Optimal. Leaf size=121 \[ \frac{2 a \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f \sqrt{a-b} \sqrt{a+b} (a c-b d)}-\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (a c-b d)} \]
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Rubi [A] time = 0.265811, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2828, 3001, 2659, 205, 208} \[ \frac{2 a \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f \sqrt{a-b} \sqrt{a+b} (a c-b d)}-\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f \sqrt{c-d} \sqrt{c+d} (a c-b d)} \]
Antiderivative was successfully verified.
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Rule 2828
Rule 3001
Rule 2659
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx &=\int \frac{\cos (e+f x)}{(a+b \cos (e+f x)) (d+c \cos (e+f x))} \, dx\\ &=\frac{a \int \frac{1}{a+b \cos (e+f x)} \, dx}{a c-b d}-\frac{d \int \frac{1}{d+c \cos (e+f x)} \, dx}{a c-b d}\\ &=\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{(a c-b 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 )}{(a c-b d) f}\\ &=\frac{2 a \tan ^{-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} (a c-b 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} (a c-b d) f}\\ \end{align*}
Mathematica [A] time = 0.212984, size = 106, normalized size = 0.88 \[ \frac{\frac{2 d \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}-\frac{2 a \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}}{a c f-b d f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.071, size = 110, normalized size = 0.9 \begin{align*} -2\,{\frac{d}{f \left ( ac-bd \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{ \left ( c-d \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}} \right ) }+2\,{\frac{a}{f \left ( ac-bd \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \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 = 11.5093, size = 2182, normalized size = 18.03 \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{1}{\left (a + b \cos{\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 = 1.33298, size = 690, normalized size = 5.7 \begin{align*} \frac{\frac{{\left (\sqrt{a^{2} - b^{2}} a c{\left | a - b \right |} - \sqrt{a^{2} - b^{2}}{\left (2 \, a - b\right )} d{\left | a - b \right |} + \sqrt{a^{2} - b^{2}}{\left | a c - b 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{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{\frac{b c - a d + \sqrt{{\left (a c + b c + a d + b d\right )}{\left (a c - b c - a d + b d\right )} +{\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )}{\left (a c - b d\right )}^{2} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} c{\left | a c - b d \right |} -{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d{\left | a c - b d \right |}} + \frac{{\left (\sqrt{-c^{2} + d^{2}} a{\left (c - 2 \, d\right )}{\left | -c + d \right |} + \sqrt{-c^{2} + d^{2}} b d{\left | -c + d \right |} - \sqrt{-c^{2} + d^{2}}{\left | a c - b 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{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{\frac{b c - a d - \sqrt{{\left (a c + b c + a d + b d\right )}{\left (a c - b c - a d + b d\right )} +{\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a c - b d\right )}^{2}{\left (c^{2} - 2 \, c d + d^{2}\right )} +{\left (c^{2} d - 2 \, c d^{2} + d^{3}\right )} a{\left | a c - b d \right |} -{\left (c^{3} - 2 \, c^{2} d + c d^{2}\right )} b{\left | a c - b d \right |}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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