Optimal. Leaf size=129 \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)}-\frac{(a-2 b (p+1)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \text{Hypergeometric2F1}\left (\frac{1}{2},-p,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )}{b f (2 p+3)} \]
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Rubi [A] time = 0.0992134, antiderivative size = 129, 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 = {4146, 388, 246, 245} \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)}-\frac{(a-2 b (p+1)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right )}{b f (2 p+3)} \]
Antiderivative was successfully verified.
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Rule 4146
Rule 388
Rule 246
Rule 245
Rubi steps
\begin{align*} \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac{(a-2 b (1+p)) \operatorname{Subst}\left (\int \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{b f (3+2 p)}\\ &=\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac{\left ((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 \left (1+\frac{b x^2}{a+b}\right )^p \, dx,x,\tan (e+f x)\right )}{b f (3+2 p)}\\ &=\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac{(a-2 b (1+p)) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}}{b f (3+2 p)}\\ \end{align*}
Mathematica [A] time = 2.26217, size = 126, normalized size = 0.98 \[ \frac{\tan (e+f x) \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p \left ((2 b (p+1)-a) \text{Hypergeometric2F1}\left (\frac{1}{2},-p,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )+\left (a+b \tan ^2(e+f x)+b\right ) \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^p\right )}{b f (2 p+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.426, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \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} \sec \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} \sec \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} \sec \left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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