Optimal. Leaf size=129 \[ \frac{2 d^2 \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{a^2 f (c-d)^{5/2} \sqrt{c+d}}+\frac{(c-4 d) \tan (e+f x)}{3 f (c-d)^2 \left (a^2 \sec (e+f x)+a^2\right )}+\frac{\tan (e+f x)}{3 f (c-d) (a \sec (e+f x)+a)^2} \]
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Rubi [A] time = 0.241377, antiderivative size = 183, normalized size of antiderivative = 1.42, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3987, 104, 152, 12, 93, 205} \[ \frac{(c-4 d) \tan (e+f x)}{3 f (c-d)^2 \left (a^2 \sec (e+f x)+a^2\right )}-\frac{2 d^2 \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{a f (c-d)^{5/2} \sqrt{c+d} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{\tan (e+f x)}{3 f (c-d) (a \sec (e+f x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 3987
Rule 104
Rule 152
Rule 12
Rule 93
Rule 205
Rubi steps
\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2 (c+d \sec (e+f x))} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (a+a x)^{5/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{3 (c-d) f (a+a \sec (e+f x))^2}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{-a^2 (c-3 d)-a^2 d x}{\sqrt{a-a x} (a+a x)^{3/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{3 a (c-d) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{3 (c-d) f (a+a \sec (e+f x))^2}+\frac{(c-4 d) \tan (e+f x)}{3 (c-d)^2 f \left (a^2+a^2 \sec (e+f x)\right )}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{3 a^4 d^2}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{3 a^4 (c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{3 (c-d) f (a+a \sec (e+f x))^2}+\frac{(c-4 d) \tan (e+f x)}{3 (c-d)^2 f \left (a^2+a^2 \sec (e+f x)\right )}-\frac{\left (d^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{(c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{3 (c-d) f (a+a \sec (e+f x))^2}+\frac{(c-4 d) \tan (e+f x)}{3 (c-d)^2 f \left (a^2+a^2 \sec (e+f x)\right )}-\frac{\left (2 d^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{a c-a d-(-a c-a d) x^2} \, dx,x,\frac{\sqrt{a+a \sec (e+f x)}}{\sqrt{a-a \sec (e+f x)}}\right )}{(c-d)^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{\tan (e+f x)}{3 (c-d) f (a+a \sec (e+f x))^2}-\frac{2 d^2 \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a+a \sec (e+f x)}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right ) \tan (e+f x)}{a (c-d)^{5/2} \sqrt{c+d} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{(c-4 d) \tan (e+f x)}{3 (c-d)^2 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end{align*}
Mathematica [C] time = 1.62861, size = 209, normalized size = 1.62 \[ \frac{\cos \left (\frac{1}{2} (e+f x)\right ) \left (\sec \left (\frac{e}{2}\right ) \left (-3 (c-2 d) \sin \left (e+\frac{f x}{2}\right )+(2 c-5 d) \sin \left (e+\frac{3 f x}{2}\right )+3 (c-3 d) \sin \left (\frac{f x}{2}\right )\right )-\frac{24 i d^2 (\cos (e)-i \sin (e)) \cos ^3\left (\frac{1}{2} (e+f x)\right ) \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{3 a^2 f (c-d)^2 (\cos (e+f x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.07, size = 122, normalized size = 1. \begin{align*}{\frac{1}{2\,f{a}^{2}} \left ( -{\frac{1}{ \left ( c-d \right ) ^{2}} \left ({\frac{c}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{d}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-c\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +3\,\tan \left ( 1/2\,fx+e/2 \right ) d \right ) }+4\,{\frac{{d}^{2}}{ \left ( c-d \right ) ^{2}\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) } \right ) } \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: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.530989, size = 1307, normalized size = 10.13 \begin{align*} \left [\frac{3 \,{\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \sqrt{c^{2} - d^{2}} \log \left (\frac{2 \, c d \cos \left (f x + e\right ) -{\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt{c^{2} - d^{2}}{\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \,{\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3} +{\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \,{\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \,{\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) +{\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )}}, \frac{3 \,{\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \sqrt{-c^{2} + d^{2}} \arctan \left (-\frac{\sqrt{-c^{2} + d^{2}}{\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) +{\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3} +{\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} + 2 \,{\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) +{\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )}}\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*} \frac{\int \frac{\sec{\left (e + f x \right )}}{c \sec ^{2}{\left (e + f x \right )} + 2 c \sec{\left (e + f x \right )} + c + d \sec ^{3}{\left (e + f x \right )} + 2 d \sec ^{2}{\left (e + f x \right )} + d \sec{\left (e + f x \right )}}\, dx}{a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.15912, size = 348, normalized size = 2.7 \begin{align*} -\frac{\frac{12 \,{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )} d^{2}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt{-c^{2} + d^{2}}} + \frac{a^{4} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a^{4} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + a^{4} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, a^{4} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 \, a^{4} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 9 \, a^{4} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6} c^{3} - 3 \, a^{6} c^{2} d + 3 \, a^{6} c d^{2} - a^{6} d^{3}}}{6 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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