Optimal. Leaf size=226 \[ \frac {\sqrt {2} (A b-a B) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}+\frac {\sqrt {2} B \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]
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Rubi [A] time = 0.18, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2756, 2665, 139, 138} \[ \frac {\sqrt {2} (A b-a B) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}+\frac {\sqrt {2} B \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]
Antiderivative was successfully verified.
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Rule 138
Rule 139
Rule 2665
Rule 2756
Rubi steps
\begin {align*} \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx &=\frac {B \int (a+b \cos (c+d x))^{2/3} \, dx}{b}+\frac {(A b-a B) \int \frac {1}{\sqrt [3]{a+b \cos (c+d x)}} \, dx}{b}\\ &=-\frac {(B \sin (c+d x)) \operatorname {Subst}\left (\int \frac {(a+b x)^{2/3}}{\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)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\cos (c+d x)\right )}{b d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}\\ &=-\frac {\left (B (a+b \cos (c+d x))^{2/3} \sin (c+d x)\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\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)} \left (-\frac {a+b \cos (c+d x)}{-a-b}\right )^{2/3}}-\frac {\left ((A b-a B) \sqrt [3]{-\frac {a+b \cos (c+d x)}{-a-b}} \sin (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}} \, dx,x,\cos (c+d x)\right )}{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} B 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 ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} (A b-a B) 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]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 189, normalized size = 0.84 \[ -\frac {3 \csc (c+d x) \sqrt {-\frac {b (\cos (c+d x)-1)}{a+b}} \sqrt {\frac {b (\cos (c+d x)+1)}{b-a}} (a+b \cos (c+d x))^{2/3} \left (5 (A b-a B) F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )+2 B (a+b \cos (c+d x)) F_1\left (\frac {5}{3};\frac {1}{2},\frac {1}{2};\frac {8}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )\right )}{10 b^2 d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}, x\right ) \]
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 \[ \int \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.25, size = 0, normalized size = 0.00 \[ \int \frac {A +B \cos \left (d x +c \right )}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]
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 \[ \int \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\,{d x} \]
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 \[ \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \]
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 \[ \int \frac {A + B \cos {\left (c + d x \right )}}{\sqrt [3]{a + b \cos {\left (c + d x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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