Optimal. Leaf size=132 \[ -\frac{2 a^5 c \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}-\frac{2 a^4 c \tan ^3(e+f x)}{f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{5/2} c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^3 c \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]
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Rubi [A] time = 0.140682, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3904, 3887, 461, 203} \[ -\frac{2 a^5 c \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}-\frac{2 a^4 c \tan ^3(e+f x)}{f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{5/2} c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^3 c \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
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Rule 3904
Rule 3887
Rule 461
Rule 203
Rubi steps
\begin{align*} \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x)) \, dx &=-\left ((a c) \int (a+a \sec (e+f x))^{3/2} \tan ^2(e+f x) \, dx\right )\\ &=\frac{\left (2 a^4 c\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (2+a x^2\right )^2}{1+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}\\ &=\frac{\left (2 a^4 c\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a}+3 x^2+a x^4-\frac{1}{a \left (1+a x^2\right )}\right ) \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}\\ &=-\frac{2 a^3 c \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}-\frac{2 a^4 c \tan ^3(e+f x)}{f (a+a \sec (e+f x))^{3/2}}-\frac{2 a^5 c \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}-\frac{\left (2 a^3 c\right ) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}\\ &=\frac{2 a^{5/2} c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}-\frac{2 a^3 c \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}-\frac{2 a^4 c \tan ^3(e+f x)}{f (a+a \sec (e+f x))^{3/2}}-\frac{2 a^5 c \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.802766, size = 110, normalized size = 0.83 \[ -\frac{a^2 c \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{a (\sec (e+f x)+1)} \left ((6 \cos (e+f x)+\cos (2 (e+f x))+3) \sqrt{\sec (e+f x)-1}-10 \cos ^2(e+f x) \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )\right )}{5 f \sqrt{\sec (e+f x)-1}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.229, size = 303, normalized size = 2.3 \begin{align*} -{\frac{{a}^{2}c}{20\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) }\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 5\,\sqrt{2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}+10\,\sqrt{2}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ){\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}+5\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}\sin \left ( fx+e \right ) -8\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+16\,\cos \left ( fx+e \right ) +8 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.03237, size = 1885, normalized size = 14.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.24532, size = 887, normalized size = 6.72 \begin{align*} \left [\frac{5 \,{\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) - 2 \,{\left (a^{2} c \cos \left (f x + e\right )^{2} + 3 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{5 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac{2 \,{\left (5 \,{\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right ) +{\left (a^{2} c \cos \left (f x + e\right )^{2} + 3 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{5 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\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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