Optimal. Leaf size=76 \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a f \sqrt{a-b} \sqrt{a+b}}+\frac{d \tanh ^{-1}(\sin (e+f x))}{a f} \]
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Rubi [A] time = 0.137298, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2828, 3001, 3770, 2659, 205} \[ \frac{2 (a c-b d) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a f \sqrt{a-b} \sqrt{a+b}}+\frac{d \tanh ^{-1}(\sin (e+f x))}{a f} \]
Antiderivative was successfully verified.
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Rule 2828
Rule 3001
Rule 3770
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{c+d \sec (e+f x)}{a+b \cos (e+f x)} \, dx &=\int \frac{(d+c \cos (e+f x)) \sec (e+f x)}{a+b \cos (e+f x)} \, dx\\ &=\frac{d \int \sec (e+f x) \, dx}{a}+\frac{(a c-b d) \int \frac{1}{a+b \cos (e+f x)} \, dx}{a}\\ &=\frac{d \tanh ^{-1}(\sin (e+f x))}{a f}+\frac{(2 (a c-b d)) \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 f}\\ &=\frac{2 (a c-b d) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a \sqrt{a-b} \sqrt{a+b} f}+\frac{d \tanh ^{-1}(\sin (e+f x))}{a f}\\ \end{align*}
Mathematica [A] time = 0.144669, size = 112, normalized size = 1.47 \[ \frac{\frac{(2 b d-2 a c) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+d \left (\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{a f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.054, size = 135, normalized size = 1.8 \begin{align*}{\frac{d}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{d}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+2\,{\frac{c}{f\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 ) }-2\,{\frac{bd}{fa\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 = 3.61751, size = 678, normalized size = 8.92 \begin{align*} \left [\frac{{\left (a^{2} - b^{2}\right )} d \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (a^{2} - b^{2}\right )} d \log \left (-\sin \left (f x + e\right ) + 1\right ) + \sqrt{-a^{2} + b^{2}}{\left (a c - b d\right )} \log \left (\frac{2 \, a b \cos \left (f x + e\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + a^{2}}\right )}{2 \,{\left (a^{3} - a b^{2}\right )} f}, \frac{{\left (a^{2} - b^{2}\right )} d \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (a^{2} - b^{2}\right )} d \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, \sqrt{a^{2} - b^{2}}{\left (a c - b d\right )} \arctan \left (-\frac{a \cos \left (f x + e\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (f x + e\right )}\right )}{2 \,{\left (a^{3} - a b^{2}\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*} \int \frac{c + d \sec{\left (e + f x \right )}}{a + b \cos{\left (e + f x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.20483, size = 180, normalized size = 2.37 \begin{align*} \frac{\frac{d \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{d \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{2 \,{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}{\left (a c - b d\right )}}{\sqrt{a^{2} - b^{2}} a}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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