Optimal. Leaf size=326 \[ -\frac{\left (-a^2 b^2 (7 A-6 C)+a^4 (-(2 A+3 C))+12 A b^4\right ) \sin (c+d x)}{3 a^4 d \left (a^2-b^2\right )}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \sin (c+d x) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{2 b^2 \left (5 a^2 A b^2-2 a^2 b^2 C+3 a^4 C-4 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{b x \left (a^2 (A+2 C)+4 A b^2\right )}{a^5} \]
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Rubi [A] time = 1.25529, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4101, 4104, 3919, 3831, 2659, 208} \[ -\frac{\left (-a^2 b^2 (7 A-6 C)+a^4 (-(2 A+3 C))+12 A b^4\right ) \sin (c+d x)}{3 a^4 d \left (a^2-b^2\right )}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d \left (a^2-b^2\right )}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \sin (c+d x) \cos (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{2 b^2 \left (5 a^2 A b^2-2 a^2 b^2 C+3 a^4 C-4 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{b x \left (a^2 (A+2 C)+4 A b^2\right )}{a^5} \]
Antiderivative was successfully verified.
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Rule 4101
Rule 4104
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\frac{\left (A b^2+a^2 C\right ) \cos ^2(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 ^3(c+d x) \left (4 A b^2-a^2 (A-3 C)+a b (A+C) \sec (c+d x)-3 \left (A b^2+a^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \cos ^2(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 (6 b \left (2 A b^2-a^2 (A-C)\right )+a \left (A b^2+a^2 (2 A+3 C)\right ) \sec (c+d x)-2 b \left (4 A b^2-a^2 (A-3 C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac{b \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \cos ^2(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 (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right )+2 a b \left (2 A b^2+a^2 (A+3 C)\right ) \sec (c+d x)-6 b^2 \left (2 A b^2-a^2 (A-C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-6 b \left (a^2-b^2\right ) \left (4 A b^2+a^2 (A+2 C)\right )+6 a b^2 \left (2 A b^2-a^2 (A-C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac{b \left (4 A b^2+a^2 (A+2 C)\right ) x}{a^5}-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \cos ^2(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 b^4-a^2 b^2 (5 A-2 C)-3 a^4 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{b \left (4 A b^2+a^2 (A+2 C)\right ) x}{a^5}-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \cos ^2(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 b^4-a^2 b^2 (5 A-2 C)-3 a^4 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^5 \left (a^2-b^2\right )}\\ &=-\frac{b \left (4 A b^2+a^2 (A+2 C)\right ) x}{a^5}-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \cos ^2(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 b^4-a^2 b^2 (5 A-2 C)-3 a^4 C\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^5 \left (a^2-b^2\right ) d}\\ &=-\frac{b \left (4 A b^2+a^2 (A+2 C)\right ) x}{a^5}+\frac{2 b^2 \left (5 a^2 A b^2-4 A b^4+3 a^4 C-2 a^2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 (a-b)^{3/2} (a+b)^{3/2} d}-\frac{\left (12 A b^4-a^2 b^2 (7 A-6 C)-a^4 (2 A+3 C)\right ) \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (4 A b^2-a^2 (A-3 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \cos ^2(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 = 1.16316, size = 212, normalized size = 0.65 \[ \frac{-12 b (c+d x) \left (a^2 (A+2 C)+4 A b^2\right )+3 a \left (a^2 (3 A+4 C)+12 A b^2\right ) \sin (c+d x)-\frac{12 a b^3 \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}+\frac{24 b^2 \left (a^2 b^2 (2 C-5 A)-3 a^4 C+4 A b^4\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-6 a^2 A b \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^5 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.128, size = 836, 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 [A] time = 0.785335, size = 2187, normalized size = 6.71 \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 = 1.25442, size = 595, normalized size = 1.83 \begin{align*} \frac{\frac{6 \,{\left (3 \, C a^{4} b^{2} + 5 \, A a^{2} b^{4} - 2 \, C a^{2} b^{4} - 4 \, A b^{6}\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^{7} - a^{5} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{6 \,{\left (C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{6} - a^{4} 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{3 \,{\left (A a^{2} b + 2 \, C a^{2} b + 4 \, A b^{3}\right )}{\left (d x + c\right )}}{a^{5}} + \frac{2 \,{\left (3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, A b^{2} \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} a^{4}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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