Integrand size = 29, antiderivative size = 167 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a \left (a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}+\frac {\left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a b^4 d (1+n)}+\frac {\left (a^2-2 b^2\right ) \sin ^{2+n}(c+d x)}{b^3 d (2+n)}-\frac {a \sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\sin ^{4+n}(c+d x)}{b d (4+n)} \]
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Time = 0.25 (sec) , antiderivative size = 167, 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, 966, 1634, 66} \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right )^2 \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,-\frac {b \sin (c+d x)}{a}\right )}{a b^4 d (n+1)}-\frac {a \left (a^2-2 b^2\right ) \sin ^{n+1}(c+d x)}{b^4 d (n+1)}+\frac {\left (a^2-2 b^2\right ) \sin ^{n+2}(c+d x)}{b^3 d (n+2)}-\frac {a \sin ^{n+3}(c+d x)}{b^2 d (n+3)}+\frac {\sin ^{n+4}(c+d x)}{b d (n+4)} \]
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Rule 66
Rule 966
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} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\sin ^{4+n}(c+d x)}{b d (4+n)}+\frac {\text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (4+n-\frac {2 (4+n) x^2}{b^2}-\frac {a (4+n) x^3}{b^4}\right )}{a+x} \, dx,x,b \sin (c+d x)\right )}{b d (4+n)} \\ & = \frac {\sin ^{4+n}(c+d x)}{b d (4+n)}+\frac {\text {Subst}\left (\int \left (-\frac {a \left (a^2-2 b^2\right ) (4+n) \left (\frac {x}{b}\right )^n}{b^4}-\frac {\left (-a^2+2 b^2\right ) (4+n) \left (\frac {x}{b}\right )^{1+n}}{b^3}-\frac {a (4+n) \left (\frac {x}{b}\right )^{2+n}}{b^2}+\frac {\left (-a^2+b^2\right )^2 (4+n) \left (\frac {x}{b}\right )^n}{b^4 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{b d (4+n)} \\ & = -\frac {a \left (a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}+\frac {\left (a^2-2 b^2\right ) \sin ^{2+n}(c+d x)}{b^3 d (2+n)}-\frac {a \sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\sin ^{4+n}(c+d x)}{b d (4+n)}+\frac {\left (a^2-b^2\right )^2 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = -\frac {a \left (a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}+\frac {\left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a b^4 d (1+n)}+\frac {\left (a^2-2 b^2\right ) \sin ^{2+n}(c+d x)}{b^3 d (2+n)}-\frac {a \sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\sin ^{4+n}(c+d x)}{b d (4+n)} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sin ^{1+n}(c+d x) \left (-\frac {a^3-2 a b^2}{1+n}+\frac {\left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{a (1+n)}+\frac {b \left (a^2-2 b^2\right ) \sin (c+d x)}{2+n}-\frac {a b^2 \sin ^2(c+d x)}{3+n}+\frac {b^3 \sin ^3(c+d x)}{4+n}\right )}{b^4 d} \]
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\[\int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]
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\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{a+b\,\sin \left (c+d\,x\right )} \,d x \]
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