Optimal. Leaf size=270 \[ -\frac{2 b^2 \left (4 a^2 A b-3 a^3 B+2 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (2 a^2 A b+a^3 (-B)+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 A-4 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac{b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.975468, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3000, 3055, 3001, 3770, 2659, 205} \[ -\frac{2 b^2 \left (4 a^2 A b-3 a^3 B+2 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (2 a^2 A b+a^3 (-B)+2 a b^2 B-3 A b^3\right ) \tan (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 A-4 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac{b (A b-a B) \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 3000
Rule 3055
Rule 3001
Rule 3770
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac{b (A b-a B) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (a^2 A-3 A b^2+2 a b B-a (A b-a B) \cos (c+d x)+2 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A-3 A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 \left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right )+a \left (a^2 A+A b^2-2 a b B\right ) \cos (c+d x)+b \left (a^2 A-3 A b^2+2 a b B\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-3 A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (\left (a^2-b^2\right ) \left (a^2 A+6 A b^2-4 a b B\right )+a b \left (a^2 A-3 A b^2+2 a b B\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-3 A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a^2 A+6 A b^2-4 a b B\right ) \int \sec (c+d x) \, dx}{2 a^4}-\frac{\left (b^2 \left (4 a^2 A b-3 A b^3-3 a^3 B+2 a b^2 B\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A+6 A b^2-4 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-3 A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (2 b^2 \left (4 a^2 A b-3 A b^3-3 a^3 B+2 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 )}{a^4 \left (a^2-b^2\right ) d}\\ &=-\frac{2 b^2 \left (4 a^2 A b-3 A b^3-3 a^3 B+2 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^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (a^2 A+6 A b^2-4 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2 A-3 A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 6.24621, size = 438, normalized size = 1.62 \[ \frac{A b^4 \sin (c+d x)-a b^3 B \sin (c+d x)}{a^3 d (a-b) (a+b) (a+b \cos (c+d x))}-\frac{2 b^2 \left (-4 a^2 A b+3 a^3 B-2 a b^2 B+3 A b^3\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{a^4 d \left (a^2-b^2\right ) \sqrt{b^2-a^2}}+\frac{\left (a^2 (-A)+4 a b B-6 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac{\left (a^2 A-4 a b B+6 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac{a B \sin \left (\frac{1}{2} (c+d x)\right )-2 A b \sin \left (\frac{1}{2} (c+d x)\right )}{a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{a B \sin \left (\frac{1}{2} (c+d x)\right )-2 A b \sin \left (\frac{1}{2} (c+d x)\right )}{a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{A}{4 a^2 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{4 a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.198, size = 690, normalized size = 2.6 \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 = 109.609, size = 2952, normalized size = 10.93 \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 = 1.5163, size = 510, normalized size = 1.89 \begin{align*} -\frac{\frac{4 \,{\left (3 \, B a^{3} b^{2} - 4 \, A a^{2} b^{3} - 2 \, B a b^{4} + 3 \, A b^{5}\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^{6} - a^{4} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{4 \,{\left (B a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\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 (A a^{2} - 4 \, B a b + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} + \frac{{\left (A a^{2} - 4 \, B a b + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, 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 )}^{2} a^{3}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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