Optimal. Leaf size=145 \[ -\frac{d \tan (e+f x)}{f \left (c^2-d^2\right ) (a \sec (e+f x)+a) (c+d \sec (e+f x))}-\frac{2 d (2 c+d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{a f (c-d)^{5/2} (c+d)^{3/2}}+\frac{(c+2 d) \tan (e+f x)}{f (c-d)^2 (c+d) (a \sec (e+f x)+a)} \]
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Rubi [A] time = 0.252992, antiderivative size = 196, normalized size of antiderivative = 1.35, 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, 103, 152, 12, 93, 205} \[ -\frac{d \tan (e+f x)}{f \left (c^2-d^2\right ) (a \sec (e+f x)+a) (c+d \sec (e+f x))}+\frac{2 d (2 c+d) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{f (c-d)^{5/2} (c+d)^{3/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(c+2 d) \tan (e+f x)}{f (c-d)^2 (c+d) (a \sec (e+f x)+a)} \]
Antiderivative was successfully verified.
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Rule 3987
Rule 103
Rule 152
Rule 12
Rule 93
Rule 205
Rubi steps
\begin{align*} \int \frac{\sec (e+f x)}{(a+a \sec (e+f x)) (c+d \sec (e+f x))^2} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} (a+a x)^{3/2} (c+d x)^2} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x)) (c+d \sec (e+f x))}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{a^2 (c+d)-a^2 d x}{\sqrt{a-a x} (a+a x)^{3/2} (c+d x)} \, dx,x,\sec (e+f x)\right )}{\left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c+2 d) \tan (e+f x)}{(c-d)^2 (c+d) f (a+a \sec (e+f x))}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x)) (c+d \sec (e+f x))}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{a^4 d (2 c+d)}{\sqrt{a-a x} \sqrt{a+a x} (c+d x)} \, dx,x,\sec (e+f x)\right )}{a^3 (c-d) \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c+2 d) \tan (e+f x)}{(c-d)^2 (c+d) f (a+a \sec (e+f x))}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x)) (c+d \sec (e+f x))}+\frac{(a d (2 c+d) \tan (e+f x)) \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) \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c+2 d) \tan (e+f x)}{(c-d)^2 (c+d) f (a+a \sec (e+f x))}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x)) (c+d \sec (e+f x))}+\frac{(2 a d (2 c+d) \tan (e+f x)) \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) \left (c^2-d^2\right ) f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c+2 d) \tan (e+f x)}{(c-d)^2 (c+d) f (a+a \sec (e+f x))}+\frac{2 d (2 c+d) \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)}{(c-d)^{5/2} (c+d)^{3/2} f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{d \tan (e+f x)}{\left (c^2-d^2\right ) f (a+a \sec (e+f x)) (c+d \sec (e+f x))}\\ \end{align*}
Mathematica [C] time = 3.3504, size = 286, normalized size = 1.97 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) (c \cos (e+f x)+d) \left (\frac{2 d (2 c+d) (\sin (e)+i \cos (e)) \cos \left (\frac{1}{2} (e+f x)\right ) (c \cos (e+f x)+d) \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{(c+d) \sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}+\frac{d^2 \cos \left (\frac{1}{2} (e+f x)\right ) (c \sin (f x)-d \sin (e))}{c (c+d) \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right )}+\sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) (c \cos (e+f x)+d)\right )}{a f (c-d)^2 (\sec (e+f x)+1) (c+d \sec (e+f x))^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.082, size = 146, normalized size = 1. \begin{align*}{\frac{1}{fa} \left ({\frac{1}{{c}^{2}-2\,cd+{d}^{2}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+4\,{\frac{d}{ \left ( c-d \right ) ^{2}} \left ( -1/2\,{\frac{d\tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) }}-1/2\,{\frac{2\,c+d}{ \left ( c+d \right ) \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 ) } \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 = 0.563112, size = 1497, normalized size = 10.32 \begin{align*} \left [\frac{{\left (2 \, c d^{2} + d^{3} +{\left (2 \, c^{2} d + c d^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (2 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\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} d + 2 \, c^{2} d^{2} - c d^{3} - 2 \, d^{4} +{\left (c^{4} + c^{3} d - c d^{3} - d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (a c^{6} - a c^{5} d - 2 \, a c^{4} d^{2} + 2 \, a c^{3} d^{3} + a c^{2} d^{4} - a c d^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f \cos \left (f x + e\right ) +{\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} f\right )}}, -\frac{{\left (2 \, c d^{2} + d^{3} +{\left (2 \, c^{2} d + c d^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (2 \, c^{2} d + 3 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\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} d + 2 \, c^{2} d^{2} - c d^{3} - 2 \, d^{4} +{\left (c^{4} + c^{3} d - c d^{3} - d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{{\left (a c^{6} - a c^{5} d - 2 \, a c^{4} d^{2} + 2 \, a c^{3} d^{3} + a c^{2} d^{4} - a c d^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f \cos \left (f x + e\right ) +{\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} 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*} \frac{\int \frac{\sec{\left (e + f x \right )}}{c^{2} \sec{\left (e + f x \right )} + c^{2} + 2 c d \sec ^{2}{\left (e + f x \right )} + 2 c d \sec{\left (e + f x \right )} + d^{2} \sec ^{3}{\left (e + f x \right )} + d^{2} \sec ^{2}{\left (e + f x \right )}}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.30121, size = 308, normalized size = 2.12 \begin{align*} -\frac{\frac{2 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c - d\right )}} - \frac{2 \,{\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 )}{\left (2 \, c d + d^{2}\right )}}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} \sqrt{-c^{2} + d^{2}}} - \frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a c^{2} - 2 \, a c d + a d^{2}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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