Integrand size = 35, antiderivative size = 164 \[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {(B+A n+B n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-n,-n,1-n,-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right ) \sec ^{1-n}(c+d x) \left (\frac {1+\sec (c+d x)}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d n (1+n) (1+\sec (c+d x))} \]
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Time = 0.27 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {4098, 3913, 3910, 134} \[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {(A n+B n+B) \sin (c+d x) \sec ^{1-n}(c+d x) \left (\frac {\sec (c+d x)+1}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a \sec (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-n,-n,1-n,-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right )}{d n (n+1) (\sec (c+d x)+1)}+\frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)} \]
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Rule 134
Rule 3910
Rule 3913
Rule 4098
Rubi steps \begin{align*} \text {integral}& = \frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {(B+A n+B n) \int \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \, dx}{1+n} \\ & = \frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {\left ((B+A n+B n) (1+\sec (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \sec ^{-n}(c+d x) (1+\sec (c+d x))^n \, dx}{1+n} \\ & = \frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {\left ((B+A n+B n) (1+\sec (c+d x))^{-\frac {1}{2}-n} (a+a \sec (c+d x))^n \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1-x)^{-1-n} (2-x)^{-\frac {1}{2}+n}}{\sqrt {x}} \, dx,x,1-\sec (c+d x)\right )}{d (1+n) \sqrt {1-\sec (c+d x)}} \\ & = \frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {(B+A n+B n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-n,-n,1-n,-\frac {2 \sec (c+d x)}{1-\sec (c+d x)}\right ) \sec ^{1-n}(c+d x) \left (\frac {1+\sec (c+d x)}{1-\sec (c+d x)}\right )^{\frac {1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d n (1+n) (1+\sec (c+d x))} \\ \end{align*}
Time = 0.84 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.68 \[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\left (A+\frac {(B+A n+B n) \left (-\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )^{\frac {1}{2}-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-n,-n,1-n,\csc ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n (1+\cos (c+d x))}\right ) \sec ^{-n}(c+d x) (a (1+\sec (c+d x)))^n \sin (c+d x)}{d (1+n)} \]
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\[\int \sec \left (d x +c \right )^{-1-n} \left (a +a \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )d x\]
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\[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{-n - 1} \,d x } \]
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\[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{- n - 1}{\left (c + d x \right )}\, dx \]
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\[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{-n - 1} \,d x } \]
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\[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{-n - 1} \,d x } \]
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Timed out. \[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{n+1}} \,d x \]
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