Integrand size = 25, antiderivative size = 229 \[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} (a+b) B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {2} (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}} \]
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Time = 0.24 (sec) , antiderivative size = 229, 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} \[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} (A b-a B) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {2} B (a+b) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}} \]
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Rule 143
Rule 144
Rule 2744
Rule 2835
Rubi steps \begin{align*} \text {integral}& = \frac {B \int (a+b \cos (c+d x))^{4/3} \, dx}{b}+\frac {(A b-a B) \int \sqrt [3]{a+b \cos (c+d x)} \, dx}{b} \\ & = -\frac {(B \sin (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{4/3}}{\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 {\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 {\left ((-a-b) B \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{4/3}}{\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) \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 {\sqrt {2} (a+b) 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 ) \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 {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}}} \\ \end{align*}
Time = 1.41 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.10 \[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 \sqrt [3]{a+b \cos (c+d x)} \csc (c+d x) \left (4 \left (-a^2+b^2\right ) B \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}}+(4 A b+a B) \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))-4 b^2 B \sin ^2(c+d x)\right )}{16 b^2 d} \]
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\[\int \left (a +\cos \left (d x +c \right ) b \right )^{\frac {1}{3}} \left (A +B \cos \left (d x +c \right )\right )d x\]
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\[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
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\[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \left (A + B \cos {\left (c + d x \right )}\right ) \sqrt [3]{a + b \cos {\left (c + d x \right )}}\, dx \]
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\[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
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\[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]
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Timed out. \[ \int \sqrt [3]{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{1/3} \,d x \]
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