Optimal. Leaf size=226 \[ \frac{\sqrt{2} (A b-a B) \sin (c+d x) \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}+\frac{\sqrt{2} B \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}} \]
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Rubi [A] time = 0.185021, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2756, 2665, 139, 138} \[ \frac{\sqrt{2} (A b-a B) \sin (c+d x) \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}+\frac{\sqrt{2} B \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 2756
Rule 2665
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx &=\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)) \operatorname{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)) \operatorname{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 ) \operatorname{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 ) \operatorname{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 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)}{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) 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)}{b d \sqrt{1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}}\\ \end{align*}
Mathematica [A] time = 0.418399, size = 188, normalized size = 0.83 \[ -\frac{3 \csc (c+d x) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} \sqrt [3]{a+b \cos (c+d x)} \left (4 (A b-a B) F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )+B (a+b \cos (c+d x)) F_1\left (\frac{4}{3};\frac{1}{2},\frac{1}{2};\frac{7}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )\right )}{4 b^2 d} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.321, size = 0, normalized size = 0. \begin{align*} \int{(A+B\cos \left ( dx+c \right ) ) \left ( a+b\cos \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{2}{3}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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