Optimal. Leaf size=230 \[ -\frac{(4 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 \sqrt{a} d}+\frac{(4 A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{B \sin (c+d x)}{2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}} \]
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Rubi [A] time = 0.699175, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2961, 2983, 2982, 2782, 205, 2774, 216} \[ -\frac{(4 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 \sqrt{a} d}+\frac{(4 A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{B \sin (c+d x)}{2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 2961
Rule 2983
Rule 2982
Rule 2782
Rule 205
Rule 2774
Rule 216
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)}{\sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x))}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{B \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{3 a B}{2}+\frac{1}{2} a (4 A-B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a}\\ &=\frac{B \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(4 A-B) \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{1}{4} a^2 (4 A-B)-\frac{1}{4} a^2 (4 A-7 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{B \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(4 A-B) \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{\left ((4 A-7 B) \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}+\left ((A-B) \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{B \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(4 A-B) \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{\left ((4 A-7 B) \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{\left (2 a (A-B) \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{(4 A-7 B) \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-B) \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{B \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(4 A-B) \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.41316, size = 412, normalized size = 1.79 \[ -\frac{i e^{-3 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt{\sec (c+d x)} \left (-(4 A-7 B) e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-8 \sqrt{2} (A-B) e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-4 A e^{i (c+d x)}+4 A e^{2 i (c+d x)}-4 A e^{3 i (c+d x)}+4 A e^{4 i (c+d x)}+4 A e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+2 B e^{i (c+d x)}-3 B e^{2 i (c+d x)}+3 B e^{3 i (c+d x)}-2 B e^{4 i (c+d x)}+B e^{5 i (c+d x)}-7 B e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-B\right )}{16 d \sqrt{a (\cos (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.786, size = 300, normalized size = 1.3 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) ^{3}}{8\,da \left ( \sin \left ( dx+c \right ) \right ) ^{6}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( -2\,B\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -4\,A\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) +B\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) +4\,A\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 ) \sqrt{2}-7\,B\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\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{-{\frac{3}{2}}} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \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{B \cos \left (d x + c\right ) + A}{\sqrt{a \cos \left (d x + c\right ) + a} \sec \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 [A] time = 25.8801, size = 539, normalized size = 2.34 \begin{align*} \frac{{\left ({\left (4 \, A - 7 \, B\right )} \cos \left (d x + c\right ) + 4 \, A - 7 \, B\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 - B\right )} a \cos \left (d x + c\right ) +{\left (A - B\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 \, B \cos \left (d x + c\right )^{2} +{\left (4 \, A - B\right )} \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(-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{B \cos \left (d x + c\right ) + A}{\sqrt{a \cos \left (d x + c\right ) + a} \sec \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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