Optimal. Leaf size=119 \[ \frac{2 c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{3/2} f}-\frac{\sqrt{2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{a^{3/2} f}-\frac{2 c^2 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2}} \]
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Rubi [A] time = 0.191734, antiderivative size = 134, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3904, 3887, 470, 12, 391, 203} \[ \frac{2 c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{3/2} f}-\frac{\sqrt{2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{a^{3/2} f}-\frac{c^2 \sin (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{a f \sqrt{a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
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Rule 3904
Rule 3887
Rule 470
Rule 12
Rule 391
Rule 203
Rubi steps
\begin{align*} \int \frac{(c-c \sec (e+f x))^2}{(a+a \sec (e+f x))^{3/2}} \, dx &=\left (a^2 c^2\right ) \int \frac{\tan ^4(e+f x)}{(a+a \sec (e+f x))^{7/2}} \, dx\\ &=-\frac{\left (2 a c^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}\\ &=-\frac{c^2 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}-\frac{c^2 \operatorname{Subst}\left (\int \frac{2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a f}\\ &=-\frac{c^2 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}-\frac{\left (2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a f}\\ &=-\frac{c^2 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}-\frac{\left (2 c^2\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 )}{a f}+\frac{\left (2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{2+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a f}\\ &=\frac{2 c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^{3/2} f}-\frac{\sqrt{2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{a^{3/2} f}-\frac{c^2 \sec ^2\left (\frac{1}{2} (e+f x)\right ) \sin (e+f x)}{a f \sqrt{a+a \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.06782, size = 128, normalized size = 1.08 \[ \frac{c^2 \cot \left (\frac{1}{2} (e+f x)\right ) \left (\sec ^2\left (\frac{1}{2} (e+f x)\right ) \left (\cos (e+f x)+(\cos (e+f x)+1) \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )-1\right )-\sqrt{2} \sqrt{\sec (e+f x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (e+f x)-1}}{\sqrt{2}}\right )\right )}{a f \sqrt{a (\sec (e+f x)+1)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.189, size = 369, normalized size = 3.1 \begin{align*} -{\frac{{c}^{2}}{f{a}^{2} \left ( 1+\cos \left ( fx+e \right ) \right ) \sin \left ( fx+e \right ) }\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( \sqrt{2}\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( fx+e \right ) }{2\,\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) +\sqrt{2}\sin \left ( fx+e \right ){\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( fx+e \right ) }{2\,\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}+\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -\cos \left ( fx+e \right ) +1 \right ) } \right ) +\sin \left ( fx+e \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) -\cos \left ( fx+e \right ) +1 \right ) } \right ) -2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\cos \left ( fx+e \right ) \right ) } \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 )}^{2}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 3.8022, size = 1389, normalized size = 11.67 \begin{align*} \left [-\frac{4 \, c^{2} \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 ) - \sqrt{2}{\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \sqrt{-\frac{1}{a}} \log \left (\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{-\frac{1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, \cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 2 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \cos \left (f x + e\right ) + c^{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} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}}, -\frac{2 \, c^{2} \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 ) + 2 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \cos \left (f x + e\right ) + c^{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 ) - \frac{\sqrt{2}{\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \arctan \left (\frac{\sqrt{2} \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 )}{\sqrt{a}}}{a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f}\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*} c^{2} \left (\int - \frac{2 \sec{\left (e + f x \right )}}{a \sqrt{a \sec{\left (e + f x \right )} + a} \sec{\left (e + f x \right )} + a \sqrt{a \sec{\left (e + f x \right )} + a}}\, dx + \int \frac{\sec ^{2}{\left (e + f x \right )}}{a \sqrt{a \sec{\left (e + f x \right )} + a} \sec{\left (e + f x \right )} + a \sqrt{a \sec{\left (e + f x \right )} + a}}\, dx + \int \frac{1}{a \sqrt{a \sec{\left (e + f x \right )} + a} \sec{\left (e + f x \right )} + a \sqrt{a \sec{\left (e + f x \right )} + a}}\, dx\right ) \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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