Optimal. Leaf size=100 \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^m}{f \left (4 m^2+8 m+3\right ) \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) \sqrt{c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f (2 m+3)} \]
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Rubi [A] time = 0.221745, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^m}{f \left (4 m^2+8 m+3\right ) \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) \sqrt{c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f (2 m+3)} \]
Antiderivative was successfully verified.
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Rule 3955
Rule 3953
Rubi steps
\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \, dx &=-\frac{2 c (a+a \sec (e+f x))^m \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{f (3+2 m)}+\frac{(4 c) \int \sec (e+f x) (a+a \sec (e+f x))^m \sqrt{c-c \sec (e+f x)} \, dx}{3+2 m}\\ &=-\frac{8 c^2 (a+a \sec (e+f x))^m \tan (e+f x)}{f \left (3+8 m+4 m^2\right ) \sqrt{c-c \sec (e+f x)}}-\frac{2 c (a+a \sec (e+f x))^m \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{f (3+2 m)}\\ \end{align*}
Mathematica [F] time = 38.3734, size = 0, normalized size = 0. \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.261, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( fx+e \right ) \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sec \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.56104, size = 231, normalized size = 2.31 \begin{align*} -\frac{2 \,{\left (\sqrt{2} 2^{m + 2} \left (-a\right )^{m} c^{\frac{3}{2}} - \frac{\sqrt{2}{\left (2^{m + 2} m + 3 \cdot 2^{m + 1}\right )} \left (-a\right )^{m} c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} e^{\left (-m \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )\right )}}{{\left (4 \, m^{2} + 8 \, m + 3\right )} f{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac{3}{2}}{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac{3}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.498297, size = 263, normalized size = 2.63 \begin{align*} \frac{2 \,{\left ({\left (2 \, c m + 5 \, c\right )} \cos \left (f x + e\right )^{2} - 2 \, c m + 4 \, c \cos \left (f x + e\right ) - c\right )} \left (\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \cos \left (f x + e\right ) \sin \left (f x + e\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 (-c \sec \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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