Optimal. Leaf size=109 \[ -\frac{x \left (2 a^2-b^2\right )}{2 b^3}+\frac{a \sin (c+d x)}{b^2 d}+\frac{2 a \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 b d} \]
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Rubi [A] time = 0.200214, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3050, 3023, 2735, 2659, 205} \[ -\frac{x \left (2 a^2-b^2\right )}{2 b^3}+\frac{a \sin (c+d x)}{b^2 d}+\frac{2 a \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d}-\frac{\sin (c+d x) \cos (c+d x)}{2 b d} \]
Antiderivative was successfully verified.
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Rule 3050
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \frac{-a+b \cos (c+d x)+2 a \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b}\\ &=\frac{a \sin (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \frac{-a b-\left (2 a^2-b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2}\\ &=-\frac{\left (2 a^2-b^2\right ) x}{2 b^3}+\frac{a \sin (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\left (a \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^3}\\ &=-\frac{\left (2 a^2-b^2\right ) x}{2 b^3}+\frac{a \sin (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\left (2 a \left (a^2-b^2\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 (2 a^2-b^2\right ) x}{2 b^3}+\frac{2 a \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d}+\frac{a \sin (c+d x)}{b^2 d}-\frac{\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}
Mathematica [A] time = 0.303774, size = 98, normalized size = 0.9 \[ \frac{-2 \left (2 a^2-b^2\right ) (c+d x)+8 a \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )+4 a b \sin (c+d x)+b^2 (-\sin (2 (c+d x)))}{4 b^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.024, size = 269, normalized size = 2.5 \begin{align*} 2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}a}{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{1}{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{\tan \left ( 1/2\,dx+c/2 \right ) a}{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{1}{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}^{2}}{d{b}^{3}}}+{\frac{1}{db}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{{a}^{3}}{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 ) }-2\,{\frac{a}{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 ) } \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 = 1.467, size = 586, normalized size = 5.38 \begin{align*} \left [-\frac{{\left (2 \, a^{2} - b^{2}\right )} d x - \sqrt{-a^{2} + b^{2}} a \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 (b^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, b^{3} d}, -\frac{{\left (2 \, a^{2} - b^{2}\right )} d x - 2 \, \sqrt{a^{2} - b^{2}} a \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) +{\left (b^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, b^{3} 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 = 1.35877, size = 251, normalized size = 2.3 \begin{align*} -\frac{\frac{{\left (2 \, a^{2} - b^{2}\right )}{\left (d x + c\right )}}{b^{3}} + \frac{4 \,{\left (a^{3} - 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 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, 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 )\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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