Optimal. Leaf size=118 \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a+b)}-\frac{2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{2 A \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d} \]
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Rubi [A] time = 0.527507, antiderivative size = 118, 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} \[ -\frac{2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a+b)}-\frac{2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{2 A \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d} \]
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)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx &=\frac{2 A \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 \int \frac{\frac{1}{2} (-A b+a B)-\frac{1}{2} a (A-C) \cos (c+d x)-\frac{1}{2} A b \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a}\\ &=\frac{2 A \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}-\frac{A \int \sqrt{\cos (c+d x)} \, dx}{a}-\frac{2 \int \frac{\frac{1}{2} b (A b-a B)-\frac{1}{2} a b C \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a b}\\ &=-\frac{2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{2 A \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{C \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{b}-\left (\frac{A b}{a}-B+\frac{a C}{b}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx\\ &=-\frac{2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d}-\frac{2 \left (\frac{A b}{a}-B+\frac{a C}{b}\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{(a+b) d}+\frac{2 A \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.23216, size = 216, normalized size = 1.83 \[ \frac{-\frac{2 A \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}+\frac{2 (2 a B-3 A b) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}-\frac{4 a (A-C) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}+\frac{4 A \sin (c+d x)}{\sqrt{\cos (c+d x)}}}{2 a d} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.899, size = 411, normalized size = 3.5 \begin{align*} -{\frac{1}{d}\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 2\,{\frac{C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{b\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}-4\,{\frac{ \left ( -A{b}^{2}+abB-{a}^{2}C \right ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}}{a \left ( -2\,ab+2\,{b}^{2} \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{\it EllipticPi} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,-2\,{\frac{b}{a-b}},\sqrt{2} \right ) }+2\,{\frac{A \left ( -\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +2\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{3}{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{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 )} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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