Integrand size = 12, antiderivative size = 60 \[ \int (2+2 \sin (c+d x))^n \, dx=-\frac {2^{\frac {1}{2}+2 n} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d \sqrt {1+\sin (c+d x)}} \]
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Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2730} \[ \int (2+2 \sin (c+d x))^n \, dx=-\frac {2^{2 n+\frac {1}{2}} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d \sqrt {\sin (c+d x)+1}} \]
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Rule 2730
Rubi steps \begin{align*} \text {integral}& = -\frac {2^{\frac {1}{2}+2 n} \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d \sqrt {1+\sin (c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int (2+2 \sin (c+d x))^n \, dx=\frac {4^n B_{\frac {1}{2} (1+\sin (c+d x))}\left (\frac {1}{2}+n,\frac {1}{2}\right ) \sqrt {\cos ^2(c+d x)} \sec (c+d x)}{d} \]
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\[\int \left (2+2 \sin \left (d x +c \right )\right )^{n}d x\]
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\[ \int (2+2 \sin (c+d x))^n \, dx=\int { {\left (2 \, \sin \left (d x + c\right ) + 2\right )}^{n} \,d x } \]
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\[ \int (2+2 \sin (c+d x))^n \, dx=2^{n} \int \left (\sin {\left (c + d x \right )} + 1\right )^{n}\, dx \]
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\[ \int (2+2 \sin (c+d x))^n \, dx=\int { {\left (2 \, \sin \left (d x + c\right ) + 2\right )}^{n} \,d x } \]
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\[ \int (2+2 \sin (c+d x))^n \, dx=\int { {\left (2 \, \sin \left (d x + c\right ) + 2\right )}^{n} \,d x } \]
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Timed out. \[ \int (2+2 \sin (c+d x))^n \, dx=\int {\left (2\,\sin \left (c+d\,x\right )+2\right )}^n \,d x \]
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