Integrand size = 14, antiderivative size = 110 \[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{(a+b) d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \]
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Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2744, 144, 143} \[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\frac {\sqrt {2} \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{d (a+b) \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}} \]
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Rule 143
Rule 144
Rule 2744
Rubi steps \begin{align*} \text {integral}& = -\frac {\sin (c+d x) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{4/3}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}} \\ & = -\frac {\left (\sqrt [3]{-\frac {a+b \cos (c+d x)}{-a-b}} \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 )^{4/3}} \, dx,x,\cos (c+d x)\right )}{(a+b) d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \\ & = \frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{(a+b) d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(110)=220\).
Time = 2.27 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.44 \[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\frac {15 a \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} (a+b \cos (c+d x)) \csc (c+d x)-6 \left (5 b^2+2 \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))^2 \csc ^2(c+d x)\right ) \sin (c+d x)}{10 b \left (a^2-b^2\right ) d \sqrt [3]{a+b \cos (c+d x)}} \]
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\[\int \frac {1}{\left (a +\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]
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\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
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\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^{4/3}} \, dx=\int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \]
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