Integrand size = 10, antiderivative size = 72 \[ \int (c \csc (a+b x))^n \, dx=\frac {c \cos (a+b x) (c \csc (a+b x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(a+b x)\right )}{b (1-n) \sqrt {\cos ^2(a+b x)}} \]
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Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3857, 2722} \[ \int (c \csc (a+b x))^n \, dx=\frac {c \cos (a+b x) (c \csc (a+b x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(a+b x)\right )}{b (1-n) \sqrt {\cos ^2(a+b x)}} \]
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Rule 2722
Rule 3857
Rubi steps \begin{align*} \text {integral}& = (c \csc (a+b x))^n \left (\frac {\sin (a+b x)}{c}\right )^n \int \left (\frac {\sin (a+b x)}{c}\right )^{-n} \, dx \\ & = \frac {\cos (a+b x) (c \csc (a+b x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(a+b x)\right ) \sin (a+b x)}{b (1-n) \sqrt {\cos ^2(a+b x)}} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90 \[ \int (c \csc (a+b x))^n \, dx=-\frac {\cos (a+b x) (c \csc (a+b x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3}{2},\cos ^2(a+b x)\right ) \sin (a+b x) \sin ^2(a+b x)^{\frac {1}{2} (-1+n)}}{b} \]
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\[\int \left (c \csc \left (x b +a \right )\right )^{n}d x\]
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\[ \int (c \csc (a+b x))^n \, dx=\int { \left (c \csc \left (b x + a\right )\right )^{n} \,d x } \]
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\[ \int (c \csc (a+b x))^n \, dx=\int \left (c \csc {\left (a + b x \right )}\right )^{n}\, dx \]
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\[ \int (c \csc (a+b x))^n \, dx=\int { \left (c \csc \left (b x + a\right )\right )^{n} \,d x } \]
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\[ \int (c \csc (a+b x))^n \, dx=\int { \left (c \csc \left (b x + a\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (c \csc (a+b x))^n \, dx=\int {\left (\frac {c}{\sin \left (a+b\,x\right )}\right )}^n \,d x \]
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