Optimal. Leaf size=281 \[ \frac {3 C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{4 b d}+\frac {(4 b B-3 a C) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{2 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (4 a b B-3 a^2 C-b^2 C\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \sin (c+d x)}{2 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}} \]
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Rubi [A]
time = 0.22, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3102, 2835,
2744, 144, 143} \begin {gather*} -\frac {\left (-3 a^2 C+4 a b B-b^2 C\right ) \sin (c+d x) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\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 )}{2 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}+\frac {(4 b B-3 a C) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\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 )}{2 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 2744
Rule 2835
Rule 3102
Rubi steps
\begin {align*} \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx &=\frac {3 C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{4 b d}+\frac {3 \int \frac {\frac {b C}{3}+\frac {1}{3} (4 b B-3 a C) \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx}{4 b}\\ &=\frac {3 C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{4 b d}+\frac {(4 b B-3 a C) \int \sqrt [3]{a+b \cos (c+d x)} \, dx}{4 b^2}-\frac {\left (4 a b B-3 a^2 C-b^2 C\right ) \int \frac {1}{(a+b \cos (c+d x))^{2/3}} \, dx}{4 b^2}\\ &=\frac {3 C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{4 b d}-\frac {((4 b B-3 a C) \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 )}{4 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}+\frac {\left (\left (4 a b B-3 a^2 C-b^2 C\right ) \sin (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{2/3}} \, dx,x,\cos (c+d x)\right )}{4 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}\\ &=\frac {3 C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{4 b d}-\frac {\left ((4 b B-3 a C) \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 )}{4 b^2 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 (\left (4 a b B-3 a^2 C-b^2 C\right ) \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 )}{4 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}}\\ &=\frac {3 C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{4 b d}+\frac {(4 b B-3 a C) F_1\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)}{2 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (4 a b B-3 a^2 C-b^2 C\right ) F_1\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)}{2 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}}\\ \end {align*}
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Mathematica [A]
time = 2.18, size = 261, normalized size = 0.93 \begin {gather*} -\frac {3 \sqrt [3]{a+b \cos (c+d x)} \csc (c+d x) \left (4 \left (-4 a b B+3 a^2 C+b^2 C\right ) F_1\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 b B-3 a C) F_1\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 C \sin ^2(c+d x)\right )}{16 b^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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