Optimal. Leaf size=128 \[ -\frac{(3 a+2 b (p+1)) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )}{3 f (a+b)}-\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{3 f (a+b)} \]
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Rubi [A] time = 0.106278, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4132, 453, 365, 364} \[ -\frac{(3 a+2 b (p+1)) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right )}{3 f (a+b)}-\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{3 f (a+b)} \]
Antiderivative was successfully verified.
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Rule 4132
Rule 453
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right ) \left (a+b+b x^2\right )^p}{x^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{3 (a+b) f}+\frac{(3 a+2 b (1+p)) \operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{3 (a+b) f}\\ &=-\frac{\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{3 (a+b) f}+\frac{\left ((3 a+2 b (1+p)) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a+b}\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{3 (a+b) f}\\ &=-\frac{\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{3 (a+b) f}-\frac{(3 a+2 b (1+p)) \cot (e+f x) \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}}{3 (a+b) f}\\ \end{align*}
Mathematica [A] time = 2.28075, size = 132, normalized size = 1.03 \[ -\frac{\cot (e+f x) \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p \left ((3 a+2 b (p+1)) \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )+\cot ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right ) \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^p\right )}{3 f (a+b)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.398, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( fx+e \right ) \right ) ^{4} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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