Optimal. Leaf size=343 \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (3 A-16 C)+9 a^3 b B-15 a^4 C-12 a b^3 B+2 b^4 (3 A+C)\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac{a \left (-a^2 b^2 (A-7 C)+3 a^3 b B-5 a^4 C-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a-b) (a+b)^2}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )} \]
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Rubi [A] time = 1.12625, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3047, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (3 A-16 C)+9 a^3 b B-15 a^4 C-12 a b^3 B+2 b^4 (3 A+C)\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac{a \left (-a^2 b^2 (A-7 C)+3 a^3 b B-5 a^4 C-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a-b) (a+b)^2}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 3047
Rule 3049
Rule 3059
Rule 2639
Rule 3002
Rule 2641
Rule 2805
Rubi steps
\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \cos (c+d x)-\frac{1}{2} \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{2 \int \frac{-\frac{1}{4} a \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right )+\frac{1}{2} b \left (3 A b^2-3 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)-\frac{3}{4} \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{2 \int \frac{\frac{1}{4} a b \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right )-\frac{1}{4} \left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}+\frac{\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a \left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^4 \left (a^2-b^2\right )}-\frac{\left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 \left (a^2-b^2\right ) d}+\frac{a \left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{(a-b) b^4 (a+b)^2 d}+\frac{\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 3.62617, size = 343, normalized size = 1. \[ \frac{4 \sin (c+d x) \sqrt{\cos (c+d x)} \left (\frac{3 a \left (a (a C-b B)+A b^2\right )}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+2 C\right )-\frac{\frac{2 \left (-3 a^2 b B+5 a^3 C-a b^2 (3 A+8 C)+6 b^3 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{8 \left (2 a^2 C-3 a b B+3 A b^2+b^2 C\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac{6 \sin (c+d x) \left (-3 a^2 b B+5 a^3 C+a b^2 (A-4 C)+2 b^3 B\right ) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{\sin ^2(c+d x)}}}{(a-b) (a+b)}}{12 b^2 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.989, size = 1129, normalized size = 3.3 \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 (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\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 (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\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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