Integrand size = 25, antiderivative size = 105 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {2^{5/6} (A-2 B) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{a d (1+\cos (c+d x))^{5/6}} \]
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Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2829, 2731, 2730} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\frac {3 (A-B) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-\frac {2^{5/6} (A-2 B) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{a d (\cos (c+d x)+1)^{5/6}} \]
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Rule 2730
Rule 2731
Rule 2829
Rubi steps \begin{align*} \text {integral}& = \frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {(A-2 B) \int \sqrt [3]{a+a \cos (c+d x)} \, dx}{a} \\ & = \frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {\left ((A-2 B) \sqrt [3]{a+a \cos (c+d x)}\right ) \int \sqrt [3]{1+\cos (c+d x)} \, dx}{a \sqrt [3]{1+\cos (c+d x)}} \\ & = \frac {3 (A-B) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}-\frac {2^{5/6} (A-2 B) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{a d (1+\cos (c+d x))^{5/6}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(254\) vs. \(2(105)=210\).
Time = 1.99 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.42 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (4 (A-2 B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\cos ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )-\csc \left (\frac {c}{2}\right ) \left (5 (A-2 B) \cos \left (\frac {1}{2} \left (c-d x-2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right )+(A-2 B) \cos \left (\frac {1}{2} \left (c+d x+2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right )+3 \left ((-2 A+3 B) \cos \left (\frac {d x}{2}\right )+B \cos \left (c+\frac {d x}{2}\right )\right ) \sqrt {\sec ^2\left (\frac {c}{2}\right )}\right ) \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}\right )}{d (a (1+\cos (c+d x)))^{2/3} \sqrt {\sec ^2\left (\frac {c}{2}\right )} \sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}} \]
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\[\int \frac {A +B \cos \left (d x +c \right )}{\left (a +\cos \left (d x +c \right ) a \right )^{\frac {2}{3}}}d x\]
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\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}}}\, dx \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]
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