Optimal. Leaf size=261 \[ -\frac{\left (2 a^2 A b+a^3 (-B)+2 a b^2 B-3 A b^3\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 A+2 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )}-\frac{2 b^2 \left (4 a^2 A b-3 a^3 B+2 a b^2 B-3 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \sin (c+d x) \cos (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x \left (a^2 A-4 a b B+6 A b^2\right )}{2 a^4} \]
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Rubi [A] time = 0.890839, antiderivative size = 261, 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 = {4030, 4104, 3919, 3831, 2659, 208} \[ -\frac{\left (2 a^2 A b+a^3 (-B)+2 a b^2 B-3 A b^3\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 A+2 a b B-3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )}-\frac{2 b^2 \left (4 a^2 A b-3 a^3 B+2 a b^2 B-3 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \sin (c+d x) \cos (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x \left (a^2 A-4 a b B+6 A b^2\right )}{2 a^4} \]
Antiderivative was successfully verified.
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Rule 4030
Rule 4104
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\frac{b (A b-a B) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos ^2(c+d x) \left (-a^2 A+3 A b^2-2 a b B+a (A b-a B) \sec (c+d x)-2 b (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (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 ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\cos (c+d x) \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 ) \sec (c+d x)+b \left (a^2 A-3 A b^2+2 a b B\right ) \sec ^2(c+d x)\right )}{a+b \sec (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 ) \sin (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 ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{-\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 ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A+6 A b^2-4 a b B\right ) x}{2 a^4}-\frac{\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \sin (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 ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\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{\sec (c+d x)}{a+b \sec (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 ) x}{2 a^4}-\frac{\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \sin (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 ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (b \left (4 a^2 A b-3 A b^3-3 a^3 B+2 a b^2 B\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A+6 A b^2-4 a b B\right ) x}{2 a^4}-\frac{\left (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \sin (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 ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 b \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}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right ) d}\\ &=\frac{\left (a^2 A+6 A b^2-4 a b B\right ) x}{2 a^4}-\frac{2 b^2 \left (4 a^2 A b-3 A b^3-3 a^3 B+2 a b^2 B\right ) \tanh ^{-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 (2 a^2 A b-3 A b^3-a^3 B+2 a b^2 B\right ) \sin (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 ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.07366, size = 184, normalized size = 0.7 \[ \frac{2 (c+d x) \left (a^2 A-4 a b B+6 A b^2\right )-\frac{8 b^2 \left (-4 a^2 A b+3 a^3 B-2 a b^2 B+3 A b^3\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+a^2 A \sin (2 (c+d x))-\frac{4 a b^3 (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}+4 a (a B-2 A b) \sin (c+d x)}{4 a^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.114, size = 651, normalized size = 2.5 \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 = 0.740963, size = 2136, normalized size = 8.18 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.50577, size = 459, normalized size = 1.76 \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 )}{\left (d x + c\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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