Optimal. Leaf size=144 \[ -\frac{2 a \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (b^2 (2 A+C)-2 a (b B-a C)\right )}{2 b^3}+\frac{(b B-a C) \sin (c+d x)}{b^2 d}+\frac{C \sin (c+d x) \cos (c+d x)}{2 b d} \]
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Rubi [A] time = 0.334223, antiderivative size = 142, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3049, 3023, 2735, 2659, 205} \[ -\frac{2 a \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (-\frac{2 a (b B-a C)}{b^2}+2 A+C\right )}{2 b}+\frac{(b B-a C) \sin (c+d x)}{b^2 d}+\frac{C \sin (c+d x) \cos (c+d x)}{2 b d} \]
Antiderivative was successfully verified.
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Rule 3049
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \frac{a C+b (2 A+C) \cos (c+d x)+2 (b B-a C) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b}\\ &=\frac{(b B-a C) \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \frac{a b C+\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2}\\ &=\frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) x}{2 b^3}+\frac{(b B-a C) \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (a \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^3}\\ &=\frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) x}{2 b^3}+\frac{(b B-a C) \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (2 a \left (A b^2-a (b B-a C)\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 )}{b^3 d}\\ &=\frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) x}{2 b^3}-\frac{2 a \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 )}{\sqrt{a-b} b^3 \sqrt{a+b} d}+\frac{(b B-a C) \sin (c+d x)}{b^2 d}+\frac{C \cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}
Mathematica [A] time = 0.404276, size = 133, normalized size = 0.92 \[ \frac{2 (c+d x) \left (2 a^2 C-2 a b B+2 A b^2+b^2 C\right )+\frac{8 a \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+4 b (b B-a C) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.034, size = 434, normalized size = 3. \begin{align*} 2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{db \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}aC}{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{C}{db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{db \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) aC}{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{C}{db}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{db}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) aB}{d{b}^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}C}{d{b}^{3}}}+{\frac{C}{db}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{aA}{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 ) }+2\,{\frac{{a}^{2}B}{d{b}^{2}\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{{a}^{3}C}{d{b}^{3}\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 ) } \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 = 1.94794, size = 988, normalized size = 6.86 \begin{align*} \left [\frac{{\left (2 \, C a^{4} - 2 \, B a^{3} b +{\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} -{\left (2 \, A + C\right )} b^{4}\right )} d x -{\left (C a^{3} - B a^{2} b + A 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 (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} -{\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} d}, \frac{{\left (2 \, C a^{4} - 2 \, B a^{3} b +{\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} -{\left (2 \, A + C\right )} b^{4}\right )} d x - 2 \,{\left (C a^{3} - B a^{2} b + A 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 (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} -{\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.24016, size = 323, normalized size = 2.24 \begin{align*} \frac{\frac{{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )}{\left (d x + c\right )}}{b^{3}} + \frac{4 \,{\left (C a^{3} - B a^{2} b + A 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}} b^{3}} - \frac{2 \,{\left (2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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