Optimal. Leaf size=140 \[ -\frac{a^2 c \tan ^3(e+f x) \cos ^2(e+f x)^{\frac{p+3}{2}} (g \sec (e+f x))^p \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{p+3}{2},\frac{5}{2},\sin ^2(e+f x)\right )}{3 f}-\frac{a^2 c \tan ^3(e+f x) \cos ^2(e+f x)^{\frac{p+4}{2}} (g \sec (e+f x))^{p+1} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{p+4}{2},\frac{5}{2},\sin ^2(e+f x)\right )}{3 f g} \]
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Rubi [A] time = 0.198732, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3962, 2617, 16} \[ -\frac{a^2 c \tan ^3(e+f x) \cos ^2(e+f x)^{\frac{p+3}{2}} (g \sec (e+f x))^p \, _2F_1\left (\frac{3}{2},\frac{p+3}{2};\frac{5}{2};\sin ^2(e+f x)\right )}{3 f}-\frac{a^2 c \tan ^3(e+f x) \cos ^2(e+f x)^{\frac{p+4}{2}} (g \sec (e+f x))^{p+1} \, _2F_1\left (\frac{3}{2},\frac{p+4}{2};\frac{5}{2};\sin ^2(e+f x)\right )}{3 f g} \]
Antiderivative was successfully verified.
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Rule 3962
Rule 2617
Rule 16
Rubi steps
\begin{align*} \int (g \sec (e+f x))^p (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx &=-\left ((a c) \int \left (a (g \sec (e+f x))^p \tan ^2(e+f x)+a \sec (e+f x) (g \sec (e+f x))^p \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^2 c\right ) \int (g \sec (e+f x))^p \tan ^2(e+f x) \, dx\right )-\left (a^2 c\right ) \int \sec (e+f x) (g \sec (e+f x))^p \tan ^2(e+f x) \, dx\\ &=-\frac{a^2 c \cos ^2(e+f x)^{\frac{3+p}{2}} \, _2F_1\left (\frac{3}{2},\frac{3+p}{2};\frac{5}{2};\sin ^2(e+f x)\right ) (g \sec (e+f x))^p \tan ^3(e+f x)}{3 f}-\frac{\left (a^2 c\right ) \int (g \sec (e+f x))^{1+p} \tan ^2(e+f x) \, dx}{g}\\ &=-\frac{a^2 c \cos ^2(e+f x)^{\frac{3+p}{2}} \, _2F_1\left (\frac{3}{2},\frac{3+p}{2};\frac{5}{2};\sin ^2(e+f x)\right ) (g \sec (e+f x))^p \tan ^3(e+f x)}{3 f}-\frac{a^2 c \cos ^2(e+f x)^{\frac{4+p}{2}} \, _2F_1\left (\frac{3}{2},\frac{4+p}{2};\frac{5}{2};\sin ^2(e+f x)\right ) (g \sec (e+f x))^{1+p} \tan ^3(e+f x)}{3 f g}\\ \end{align*}
Mathematica [C] time = 55.6888, size = 13374, normalized size = 95.53 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.695, size = 0, normalized size = 0. \begin{align*} \int \left ( g\sec \left ( fx+e \right ) \right ) ^{p} \left ( a+a\sec \left ( fx+e \right ) \right ) ^{2} \left ( c-c\sec \left ( fx+e \right ) \right ) \, 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 (a \sec \left (f x + e\right ) + a\right )}^{2}{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}\,{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 (a^{2} c \sec \left (f x + e\right )^{3} + a^{2} c \sec \left (f x + e\right )^{2} - a^{2} c \sec \left (f x + e\right ) - a^{2} c\right )} \left (g \sec \left (f x + e\right )\right )^{p}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - a^{2} c \left (\int - \left (g \sec{\left (e + f x \right )}\right )^{p}\, dx + \int - \left (g \sec{\left (e + f x \right )}\right )^{p} \sec{\left (e + f x \right )}\, dx + \int \left (g \sec{\left (e + f x \right )}\right )^{p} \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (g \sec{\left (e + f x \right )}\right )^{p} \sec ^{3}{\left (e + f x \right )}\, dx\right ) \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 (a \sec \left (f x + e\right ) + a\right )}^{2}{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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