Optimal. Leaf size=262 \[ -\frac{2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (-8 a^2 B+6 a A b-b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (6 a^2 A b-8 a^3 B+5 a b^2 B-3 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 B \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 b^2 d} \]
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Rubi [A] time = 0.486215, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2988, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (-8 a^2 B+6 a A b-b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (6 a^2 A b-8 a^3 B+5 a b^2 B-3 A b^3\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 B \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 b^2 d} \]
Antiderivative was successfully verified.
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Rule 2988
Rule 3023
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 \int \frac{\frac{1}{2} a b (A b-a B)+\frac{1}{2} \left (2 a^2-b^2\right ) (A b-a B) \cos (c+d x)+\frac{1}{2} b \left (a^2-b^2\right ) B \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac{4 \int \frac{\frac{1}{4} b^2 \left (3 a A b-2 a^2 B-b^2 B\right )+\frac{1}{4} b \left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}-\frac{\left (6 a A b-8 a^2 B-b^2 B\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^3}+\frac{\left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac{\left (\left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{3 b^3 \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (6 a A b-8 a^2 B-b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{3 b^3 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right ) d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (6 a A b-8 a^2 B-b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \sqrt{a+b \cos (c+d x)}}-\frac{2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}\\ \end{align*}
Mathematica [A] time = 1.38617, size = 189, normalized size = 0.72 \[ \frac{2 \left (b \sin (c+d x) \left (\frac{a \left (-4 a^2 B+3 a A b+b^2 B\right )}{b^2-a^2}+b B \cos (c+d x)\right )+\frac{\sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left ((a-b) \left (8 a^2 B-6 a A b+b^2 B\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (6 a^2 A b-8 a^3 B+5 a b^2 B-3 A b^3\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{a-b}\right )}{3 b^3 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 10.478, size = 954, normalized size = 3.6 \begin{align*} \text{result too large to display} \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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{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{{\left (B \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right ) + a}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, 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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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