Optimal. Leaf size=94 \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b d \sqrt{a-b} \sqrt{a+b}}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{b} \]
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Rubi [A] time = 0.137051, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3057, 2659, 205, 3770} \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a b d \sqrt{a-b} \sqrt{a+b}}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{b} \]
Antiderivative was successfully verified.
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Rule 3057
Rule 2659
Rule 205
Rule 3770
Rubi steps
\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx &=\frac{C x}{b}+\frac{A \int \sec (c+d x) \, dx}{a}-\left (\frac{A b}{a}-B+\frac{a C}{b}\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx\\ &=\frac{C x}{b}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}-\frac{\left (2 \left (\frac{A b}{a}-B+\frac{a C}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{C x}{b}-\frac{2 \left (\frac{A b}{a}-B+\frac{a C}{b}\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b} d}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}\\ \end{align*}
Mathematica [C] time = 0.608345, size = 256, normalized size = 2.72 \[ \frac{2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (2 (\sin (c)+i \cos (c)) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}\right )+\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2} \left (a C d x-A b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+A b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a b d \sqrt{\left (b^2-a^2\right ) (\cos (2 c)-i \sin (2 c))} (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.067, size = 202, normalized size = 2.2 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{db}}-{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-2\,{\frac{Ab}{ad\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{B}{d\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{aC}{db\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{A}{ad}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \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 = 19.9063, size = 807, normalized size = 8.59 \begin{align*} \left [\frac{2 \,{\left (C a^{3} - C a b^{2}\right )} d x -{\left (C a^{2} - B a b + A b^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) +{\left (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} b - a b^{3}\right )} d}, \frac{2 \,{\left (C a^{3} - C a b^{2}\right )} d x - 2 \,{\left (C a^{2} - B a b + A b^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) +{\left (A a^{2} b - A b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{2} b - A b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} b - a b^{3}\right )} d}\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{\left (A + B \cos{\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{a + b \cos{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.23375, size = 200, normalized size = 2.13 \begin{align*} \frac{\frac{{\left (d x + c\right )} C}{b} + \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{2 \,{\left (C a^{2} - B a b + A b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a b}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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