Optimal. Leaf size=146 \[ \frac{3 \csc (e+f x)}{8 a^2 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{8 a^2 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f (a \sec (e+f x)+a)^{5/2} (c-c \sec (e+f x))^{3/2}} \]
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Rubi [A] time = 0.344092, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3960, 3959, 2611, 3770} \[ \frac{3 \csc (e+f x)}{8 a^2 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{3 \tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{8 a^2 c f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f (a \sec (e+f x)+a)^{5/2} (c-c \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3960
Rule 3959
Rule 2611
Rule 3770
Rubi steps
\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx &=\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}+\frac{3 \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx}{4 a}\\ &=\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}-\frac{(3 \tan (e+f x)) \int \cot ^2(e+f x) \csc (e+f x) \, dx}{4 a^2 c \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{3 \csc (e+f x)}{8 a^2 c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}+\frac{(3 \tan (e+f x)) \int \csc (e+f x) \, dx}{8 a^2 c \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{3 \csc (e+f x)}{8 a^2 c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}-\frac{3 \tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{8 a^2 c f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}
Mathematica [C] time = 1.39315, size = 130, normalized size = 0.89 \[ -\frac{\tan (e+f x) \left (2 \cos (e+f x)+5 \cos (2 (e+f x))-3 (\cos (e+f x)-2 \cos (2 (e+f x))-\cos (3 (e+f x))+2) \tanh ^{-1}\left (e^{i (e+f x)}\right )+1\right )}{16 a^2 c f (\cos (e+f x)-1) (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.265, size = 204, normalized size = 1.4 \begin{align*} -{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{32\,f{a}^{3}{c}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( 12\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +5\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}+12\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-15\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-12\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -9\,\cos \left ( fx+e \right ) -12\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +3 \right ) \sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.739696, size = 1315, normalized size = 9.01 \begin{align*} \left [-\frac{3 \,{\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sqrt{-a c} \log \left (-\frac{4 \,{\left (2 \, \sqrt{-a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} +{\left (a c \cos \left (f x + e\right )^{2} + a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \,{\left (5 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{16 \,{\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )}, \frac{3 \,{\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{a c \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) +{\left (5 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \,{\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )}\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(-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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