Integrand size = 8, antiderivative size = 69 \[ \int \csc ^n(a+b x) \, dx=\frac {\cos (a+b x) \csc ^{-1+n}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(a+b x)\right )}{b (1-n) \sqrt {\cos ^2(a+b x)}} \]
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Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3857, 2722} \[ \int \csc ^n(a+b x) \, dx=\frac {\cos (a+b x) \csc ^{n-1}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(a+b x)\right )}{b (1-n) \sqrt {\cos ^2(a+b x)}} \]
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Rule 2722
Rule 3857
Rubi steps \begin{align*} \text {integral}& = \csc ^n(a+b x) \sin ^n(a+b x) \int \sin ^{-n}(a+b x) \, dx \\ & = \frac {\cos (a+b x) \csc ^{-1+n}(a+b x) \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)}} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int \csc ^n(a+b x) \, dx=-\frac {\cos (a+b x) \csc ^{-1+n}(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3}{2},\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac {1}{2} (-1+n)}}{b} \]
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\[\int \csc \left (x b +a \right )^{n}d x\]
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\[ \int \csc ^n(a+b x) \, dx=\int { \csc \left (b x + a\right )^{n} \,d x } \]
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\[ \int \csc ^n(a+b x) \, dx=\int \csc ^{n}{\left (a + b x \right )}\, dx \]
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\[ \int \csc ^n(a+b x) \, dx=\int { \csc \left (b x + a\right )^{n} \,d x } \]
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\[ \int \csc ^n(a+b x) \, dx=\int { \csc \left (b x + a\right )^{n} \,d x } \]
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Timed out. \[ \int \csc ^n(a+b x) \, dx=\int {\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^n \,d x \]
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