Optimal. Leaf size=198 \[ \frac{\left (a^2 B+a A b-2 b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{\left (a^2 A b+a^3 B-3 a b^2 B+A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a-b) (a+b)^2}-\frac{(A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.540471, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2999, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (a^2 B+a A b-2 b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{\left (a^2 A b+a^3 B-3 a b^2 B+A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a-b) (a+b)^2}-\frac{(A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 2999
Rule 3059
Rule 2639
Rule 3002
Rule 2641
Rule 2805
Rubi steps
\begin{align*} \int \frac{\sqrt{\cos (c+d x)} (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx &=-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} (A b-a B)-(a A-b B) \cos (c+d x)-\frac{1}{2} (A b-a B) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{-a^2+b^2}\\ &=-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{-\frac{1}{2} b (A b-a B)+\frac{1}{2} \left (a A b+a^2 B-2 b^2 B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}+\frac{(A b-a B) \int \sqrt{\cos (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a A b+a^2 B-2 b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}-\frac{\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}+\frac{\left (a A b+a^2 B-2 b^2 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac{\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{(a-b) b^2 (a+b)^2 d}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.29557, size = 262, normalized size = 1.32 \[ \frac{\frac{4 (a B-A b) \sin (c+d x) \sqrt{\cos (c+d x)}}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{-\frac{2 (A b-a B) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{\sin ^2(c+d x)}}+\frac{2 (a B-A b) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{(4 a A-4 b B) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}}{(b-a) (a+b)}}{4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 8.609, size = 808, normalized size = 4.1 \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 )} \sqrt{\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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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 )} \sqrt{\cos \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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