Optimal. Leaf size=180 \[ -\frac{\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (a^2 A b+a^3 B-4 a b^2 B+2 A b^3\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
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Rubi [A] time = 0.290082, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2968, 3021, 2754, 12, 2659, 205} \[ -\frac{\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (a^2 A b+a^3 B-4 a b^2 B+2 A b^3\right ) \sin (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 2968
Rule 3021
Rule 2754
Rule 12
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx &=\int \frac{A \cos (c+d x)+B \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\\ &=\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{2 b (A b-a B)-\left (a A b+a^2 B-2 b^2 B\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{b \left (3 a A b-a^2 B-2 b^2 B\right )}{a+b \cos (c+d x)} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (3 a A b-a^2 B-2 b^2 B\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 )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac{a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.784091, size = 172, normalized size = 0.96 \[ \frac{\frac{\left (a^2 A b+a^3 B-4 a b^2 B+2 A b^3\right ) \sin (c+d x)}{b (a-b)^2 (a+b)^2 (a+b \cos (c+d x))}-\frac{2 \left (a^2 B-3 a A b+2 b^2 B\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{a (A b-a B) \sin (c+d x)}{b (a-b) (a+b) (a+b \cos (c+d x))^2}}{2 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.108, size = 886, normalized size = 4.9 \begin{align*} \text{result too large to display} \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 [B] time = 1.39819, size = 1616, normalized size = 8.98 \begin{align*} \text{result too large to display} \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 [B] time = 1.58545, size = 528, normalized size = 2.93 \begin{align*} \frac{\frac{{\left (B a^{2} - 3 \, A a b + 2 \, B 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 )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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