Integrand size = 12, antiderivative size = 60 \[ \int (2-2 \cos (c+d x))^n \, dx=-\frac {2^{\frac {1}{2}+2 n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1+\cos (c+d x))\right ) \sin (c+d x)}{d \sqrt {1-\cos (c+d x)}} \]
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Time = 0.02 (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} \[ \int (2-2 \cos (c+d x))^n \, dx=-\frac {2^{2 n+\frac {1}{2}} \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (\cos (c+d x)+1)\right )}{d \sqrt {1-\cos (c+d x)}} \]
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Rule 2730
Rubi steps \begin{align*} \text {integral}& = -\frac {2^{\frac {1}{2}+2 n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1+\cos (c+d x))\right ) \sin (c+d x)}{d \sqrt {1-\cos (c+d x)}} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23 \[ \int (2-2 \cos (c+d x))^n \, dx=\frac {\sqrt {2} (2-2 \cos (c+d x))^n \sqrt {1+\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}+n,\frac {3}{2}+n,\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{d+2 d n} \]
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\[\int \left (-2 \cos \left (d x +c \right )+2\right )^{n}d x\]
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\[ \int (2-2 \cos (c+d x))^n \, dx=\int { {\left (-2 \, \cos \left (d x + c\right ) + 2\right )}^{n} \,d x } \]
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\[ \int (2-2 \cos (c+d x))^n \, dx=\int \left (2 - 2 \cos {\left (c + d x \right )}\right )^{n}\, dx \]
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\[ \int (2-2 \cos (c+d x))^n \, dx=\int { {\left (-2 \, \cos \left (d x + c\right ) + 2\right )}^{n} \,d x } \]
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\[ \int (2-2 \cos (c+d x))^n \, dx=\int { {\left (-2 \, \cos \left (d x + c\right ) + 2\right )}^{n} \,d x } \]
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Timed out. \[ \int (2-2 \cos (c+d x))^n \, dx=\int {\left (4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}^n \,d x \]
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