Optimal. Leaf size=145 \[ -\frac{c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{a f (a \sec (e+f x)+a)^{3/2}}+\frac{c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}} \]
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Rubi [A] time = 0.435758, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3954, 3952} \[ -\frac{c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{a f (a \sec (e+f x)+a)^{3/2}}+\frac{c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3954
Rule 3952
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=\frac{c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}}-\frac{c \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(a+a \sec (e+f x))^{3/2}} \, dx}{a}\\ &=-\frac{c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2}}+\frac{c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}}+\frac{c^2 \int \frac{\sec (e+f x) \sqrt{c-c \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac{c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{a^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{c^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2}}+\frac{c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}}\\ \end{align*}
Mathematica [C] time = 1.42275, size = 178, normalized size = 1.23 \[ \frac{c^2 \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (6 \log \left (1+e^{i (e+f x)}\right )-3 \log \left (1+e^{2 i (e+f x)}\right )+\left (8 \log \left (1+e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (e+f x)+\left (2 \log \left (1+e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right ) \cos (2 (e+f x))-4\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.261, size = 281, normalized size = 1.9 \begin{align*} -{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{2\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( 2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +4\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +4\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,\cos \left ( fx+e \right ) +3 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.50696, size = 180, normalized size = 1.24 \begin{align*} -\frac{\frac{2 \, c^{\frac{5}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt{-a} a^{2}} + \frac{2 \, c^{\frac{5}{2}} \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt{-a} a^{2}} - \frac{\frac{2 \, \sqrt{-a} c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{\sqrt{-a} c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{a^{3}}}{2 \, f} \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^{2} \sec \left (f x + e\right )^{3} - 2 \, c^{2} \sec \left (f x + e\right )^{2} + c^{2} \sec \left (f x + e\right )\right )} \sqrt{a \sec \left (f x + e\right ) + a} \sqrt{-c \sec \left (f x + e\right ) + c}}{a^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}}, x\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 [A] time = 7.3788, size = 153, normalized size = 1.06 \begin{align*} \frac{{\left (2 \, c^{3} \log \left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right ) +{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} c + 4 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{2}\right )} c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{2 \, \sqrt{-a c} a^{2} f{\left | c \right |}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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