Optimal. Leaf size=243 \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (-C)-a b B+A b^2+2 b^2 C\right )}{b^2 d \left (a^2-b^2\right )}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a b d \left (a^2-b^2\right )}-\frac{\left (-3 a^2 b^2 (A+C)+a^3 b B+a^4 C+a b^3 B+A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b^2 d (a-b) (a+b)^2}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.708097, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (-C)-a b B+A b^2+2 b^2 C\right )}{b^2 d \left (a^2-b^2\right )}-\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a b d \left (a^2-b^2\right )}-\frac{\left (-3 a^2 b^2 (A+C)+a^3 b B+a^4 C+a b^3 B+A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b^2 d (a-b) (a+b)^2}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 3055
Rule 3059
Rule 2639
Rule 3002
Rule 2641
Rule 2805
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (-A b^2-a b B+a^2 (2 A+C)\right )-a (A b-a B+b C) \cos (c+d x)-\frac{1}{2} \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\frac{1}{2} b \left (A b^2+a b B-a^2 (2 A+C)\right )+\frac{1}{2} a \left (A b^2-a b B-a^2 C+2 b^2 C\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a b \left (a^2-b^2\right )}+\frac{\left (-A b^2+a (b B-a C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (A b^2-a b B-a^2 C+2 b^2 C\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}-\frac{\left (A b^4+a^3 b B+a b^3 B+a^4 C-3 a^2 b^2 (A+C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a b B-a^2 C+2 b^2 C\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 b^4+a^3 b B+a b^3 B+a^4 C-3 a^2 b^2 (A+C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a (a-b) b^2 (a+b)^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 3.35444, size = 299, normalized size = 1.23 \[ \frac{\frac{4 \sin (c+d x) \sqrt{\cos (c+d x)} \left (a (a C-b B)+A b^2\right )}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\frac{2 \left (a^2 (4 A+C)-a b B-3 A b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{2 \sin (c+d x) \left (a (a C-b B)+A b^2\right ) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{\sin ^2(c+d x)}}-\frac{8 a (-a B+A b+b C) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}}{(a-b) (a+b)}}{4 a d} \]
Antiderivative was successfully verified.
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Maple [B] time = 2.797, size = 815, normalized size = 3.4 \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(-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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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{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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