Integrand size = 29, antiderivative size = 173 \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos (c+d x) \sin ^{1+n}(c+d x)}{a^2 d (2+n)}+\frac {(3+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{a^2 d (1+n) (2+n) \sqrt {\cos ^2(c+d x)}}-\frac {2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{a^2 d (2+n) \sqrt {\cos ^2(c+d x)}} \]
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Time = 0.18 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2948, 2842, 2827, 2722} \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {(2 n+3) \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{a^2 d (n+1) (n+2) \sqrt {\cos ^2(c+d x)}}-\frac {2 \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{a^2 d (n+2) \sqrt {\cos ^2(c+d x)}}-\frac {\cos (c+d x) \sin ^{n+1}(c+d x)}{a^2 d (n+2)} \]
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Rule 2722
Rule 2827
Rule 2842
Rule 2948
Rubi steps \begin{align*} \text {integral}& = \frac {\int \sin ^n(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4} \\ & = -\frac {\cos (c+d x) \sin ^{1+n}(c+d x)}{a^2 d (2+n)}+\frac {\int \sin ^n(c+d x) \left (a^2 (3+2 n)-2 a^2 (2+n) \sin (c+d x)\right ) \, dx}{a^4 (2+n)} \\ & = -\frac {\cos (c+d x) \sin ^{1+n}(c+d x)}{a^2 d (2+n)}-\frac {2 \int \sin ^{1+n}(c+d x) \, dx}{a^2}+\frac {(3+2 n) \int \sin ^n(c+d x) \, dx}{a^2 (2+n)} \\ & = -\frac {\cos (c+d x) \sin ^{1+n}(c+d x)}{a^2 d (2+n)}+\frac {(3+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{a^2 d (1+n) (2+n) \sqrt {\cos ^2(c+d x)}}-\frac {2 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{a^2 d (2+n) \sqrt {\cos ^2(c+d x)}} \\ \end{align*}
Time = 4.01 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \sec ^2\left (\frac {1}{2} (c+d x)\right )^n \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \sin ^n(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},3+n,\frac {3+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{1+n}+\tan \left (\frac {1}{2} (c+d x)\right ) \left (-\frac {4 \operatorname {Hypergeometric2F1}\left (\frac {2+n}{2},3+n,\frac {4+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{2+n}+\tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {6 \operatorname {Hypergeometric2F1}\left (\frac {3+n}{2},3+n,\frac {5+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3+n}-\frac {4 \operatorname {Hypergeometric2F1}\left (3+n,\frac {4+n}{2},\frac {6+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{4+n}+\frac {\operatorname {Hypergeometric2F1}\left (3+n,\frac {5+n}{2},\frac {7+n}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{5+n}\right )\right )\right )}{d (a+a \sin (c+d x))^2} \]
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\[\int \frac {\left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +a \sin \left (d x +c \right )\right )^{2}}d x\]
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\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^4(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]
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