Integrand size = 23, antiderivative size = 226 \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{f}-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{3 f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{5 f} \]
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Time = 0.21 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3302, 1907, 252, 251, 372, 371} \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\frac {\sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b \sin ^n(e+f x)}{a}\right )}{f}+\frac {\sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {n+5}{n},-\frac {b \sin ^n(e+f x)}{a}\right )}{5 f}-\frac {2 \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {n+3}{n},-\frac {b \sin ^n(e+f x)}{a}\right )}{3 f} \]
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Rule 251
Rule 252
Rule 371
Rule 372
Rule 1907
Rule 3302
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (1-x^2\right )^2 \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (\left (a+b x^n\right )^p-2 x^2 \left (a+b x^n\right )^p+x^4 \left (a+b x^n\right )^p\right ) \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}+\frac {\text {Subst}\left (\int x^4 \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {2 \text {Subst}\left (\int x^2 \left (a+b x^n\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \left (1+\frac {b x^n}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}+\frac {\left (\left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int x^4 \left (1+\frac {b x^n}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f}-\frac {\left (2 \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int x^2 \left (1+\frac {b x^n}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{f}-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{3 f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{5 f} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.69 \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\frac {\sin (e+f x) \left (15 \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b \sin ^n(e+f x)}{a}\right )-10 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^2(e+f x)+3 \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^4(e+f x)\right ) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{15 f} \]
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\[\int \left (\cos ^{5}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{n}\left (f x +e \right )\right )\right )}^{p}d x\]
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\[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\text {Timed out} \]
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\[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
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\[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^5\,{\left (a+b\,{\sin \left (e+f\,x\right )}^n\right )}^p \,d x \]
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