Optimal. Leaf size=226 \[ -\frac{c g (3-4 p) \sin (e+f x) (g \sec (e+f x))^{p-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-p}{2},\frac{3-p}{2},\cos ^2(e+f x)\right )}{3 a^2 f \sqrt{\sin ^2(e+f x)}}+\frac{c (5-4 p) \sin (e+f x) (g \sec (e+f x))^p \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{p}{2},\frac{2-p}{2},\cos ^2(e+f x)\right )}{3 a^2 f \sqrt{\sin ^2(e+f x)}}-\frac{c (5-4 p) \tan (e+f x) (g \sec (e+f x))^p}{3 a^2 f (\sec (e+f x)+1)}-\frac{2 c \tan (e+f x) (g \sec (e+f x))^p}{3 f (a \sec (e+f x)+a)^2} \]
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Rubi [A] time = 0.412885, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {4020, 3787, 3772, 2643} \[ -\frac{c g (3-4 p) \sin (e+f x) (g \sec (e+f x))^{p-1} \, _2F_1\left (\frac{1}{2},\frac{1-p}{2};\frac{3-p}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt{\sin ^2(e+f x)}}+\frac{c (5-4 p) \sin (e+f x) (g \sec (e+f x))^p \, _2F_1\left (\frac{1}{2},-\frac{p}{2};\frac{2-p}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt{\sin ^2(e+f x)}}-\frac{c (5-4 p) \tan (e+f x) (g \sec (e+f x))^p}{3 a^2 f (\sec (e+f x)+1)}-\frac{2 c \tan (e+f x) (g \sec (e+f x))^p}{3 f (a \sec (e+f x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4020
Rule 3787
Rule 3772
Rule 2643
Rubi steps
\begin{align*} \int \frac{(g \sec (e+f x))^p (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx &=-\frac{2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\int \frac{(g \sec (e+f x))^p (a c (3-2 p)-2 a c (1-p) \sec (e+f x))}{a+a \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac{c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac{2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\int (g \sec (e+f x))^p \left (a^2 c (3-4 p) (1-p)+a^2 c (5-4 p) p \sec (e+f x)\right ) \, dx}{3 a^4}\\ &=-\frac{c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac{2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{(c (3-4 p) (1-p)) \int (g \sec (e+f x))^p \, dx}{3 a^2}+\frac{(c (5-4 p) p) \int (g \sec (e+f x))^{1+p} \, dx}{3 a^2 g}\\ &=-\frac{c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac{2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\left (c (3-4 p) (1-p) \left (\frac{\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p\right ) \int \left (\frac{\cos (e+f x)}{g}\right )^{-p} \, dx}{3 a^2}+\frac{\left (c (5-4 p) p \left (\frac{\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p\right ) \int \left (\frac{\cos (e+f x)}{g}\right )^{-1-p} \, dx}{3 a^2 g}\\ &=-\frac{c (3-4 p) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1-p}{2};\frac{3-p}{2};\cos ^2(e+f x)\right ) (g \sec (e+f x))^p \sin (e+f x)}{3 a^2 f \sqrt{\sin ^2(e+f x)}}+\frac{c (5-4 p) \, _2F_1\left (\frac{1}{2},-\frac{p}{2};\frac{2-p}{2};\cos ^2(e+f x)\right ) (g \sec (e+f x))^p \sin (e+f x)}{3 a^2 f \sqrt{\sin ^2(e+f x)}}-\frac{c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac{2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}\\ \end{align*}
Mathematica [F] time = 3.11801, size = 0, normalized size = 0. \[ \int \frac{(g \sec (e+f x))^p (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.504, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( g\sec \left ( fx+e \right ) \right ) ^{p} \left ( c-c\sec \left ( fx+e \right ) \right ) }{ \left ( a+a\sec \left ( fx+e \right ) \right ) ^{2}}}\, 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 \frac{{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2}}\,{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 (-\frac{{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}}{a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}}, 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*} - \frac{c \left (\int - \frac{\left (g \sec{\left (e + f x \right )}\right )^{p}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\left (g \sec{\left (e + f x \right )}\right )^{p} \sec{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \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 -\frac{{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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