Optimal. Leaf size=332 \[ \frac{\left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b^2-5 a^2 b^2 C+4 a^4 C-3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )}-\frac{a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{a x \left (C \left (4 a^2+b^2\right )+2 A b^2\right )}{b^5} \]
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Rubi [A] time = 1.11783, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3049, 3023, 2735, 2659, 205} \[ \frac{\left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b^2-5 a^2 b^2 C+4 a^4 C-3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (4 a^2 C+3 A b^2-b^2 C\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )}-\frac{a \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{a x \left (C \left (4 a^2+b^2\right )+2 A b^2\right )}{b^5} \]
Antiderivative was successfully verified.
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Rule 3048
Rule 3049
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \cos ^3(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 ^2(c+d x) \left (3 \left (A b^2+a^2 C\right )-a b (A+C) \cos (c+d x)-\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(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 \left (3 A b^2+\left (4 a^2-b^2\right ) C\right )+b \left (3 A b^2+\left (a^2+2 b^2\right ) C\right ) \cos (c+d x)+6 a \left (A b^2+2 a^2 C-b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a \left (A b^2+2 a^2 C-b^2 C\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{6 a^2 \left (A b^2+2 a^2 C-b^2 C\right )-2 a b \left (3 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)-2 \left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (A b^2+2 a^2 C-b^2 C\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{6 a^2 b \left (A b^2+2 a^2 C-b^2 C\right )+6 a \left (a^2-b^2\right ) \left (2 A b^2+4 a^2 C+b^2 C\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=-\frac{a \left (2 A b^2+\left (4 a^2+b^2\right ) C\right ) x}{b^5}+\frac{\left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (A b^2+2 a^2 C-b^2 C\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(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 (3 A b^4-a^2 b^2 (2 A-5 C)-4 a^4 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^5 \left (a^2-b^2\right )}\\ &=-\frac{a \left (2 A b^2+\left (4 a^2+b^2\right ) C\right ) x}{b^5}+\frac{\left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (A b^2+2 a^2 C-b^2 C\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(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 (3 A b^4-a^2 b^2 (2 A-5 C)-4 a^4 C\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^5 \left (a^2-b^2\right ) d}\\ &=-\frac{a \left (2 A b^2+\left (4 a^2+b^2\right ) C\right ) x}{b^5}+\frac{2 a^2 \left (2 a^2 A b^2-3 A b^4+4 a^4 C-5 a^2 b^2 C\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^5 (a+b)^{3/2} d}+\frac{\left (a^2 b^2 (6 A-7 C)+12 a^4 C-b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (A b^2+2 a^2 C-b^2 C\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(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 = 1.11899, size = 215, normalized size = 0.65 \[ \frac{-12 a (c+d x) \left (C \left (4 a^2+b^2\right )+2 A b^2\right )+3 b \left (3 C \left (4 a^2+b^2\right )+4 A b^2\right ) \sin (c+d x)+\frac{12 a^3 b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+\frac{24 a^2 \left (a^2 b^2 (2 A-5 C)+4 a^4 C-3 A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-6 a b^2 C \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^5 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.043, size = 828, 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 = 2.08631, size = 2171, normalized size = 6.54 \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 = 1.46857, size = 593, normalized size = 1.79 \begin{align*} -\frac{\frac{6 \,{\left (4 \, C a^{6} + 2 \, A a^{4} b^{2} - 5 \, C a^{4} b^{2} - 3 \, A a^{2} b^{4}\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^{5} - b^{7}\right )} \sqrt{a^{2} - b^{2}}} - \frac{6 \,{\left (C a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\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 (4 \, C a^{3} + 2 \, A a b^{2} + C a b^{2}\right )}{\left (d x + c\right )}}{b^{5}} - \frac{2 \,{\left (9 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C 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} b^{4}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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