Optimal. Leaf size=181 \[ \frac{(2 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{B \sin (c+d x)}{d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]
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Rubi [A] time = 0.512543, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{(2 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{B \sin (c+d x)}{d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]
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)} \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)} (A+B \cos (c+d x))}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{B \sin (c+d x)}{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 B}{2}+\frac{1}{2} a (2 A-B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{a}\\ &=\frac{B \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\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{\left ((2 A-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}{2 a}\\ &=\frac{B \sin (c+d x)}{d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\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{\left ((2 A-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 )}{a d}\\ &=\frac{(2 A-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)}}{\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)}{d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.36029, size = 467, normalized size = 2.58 \[ \frac{i e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt{\sec (c+d x)} \left (-(2 A-B) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+2 \sqrt{2} A e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{-1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+2 A e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+B \left (-e^{i (c+d x)}\right )+B e^{2 i (c+d x)}-B e^{3 i (c+d x)}+\sqrt{2} B e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-\sqrt{2} B e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{-1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-B e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+B\right )}{4 d \sqrt{a (\cos (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.727, size = 232, normalized size = 1.3 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{4}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( B\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) +2\,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}-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 ) +2\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -2\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \right ){\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \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 = 13.4661, size = 466, normalized size = 2.57 \begin{align*} \frac{\sqrt{a \cos \left (d x + c\right ) + a} B \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) -{\left ({\left (2 \, A - B\right )} \cos \left (d x + c\right ) + 2 \, A - 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{\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}}}{a d \cos \left (d x + c\right ) + a d} \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 + B \cos{\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{B \cos \left (d x + c\right ) + 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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