Optimal. Leaf size=121 \[ \frac{2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f \sqrt{a-b} \sqrt{a+b} (b c-a d)}-\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (b c-a d)} \]
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Rubi [A] time = 0.274576, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3988, 3001, 2659, 208} \[ \frac{2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f \sqrt{a-b} \sqrt{a+b} (b c-a d)}-\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (b c-a d)} \]
Antiderivative was successfully verified.
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Rule 3988
Rule 3001
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec (e+f x)}{(a+b \sec (e+f x)) (c+d \sec (e+f x))} \, dx &=\int \frac{\cos (e+f x)}{(b+a \cos (e+f x)) (d+c \cos (e+f x))} \, dx\\ &=\frac{b \int \frac{1}{b+a \cos (e+f x)} \, dx}{b c-a d}-\frac{d \int \frac{1}{d+c \cos (e+f x)} \, dx}{b c-a d}\\ &=\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{(b c-a d) f}-\frac{(2 d) \operatorname{Subst}\left (\int \frac{1}{c+d+(-c+d) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{(b c-a d) f}\\ &=\frac{2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b} (b c-a d) f}-\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{\sqrt{c-d} \sqrt{c+d} (b c-a d) f}\\ \end{align*}
Mathematica [A] time = 0.231845, size = 119, normalized size = 0.98 \[ -\frac{2 b \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2} (b c-a d)}-\frac{2 d \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{f \sqrt{c^2-d^2} (a d-b c)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.083, size = 108, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ( -2\,{\frac{b}{ \left ( ad-bc \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{d}{ \left ( ad-bc \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 ) } \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 [A] time = 6.49626, size = 2214, normalized size = 18.3 \begin{align*} \text{result too large to display} \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*} \int \frac{\sec{\left (e + f x \right )}}{\left (a + b \sec{\left (e + f x \right )}\right ) \left (c + d \sec{\left (e + f x \right )}\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.34189, size = 710, normalized size = 5.87 \begin{align*} \frac{\frac{{\left (\sqrt{-c^{2} + d^{2}} b{\left (c - 2 \, d\right )}{\left | c - d \right |} + \sqrt{-c^{2} + d^{2}} a d{\left | c - d \right |} + \sqrt{-c^{2} + d^{2}}{\left | -b c + a d \right |}{\left | c - d \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-\frac{2 \, a c - 2 \, b d + \sqrt{-4 \,{\left (a c + b c + a d + b d\right )}{\left (a c - b c - a d + b d\right )} + 4 \,{\left (a c - b d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (b c - a d\right )}^{2}{\left (c^{2} - 2 \, c d + d^{2}\right )} +{\left (c^{3} - 2 \, c^{2} d + c d^{2}\right )} a{\left | -b c + a d \right |} -{\left (c^{2} d - 2 \, c d^{2} + d^{3}\right )} b{\left | -b c + a d \right |}} + \frac{{\left (\sqrt{-a^{2} + b^{2}} b c{\left | a - b \right |} + \sqrt{-a^{2} + b^{2}}{\left (a - 2 \, b\right )} d{\left | a - b \right |} - \sqrt{-a^{2} + b^{2}}{\left | -b c + a d \right |}{\left | a - b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-\frac{2 \, a c - 2 \, b d - \sqrt{-4 \,{\left (a c + b c + a d + b d\right )}{\left (a c - b c - a d + b d\right )} + 4 \,{\left (a c - b d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )}{\left (b c - a d\right )}^{2} -{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} c{\left | -b c + a d \right |} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} d{\left | -b c + a d \right |}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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