Optimal. Leaf size=222 \[ -\frac{b \left (3 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac{a \left (-5 a^2 b^2+2 a^4+18 b^4\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.337224, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2792, 3021, 2754, 12, 2659, 205} \[ -\frac{b \left (3 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac{a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac{a \left (-5 a^2 b^2+2 a^4+18 b^4\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 2792
Rule 3021
Rule 2754
Rule 12
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x)}{(a+b \cos (c+d x))^4} \, dx &=-\frac{a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{\int \frac{a^2-3 a b \cos (c+d x)-\left (2 a^2-3 b^2\right ) \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{\int \frac{2 a b \left (a^2-6 b^2\right )+\left (2 a^4-3 a^2 b^2+6 b^4\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\int \frac{3 b^3 \left (3 a^2+2 b^2\right )}{a+b \cos (c+d x)} \, dx}{6 b^2 \left (a^2-b^2\right )^3}\\ &=-\frac{a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\left (b \left (3 a^2+2 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac{a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac{\left (b \left (3 a^2+2 b^2\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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{b \left (3 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac{a^2 \cos (c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac{a^2 \left (2 a^2-7 b^2\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac{a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.1703, size = 158, normalized size = 0.71 \[ \frac{\frac{a \sin (c+d x) \left (\left (-5 a^2 b^2+2 a^4+18 b^4\right ) \cos ^2(c+d x)+3 a b \left (a^2+9 b^2\right ) \cos (c+d x)+11 a^2 b^2+4 a^4\right )}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}-\frac{6 b \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{7/2}}}{6 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.089, size = 776, normalized size = 3.5 \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 = 2.33069, size = 1958, normalized size = 8.82 \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 = 1.36458, size = 539, normalized size = 2.43 \begin{align*} \frac{\frac{3 \,{\left (3 \, a^{2} b + 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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 32 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, 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 )}^{3}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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