Integrand size = 21, antiderivative size = 149 \[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=\frac {3 d \cos (e+f x) (d \csc (e+f x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n) \sqrt {\cos ^2(e+f x)}}+\frac {b d^2 \cos (e+f x) (d \csc (e+f x))^{-2+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\sin ^2(e+f x)\right )}{f (2-n) \sqrt {\cos ^2(e+f x)}} \]
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Time = 0.11 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3317, 3872, 3857, 2722} \[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=\frac {a d \cos (e+f x) (d \csc (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n) \sqrt {\cos ^2(e+f x)}}+\frac {b d^2 \cos (e+f x) (d \csc (e+f x))^{n-2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\sin ^2(e+f x)\right )}{f (2-n) \sqrt {\cos ^2(e+f x)}} \]
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Rule 2722
Rule 3317
Rule 3857
Rule 3872
Rubi steps \begin{align*} \text {integral}& = d \int (d \csc (e+f x))^{-1+n} (b+a \csc (e+f x)) \, dx \\ & = a \int (d \csc (e+f x))^n \, dx+(b d) \int (d \csc (e+f x))^{-1+n} \, dx \\ & = \left (a (d \csc (e+f x))^n \left (\frac {\sin (e+f x)}{d}\right )^n\right ) \int \left (\frac {\sin (e+f x)}{d}\right )^{-n} \, dx+\left (b d (d \csc (e+f x))^n \left (\frac {\sin (e+f x)}{d}\right )^n\right ) \int \left (\frac {\sin (e+f x)}{d}\right )^{1-n} \, dx \\ & = \frac {a \cos (e+f x) (d \csc (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right ) \sin (e+f x)}{f (1-n) \sqrt {\cos ^2(e+f x)}}+\frac {b \cos (e+f x) (d \csc (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\sin ^2(e+f x)\right ) \sin ^2(e+f x)}{f (2-n) \sqrt {\cos ^2(e+f x)}} \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.70 \[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=-\frac {d \cos (e+f x) (d \csc (e+f x))^{-1+n} \sin ^2(e+f x)^{\frac {1}{2} (-1+n)} \left (3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3}{2},\cos ^2(e+f x)\right )+b \csc (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {3}{2},\cos ^2(e+f x)\right ) \sqrt {\sin ^2(e+f x)}\right )}{f} \]
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\[\int \left (d \csc \left (f x +e \right )\right )^{n} \left (a +b \sin \left (f x +e \right )\right )d x\]
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\[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \csc \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=\int \left (d \csc {\left (e + f x \right )}\right )^{n} \left (a + b \sin {\left (e + f x \right )}\right )\, dx \]
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\[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \csc \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \csc \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (d \csc (e+f x))^n (3+b \sin (e+f x)) \, dx=\int {\left (\frac {d}{\sin \left (e+f\,x\right )}\right )}^n\,\left (a+b\,\sin \left (e+f\,x\right )\right ) \,d x \]
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