Optimal. Leaf size=150 \[ -\frac{\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d}+\frac{a x \left (2 a^2-b^2\right )}{2 b^4}+\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{3 b d} \]
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Rubi [A] time = 0.389628, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3050, 3049, 3023, 2735, 2659, 205} \[ -\frac{\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d}+\frac{a x \left (2 a^2-b^2\right )}{2 b^4}+\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{3 b d} \]
Antiderivative was successfully verified.
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Rule 3050
Rule 3049
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{\cos (c+d x) \left (-2 a+b \cos (c+d x)+3 a \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b}\\ &=\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{3 a^2-a b \cos (c+d x)-2 \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^2}\\ &=-\frac{\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{3 a^2 b+3 a \left (2 a^2-b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3}\\ &=\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac{\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac{\left (a^2 \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac{\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac{\left (2 a^2 \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^4 d}\\ &=\frac{a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac{2 a^2 \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^4 d}-\frac{\left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^3 d}+\frac{a \cos (c+d x) \sin (c+d x)}{2 b^2 d}-\frac{\cos ^2(c+d x) \sin (c+d x)}{3 b d}\\ \end{align*}
Mathematica [A] time = 0.422678, size = 125, normalized size = 0.83 \[ -\frac{-6 a \left (2 a^2-b^2\right ) (c+d x)+24 a^2 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )-3 a b^2 \sin (2 (c+d x))+3 b (2 a-b) (2 a+b) \sin (c+d x)+b^3 \sin (3 (c+d x))}{12 b^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.031, size = 350, normalized size = 2.3 \begin{align*} -2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{a}^{2}}{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{a}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{a}^{2}}{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{8}{3\,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 ) ^{-3}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ){a}^{2}}{d{b}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{a}{d{b}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{3}}{d{b}^{4}}}-{\frac{a}{d{b}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{a}^{4}}{d{b}^{4}\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}}{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 ) } \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 = 1.77885, size = 703, normalized size = 4.69 \begin{align*} \left [\frac{3 \, \sqrt{-a^{2} + b^{2}} a^{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 ) + 3 \,{\left (2 \, a^{3} - a b^{2}\right )} d x -{\left (2 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a b^{2} \cos \left (d x + c\right ) + 6 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d}, -\frac{6 \, \sqrt{a^{2} - b^{2}} a^{2} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \,{\left (2 \, a^{3} - a b^{2}\right )} d x +{\left (2 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a b^{2} \cos \left (d x + c\right ) + 6 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} 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 = 1.65114, size = 309, normalized size = 2.06 \begin{align*} \frac{\frac{3 \,{\left (2 \, a^{3} - a b^{2}\right )}{\left (d x + c\right )}}{b^{4}} + \frac{12 \,{\left (a^{4} - a^{2} 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^{4}} - \frac{2 \,{\left (6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a 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 )}^{3} b^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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