Integrand size = 19, antiderivative size = 53 \[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=-\frac {b^3 (b \tan (e+f x))^{-3+n}}{f (3-n)}-\frac {b (b \tan (e+f x))^{-1+n}}{f (1-n)} \]
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Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2671, 14} \[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=-\frac {b^3 (b \tan (e+f x))^{n-3}}{f (3-n)}-\frac {b (b \tan (e+f x))^{n-1}}{f (1-n)} \]
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Rule 14
Rule 2671
Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int x^{-4+n} \left (b^2+x^2\right ) \, dx,x,b \tan (e+f x)\right )}{f} \\ & = \frac {b \text {Subst}\left (\int \left (b^2 x^{-4+n}+x^{-2+n}\right ) \, dx,x,b \tan (e+f x)\right )}{f} \\ & = -\frac {b^3 (b \tan (e+f x))^{-3+n}}{f (3-n)}-\frac {b (b \tan (e+f x))^{-1+n}}{f (1-n)} \\ \end{align*}
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.87 \[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=\frac {b (-2+n+\cos (2 (e+f x))) \csc ^2(e+f x) (b \tan (e+f x))^{-1+n}}{f (-3+n) (-1+n)} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 20.03 (sec) , antiderivative size = 5281, normalized size of antiderivative = 99.64
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Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.62 \[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=\frac {{\left (2 \, \cos \left (f x + e\right )^{3} + {\left (n - 3\right )} \cos \left (f x + e\right )\right )} \left (\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right )^{n}}{{\left (f n^{2} - {\left (f n^{2} - 4 \, f n + 3 \, f\right )} \cos \left (f x + e\right )^{2} - 4 \, f n + 3 \, f\right )} \sin \left (f x + e\right )} \]
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\[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=\int \left (b \tan {\left (e + f x \right )}\right )^{n} \csc ^{4}{\left (e + f x \right )}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=\frac {\frac {b^{n} \tan \left (f x + e\right )^{n}}{{\left (n - 1\right )} \tan \left (f x + e\right )} + \frac {b^{n} \tan \left (f x + e\right )^{n}}{{\left (n - 3\right )} \tan \left (f x + e\right )^{3}}}{f} \]
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\[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{4} \,d x } \]
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Time = 4.74 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.60 \[ \int \csc ^4(e+f x) (b \tan (e+f x))^n \, dx=-\frac {2\,{\left (-\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{2\,{\sin \left (e+f\,x\right )}^2-2}\right )}^n\,\left (9\,\sin \left (2\,e+2\,f\,x\right )-6\,\sin \left (4\,e+4\,f\,x\right )+\sin \left (6\,e+6\,f\,x\right )-4\,n\,\sin \left (2\,e+2\,f\,x\right )+2\,n\,\sin \left (4\,e+4\,f\,x\right )\right )}{f\,\left (30\,{\sin \left (e+f\,x\right )}^2-12\,{\sin \left (2\,e+2\,f\,x\right )}^2+2\,{\sin \left (3\,e+3\,f\,x\right )}^2\right )\,\left (n^2-4\,n+3\right )} \]
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