Integrand size = 19, antiderivative size = 43 \[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=-\frac {\operatorname {Hypergeometric2F1}(1,1+n,2+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{1+n}}{a d (1+n)} \]
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Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3965, 67} \[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=-\frac {(a \sec (c+d x)+a)^{n+1} \operatorname {Hypergeometric2F1}(1,n+1,n+2,\sec (c+d x)+1)}{a d (n+1)} \]
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Rule 67
Rule 3965
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+a x)^n}{x} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {\operatorname {Hypergeometric2F1}(1,1+n,2+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{1+n}}{a d (1+n)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=-\frac {\operatorname {Hypergeometric2F1}(1,1+n,2+n,1+\sec (c+d x)) (a (1+\sec (c+d x)))^{1+n}}{a d (1+n)} \]
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\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )d x\]
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\[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right ) \,d x } \]
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\[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan {\left (c + d x \right )}\, dx \]
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\[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right ) \,d x } \]
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\[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right ) \,d x } \]
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Timed out. \[ \int (a+a \sec (c+d x))^n \tan (c+d x) \, dx=\int \mathrm {tan}\left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]
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