Optimal. Leaf size=263 \[ \frac{\left (2 a^2 A b-3 a^3 B+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b-3 a^3 B+4 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b^2 d \left (a^2-b^2\right )}-\frac{x \left (-6 a^2 B+4 a A b-b^2 B\right )}{2 b^4} \]
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Rubi [A] time = 0.658914, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2989, 3049, 3023, 2735, 2659, 205} \[ \frac{\left (2 a^2 A b-3 a^3 B+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b-3 a^3 B+4 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b^2 d \left (a^2-b^2\right )}-\frac{x \left (-6 a^2 B+4 a A b-b^2 B\right )}{2 b^4} \]
Antiderivative was successfully verified.
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Rule 2989
Rule 3049
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx &=\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 a (A b-a B)+b (A b-a B) \cos (c+d x)+\left (2 a A b-3 a^2 B+b^2 B\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{a \left (2 a A b-3 a^2 B+b^2 B\right )-b \left (2 a A b-a^2 B-b^2 B\right ) \cos (c+d x)-2 \left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{a b \left (2 a A b-3 a^2 B+b^2 B\right )+\left (a^2-b^2\right ) \left (4 a A b-6 a^2 B-b^2 B\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (4 a A b-6 a^2 B-b^2 B\right ) x}{2 b^4}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (4 a A b-6 a^2 B-b^2 B\right ) x}{2 b^4}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (2 a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 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 )}{b^4 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (4 a A b-6 a^2 B-b^2 B\right ) x}{2 b^4}+\frac{2 a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a 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)^{3/2} b^4 (a+b)^{3/2} d}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.982682, size = 184, normalized size = 0.7 \[ \frac{2 (c+d x) \left (6 a^2 B-4 a A b+b^2 B\right )-\frac{8 a^2 \left (-2 a^2 A b+3 a^3 B-4 a b^2 B+3 A b^3\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{4 a^3 b (A b-a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+4 b (A b-2 a B) \sin (c+d x)+b^2 B \sin (2 (c+d x))}{4 b^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.145, size = 643, normalized size = 2.4 \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 [A] time = 1.59442, size = 2120, normalized size = 8.06 \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 [A] time = 1.62894, size = 456, normalized size = 1.73 \begin{align*} \frac{\frac{4 \,{\left (3 \, B a^{5} - 2 \, A a^{4} b - 4 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\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^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} - \frac{4 \,{\left (B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\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 )}} + \frac{{\left (6 \, B a^{2} - 4 \, A a b + B b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{2 \,{\left (4 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B 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^{3}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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