Optimal. Leaf size=193 \[ \frac{a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac{2 a^3 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d}-\frac{\left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}-\frac{x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^5}+\frac{a \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{4 b d} \]
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Rubi [A] time = 0.614796, antiderivative size = 193, 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 = {3050, 3049, 3023, 2735, 2659, 205} \[ \frac{a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}+\frac{2 a^3 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d}-\frac{\left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}-\frac{x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^5}+\frac{a \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{4 b d} \]
Antiderivative was successfully verified.
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Rule 3050
Rule 3049
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) \left (1-\cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{\cos ^2(c+d x) \left (-3 a+b \cos (c+d x)+4 a \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{4 b}\\ &=\frac{a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{\cos (c+d x) \left (8 a^2-a b \cos (c+d x)-3 \left (4 a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{12 b^2}\\ &=-\frac{\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{-3 a \left (4 a^2-b^2\right )+b \left (4 a^2+3 b^2\right ) \cos (c+d x)+8 a \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^3}\\ &=\frac{a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac{\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{-3 a b \left (4 a^2-b^2\right )-3 \left (8 a^4-4 a^2 b^2-b^4\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^4}\\ &=-\frac{\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac{a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac{\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\left (a^3 \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^5}\\ &=-\frac{\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac{a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac{\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\left (2 a^3 \left (a^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 )}{b^5 d}\\ &=-\frac{\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac{2 a^3 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d}+\frac{a \left (3 a^2-b^2\right ) \sin (c+d x)}{3 b^4 d}-\frac{\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{a \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{4 b d}\\ \end{align*}
Mathematica [A] time = 0.966634, size = 168, normalized size = 0.87 \[ \frac{-24 a^2 b^2 \sin (2 (c+d x))+24 a b \left (4 a^2-b^2\right ) \sin (c+d x)+192 a^3 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )+48 a^2 b^2 c+48 a^2 b^2 d x-96 a^4 c-96 a^4 d x+8 a b^3 \sin (3 (c+d x))-3 b^4 \sin (4 (c+d x))+12 b^4 c+12 b^4 d x}{96 b^5 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.035, size = 653, normalized size = 3.4 \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 = 1.88277, size = 842, normalized size = 4.36 \begin{align*} \left [\frac{12 \, \sqrt{-a^{2} + b^{2}} a^{3} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 3 \,{\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x -{\left (6 \, b^{4} \cos \left (d x + c\right )^{3} - 8 \, a b^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} b + 8 \, a b^{3} + 3 \,{\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}, \frac{24 \, \sqrt{a^{2} - b^{2}} a^{3} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - 3 \,{\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x -{\left (6 \, b^{4} \cos \left (d x + c\right )^{3} - 8 \, a b^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} b + 8 \, a b^{3} + 3 \,{\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}\right ] \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 [B] time = 1.48516, size = 500, normalized size = 2.59 \begin{align*} -\frac{\frac{3 \,{\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )}{\left (d x + c\right )}}{b^{5}} + \frac{48 \,{\left (a^{5} - a^{3} b^{2}\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 )}}{\sqrt{a^{2} - b^{2}} b^{5}} - \frac{2 \,{\left (24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 72 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 32 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 21 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 32 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, b^{3} \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 )}^{4} b^{4}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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