Integrand size = 29, antiderivative size = 160 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=-\frac {4 (3+2 n) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^5 d (1+n)}-\frac {\sin ^{1+n}(c+d x) (a-a \sin (c+d x))^2}{d (2+n) \left (a^7+a^7 \sin (c+d x)\right )}+\frac {\sin ^{1+n}(c+d x) \left (a \left (27+30 n+8 n^2\right )+a (7+2 n) \sin (c+d x)\right )}{d \left (2+3 n+n^2\right ) \left (a^6+a^6 \sin (c+d x)\right )} \]
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Time = 0.17 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2915, 102, 151, 66} \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=-\frac {(a-a \sin (c+d x))^2 \sin ^{n+1}(c+d x)}{d (n+2) \left (a^7 \sin (c+d x)+a^7\right )}+\frac {\sin ^{n+1}(c+d x) \left (a (2 n+7) \sin (c+d x)+a \left (8 n^2+30 n+27\right )\right )}{d \left (n^2+3 n+2\right ) \left (a^6 \sin (c+d x)+a^6\right )}-\frac {4 (2 n+3) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,-\sin (c+d x))}{a^5 d (n+1)} \]
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Rule 66
Rule 102
Rule 151
Rule 2915
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^3 \left (\frac {x}{a}\right )^n}{(a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = -\frac {\sin ^{1+n}(c+d x) (a-a \sin (c+d x))^2}{d (2+n) \left (a^7+a^7 \sin (c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {(a-x) \left (\frac {x}{a}\right )^n (a (3+2 n)+(-7-2 n) x)}{(a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a^6 d (2+n)} \\ & = -\frac {\sin ^{1+n}(c+d x) (a-a \sin (c+d x))^2}{d (2+n) \left (a^7+a^7 \sin (c+d x)\right )}+\frac {\sin ^{1+n}(c+d x) \left (a \left (27+30 n+8 n^2\right )+a (7+2 n) \sin (c+d x)\right )}{d (1+n) (2+n) \left (a^6+a^6 \sin (c+d x)\right )}-\frac {(4 (3+2 n)) \text {Subst}\left (\int \frac {\left (\frac {x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = -\frac {4 (3+2 n) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^5 d (1+n)}-\frac {\sin ^{1+n}(c+d x) (a-a \sin (c+d x))^2}{d (2+n) \left (a^7+a^7 \sin (c+d x)\right )}+\frac {\sin ^{1+n}(c+d x) \left (a \left (27+30 n+8 n^2\right )+a (7+2 n) \sin (c+d x)\right )}{d (1+n) (2+n) \left (a^6+a^6 \sin (c+d x)\right )} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=\frac {\sin ^{1+n}(c+d x) \left (26+29 n+8 n^2+(9+4 n) \sin (c+d x)-(1+n) \sin ^2(c+d x)-4 \left (6+7 n+2 n^2\right ) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) (1+\sin (c+d x))\right )}{a^5 d (1+n) (2+n) (1+\sin (c+d x))} \]
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\[\int \frac {\left (\cos ^{7}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +a \sin \left (d x +c \right )\right )^{5}}d x\]
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\[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{7}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{5}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=\text {Timed out} \]
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\[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{7}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{5}} \,d x } \]
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\[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{7}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{5}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^7(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^5} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^7\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^5} \,d x \]
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