Optimal. Leaf size=272 \[ \frac{\left (3 a^2 C+b^2 (4 A+C)\right ) \sin (c+d x) \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt{2} b^2 d \sqrt{\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}-\frac{3 a C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt{2} b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d} \]
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Rubi [A] time = 0.314045, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3024, 2756, 2665, 139, 138} \[ \frac{\left (3 a^2 C+b^2 (4 A+C)\right ) \sin (c+d x) \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt{2} b^2 d \sqrt{\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}-\frac{3 a C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x)),\frac{b (1-\cos (c+d x))}{a+b}\right )}{2 \sqrt{2} b^2 d \sqrt{\cos (c+d x)+1} \sqrt [3]{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d} \]
Antiderivative was successfully verified.
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Rule 3024
Rule 2756
Rule 2665
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \frac{A+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{1}{3} b (4 A+C)-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{(3 a C) \int \sqrt [3]{a+b \cos (c+d x)} \, dx}{4 b^2}+\frac{1}{4} \left (4 A+C+\frac{3 a^2 C}{b^2}\right ) \int \frac{1}{(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 a C \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 )}{4 b^2 d \sqrt{1-\cos (c+d x)} \sqrt{1+\cos (c+d x)}}+\frac{\left (\left (-4 A-C-\frac{3 a^2 C}{b^2}\right ) \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} (a+b x)^{2/3}} \, dx,x,\cos (c+d x)\right )}{4 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 (3 a C \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 )}{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-C-\frac{3 a^2 C}{b^2}\right ) \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 )}{4 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{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+\left (1+\frac{3 a^2}{b^2}\right ) 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} d \sqrt{1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}}\\ \end{align*}
Mathematica [A] time = 1.82696, size = 256, normalized size = 0.94 \[ -\frac{3 \csc (c+d x) \sqrt [3]{a+b \cos (c+d x)} \left (4 \left (C \left (3 a^2+b^2\right )+4 A b^2\right ) \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )+C \left (-3 a \sqrt{-\frac{b (\cos (c+d x)-1)}{a+b}} \sqrt{\frac{b (\cos (c+d x)+1)}{b-a}} (a+b \cos (c+d x)) F_1\left (\frac{4}{3};\frac{1}{2},\frac{1}{2};\frac{7}{3};\frac{a+b \cos (c+d x)}{a-b},\frac{a+b \cos (c+d x)}{a+b}\right )-4 b^2 \sin ^2(c+d x)\right )\right )}{16 b^3 d} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.271, size = 0, normalized size = 0. \begin{align*} \int{(A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2}) \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{C \cos \left (d x + c\right )^{2} + 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{C \cos \left (d x + c\right )^{2} + 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{C \cos \left (d x + c\right )^{2} + 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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