f x \right )}\right )^{n} \left (1 - \sec {\left (e + f x \right )}\right )^{m}\, dx \]
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\[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\int { \left (-\sec \left (f x + e\right )\right )^{n} {\left (-\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
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\[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\int { \left (-\sec \left (f x + e\right )\right )^{n} {\left (-\sec \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
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Timed out. \[ \int (1-\sec (e+f x))^m (-\sec (e+f x))^n \, dx=\int {\left (1-\frac {1}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]
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