Optimal. Leaf size=223 \[ \frac{(8 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{C \sin (c+d x)}{2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{C \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]
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Rubi [A] time = 0.687249, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.216, Rules used = {4221, 3046, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(8 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{C \sin (c+d x)}{2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{C \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 4221
Rule 3046
Rule 2983
Rule 2982
Rule 2782
Rule 205
Rule 2774
Rule 216
Rubi steps
\begin{align*} \int \frac{A+C \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{C \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (\frac{1}{2} a (4 A+3 C)-\frac{1}{2} a C \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a}\\ &=\frac{C \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{C \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{a^2 C}{4}+\frac{1}{4} a^2 (8 A+7 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{C \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{C \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\left ((A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx+\frac{\left ((8 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{8 a}\\ &=\frac{C \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{C \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{\left (2 a (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{d}-\frac{\left ((8 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 a d}\\ &=\frac{(8 A+7 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{4 \sqrt{a} d}-\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{\sqrt{a} d}+\frac{C \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}-\frac{C \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.21001, size = 496, normalized size = 2.22 \[ -\frac{i e^{-3 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt{\sec (c+d x)} \left ((8 A+7 C) e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-8 \sqrt{2} A e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{-1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-8 A e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+2 C e^{i (c+d x)}-3 C e^{2 i (c+d x)}+3 C e^{3 i (c+d x)}-2 C e^{4 i (c+d x)}+C e^{5 i (c+d x)}+\sqrt{2} C e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-7 \sqrt{2} C e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{-1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-7 C e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-C\right )}{16 d \sqrt{a (\cos (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.192, size = 270, normalized size = 1.2 \begin{align*}{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }{8\,ad \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( 2\,C\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -C\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) +8\,A\sqrt{2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) +7\,C\sqrt{2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) +8\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +8\,C\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \right ) \sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 21.4398, size = 529, normalized size = 2.37 \begin{align*} -\frac{{\left ({\left (8 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 7 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) - \frac{4 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right ) +{\left (A + C\right )} a\right )} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}} - \frac{{\left (2 \, C \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right )\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{4 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + C \cos ^{2}{\left (c + d x \right )}}{\sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )} \sqrt{\sec{\left (c + d x \right )}}}\, dx \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} + A}{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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