Integrand size = 25, antiderivative size = 226 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\frac {\sqrt {2} B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {2} (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}} \]
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Time = 0.21 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2835, 2744, 144, 143} \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\frac {\sqrt {2} (A b-a B) \sin (c+d x) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}+\frac {\sqrt {2} B \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}} \]
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Rule 143
Rule 144
Rule 2744
Rule 2835
Rubi steps \begin{align*} \text {integral}& = \frac {B \int \sqrt [3]{a+b \cos (c+d x)} \, dx}{b}+\frac {(A b-a B) \int \frac {1}{(a+b \cos (c+d x))^{2/3}} \, dx}{b} \\ & = -\frac {(B \sin (c+d x)) \text {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}-\frac {((A b-a B) \sin (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{2/3}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}} \\ & = -\frac {\left (B \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \sqrt [3]{-\frac {a+b \cos (c+d x)}{-a-b}}}-\frac {\left ((A b-a B) \left (-\frac {a+b \cos (c+d x)}{-a-b}\right )^{2/3} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}} \\ & = \frac {\sqrt {2} B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {2} (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}} \\ \end{align*}
Time = 0.47 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.83 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=-\frac {3 \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \sqrt [3]{a+b \cos (c+d x)} \left (4 (A b-a B) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )+B \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) (a+b \cos (c+d x))\right ) \csc (c+d x)}{4 b^2 d} \]
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\[\int \frac {A +B \cos \left (d x +c \right )}{\left (a +\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}}d x\]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
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Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]
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