Integrand size = 21, antiderivative size = 180 \[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=-\frac {(3-n) (8-n) (16-n) \operatorname {Hypergeometric2F1}(6,4+n,5+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{4+n}}{42 a^4 d (1-n) (4+n)}-\frac {\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}+\frac {\cos ^7(c+d x) (a+a \sec (c+d x))^{4+n} \left (6 (8-n)-\left (108-25 n+n^2\right ) \sec (c+d x)\right )}{42 a^4 d (1-n)} \]
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Time = 0.20 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3958, 102, 150, 67} \[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=-\frac {(3-n) (8-n) (16-n) (a \sec (c+d x)+a)^{n+4} \operatorname {Hypergeometric2F1}(6,n+4,n+5,\sec (c+d x)+1)}{42 a^4 d (1-n) (n+4)}+\frac {\cos ^7(c+d x) \left (6 (8-n)-\left (n^2-25 n+108\right ) \sec (c+d x)\right ) (a \sec (c+d x)+a)^{n+4}}{42 a^4 d (1-n)}-\frac {\cos ^7(c+d x) (1-\sec (c+d x))^2 (a \sec (c+d x)+a)^{n+4}}{a^4 d (1-n)} \]
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Rule 67
Rule 102
Rule 150
Rule 3958
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {(-a-a x)^3 (a-a x)^{3+n}}{x^8} \, dx,x,-\sec (c+d x)\right )}{a^6 d} \\ & = -\frac {\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}-\frac {\text {Subst}\left (\int \frac {(-a-a x) (a-a x)^{3+n} \left (a^3 (8-n)+a^3 (4-n) x\right )}{x^8} \, dx,x,-\sec (c+d x)\right )}{a^7 d (1-n)} \\ & = -\frac {\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}+\frac {\cos ^7(c+d x) (a+a \sec (c+d x))^{4+n} \left (6 (8-n)-\left (108-25 n+n^2\right ) \sec (c+d x)\right )}{42 a^4 d (1-n)}+\frac {((3-n) (8-n) (16-n)) \text {Subst}\left (\int \frac {(a-a x)^{3+n}}{x^6} \, dx,x,-\sec (c+d x)\right )}{42 a^3 d (1-n)} \\ & = -\frac {(3-n) (8-n) (16-n) \operatorname {Hypergeometric2F1}(6,4+n,5+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{4+n}}{42 a^4 d (1-n) (4+n)}-\frac {\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}+\frac {\cos ^7(c+d x) (a+a \sec (c+d x))^{4+n} \left (6 (8-n)-\left (108-25 n+n^2\right ) \sec (c+d x)\right )}{42 a^4 d (1-n)} \\ \end{align*}
Time = 0.84 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.64 \[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=\frac {\left ((4+n) \cos ^5(c+d x) \left (\left (24-25 n+n^2\right ) \cos (c+d x)+3 (13+n+(-1+n) \cos (2 (c+d x)))\right )-\left (-384+200 n-27 n^2+n^3\right ) \operatorname {Hypergeometric2F1}(6,4+n,5+n,1+\sec (c+d x))\right ) (1+\sec (c+d x))^4 (a (1+\sec (c+d x)))^n}{42 d (-1+n) (4+n)} \]
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\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \sin \left (d x +c \right )^{7}d x\]
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\[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{7} \,d x } \]
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Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=\text {Timed out} \]
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\[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{7} \,d x } \]
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\[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{7} \,d x } \]
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Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx=\int {\sin \left (c+d\,x\right )}^7\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]
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