Integrand size = 29, antiderivative size = 191 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (3 a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}+\frac {\left (a^2-b^2\right ) \left (b^2 n-a^2 (4+n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a^2 b^4 d (1+n)}-\frac {2 a \sin ^{2+n}(c+d x)}{b^3 d (2+n)}+\frac {\sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\left (a^2-b^2\right )^2 \sin ^{1+n}(c+d x)}{a b^4 d (a+b \sin (c+d x))} \]
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Time = 0.26 (sec) , antiderivative size = 191, 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 = {2916, 964, 1634, 66} \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (a^2-b^2\right ) \left (b^2 n-a^2 (n+4)\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {b \sin (c+d x)}{a}\right )}{a^2 b^4 d (n+1)}+\frac {\left (3 a^2-2 b^2\right ) \sin ^{n+1}(c+d x)}{b^4 d (n+1)}+\frac {\left (a^2-b^2\right )^2 \sin ^{n+1}(c+d x)}{a b^4 d (a+b \sin (c+d x))}-\frac {2 a \sin ^{n+2}(c+d x)}{b^3 d (n+2)}+\frac {\sin ^{n+3}(c+d x)}{b^2 d (n+3)} \]
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Rule 66
Rule 964
Rule 1634
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (b^2-x^2\right )^2}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\left (a^2-b^2\right )^2 \sin ^{1+n}(c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac {\text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (-b^3 n-\frac {a^4 (1+n)}{b}+2 a^2 b (1+n)+a \left (\frac {a^2}{b}-2 b\right ) x-\frac {a^2 x^2}{b}+\frac {a x^3}{b}\right )}{a+x} \, dx,x,b \sin (c+d x)\right )}{a b^4 d} \\ & = \frac {\left (a^2-b^2\right )^2 \sin ^{1+n}(c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac {\text {Subst}\left (\int \left (\frac {\left (3 a^3-2 a b^2\right ) \left (\frac {x}{b}\right )^n}{b}-2 a^2 \left (\frac {x}{b}\right )^{1+n}+a b \left (\frac {x}{b}\right )^{2+n}+\frac {\left (a^2-b^2\right ) \left (b^2 n-a^2 (4+n)\right ) \left (\frac {x}{b}\right )^n}{b (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{a b^4 d} \\ & = \frac {\left (3 a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}-\frac {2 a \sin ^{2+n}(c+d x)}{b^3 d (2+n)}+\frac {\sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\left (a^2-b^2\right )^2 \sin ^{1+n}(c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac {\left (\left (a^2-b^2\right ) \left (b^2 n-a^2 (4+n)\right )\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{a+x} \, dx,x,b \sin (c+d x)\right )}{a b^5 d} \\ & = \frac {\left (3 a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}+\frac {\left (a^2-b^2\right ) \left (b^2 n-a^2 (4+n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a^2 b^4 d (1+n)}-\frac {2 a \sin ^{2+n}(c+d x)}{b^3 d (2+n)}+\frac {\sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\left (a^2-b^2\right )^2 \sin ^{1+n}(c+d x)}{a b^4 d (a+b \sin (c+d x))} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x) \left (\frac {3 a^2-2 b^2}{1+n}-\frac {4 \left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{1+n}+\frac {\left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{a^2 (1+n)}-\frac {2 a b \sin (c+d x)}{2+n}+\frac {b^2 \sin ^2(c+d x)}{3+n}\right )}{b^4 d} \]
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\[\int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +b \sin \left (d x +c \right )\right )^{2}}d x\]
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\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]
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