Optimal. Leaf size=115 \[ \frac{A \sin (c+d x)}{d \sqrt{a-a \sec (c+d x)}}+\frac{3 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d} \]
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Rubi [A] time = 0.220891, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {4022, 3920, 3774, 203, 3795} \[ \frac{A \sin (c+d x)}{d \sqrt{a-a \sec (c+d x)}}+\frac{3 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 4022
Rule 3920
Rule 3774
Rule 203
Rule 3795
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) (A+A \sec (c+d x))}{\sqrt{a-a \sec (c+d x)}} \, dx &=\frac{A \sin (c+d x)}{d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{-\frac{3 a A}{2}-\frac{1}{2} a A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{a}\\ &=\frac{A \sin (c+d x)}{d \sqrt{a-a \sec (c+d x)}}+(2 A) \int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx+\frac{(3 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{2 a}\\ &=\frac{A \sin (c+d x)}{d \sqrt{a-a \sec (c+d x)}}+\frac{(3 A) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{d}-\frac{(4 A) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{d}\\ &=\frac{3 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{A \sin (c+d x)}{d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.49979, size = 269, normalized size = 2.34 \[ \frac{A e^{-\frac{1}{2} i (c+d x)} \sin \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos \left (\frac{1}{2} (c+d x)\right )+i \sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (3 e^{-\frac{1}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+e^{-\frac{1}{2} i (c+d x)} \left (e^{-i (c+d x)}+e^{i (c+d x)}+e^{2 i (c+d x)}-4 \sqrt{2} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+3 \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+1\right )\right )}{2 d \sqrt{a-a \sec (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.299, size = 155, normalized size = 1.4 \begin{align*}{\frac{A\sqrt{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }{2\,d\sin \left ( dx+c \right ) a} \left ( 2\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{2}+3\,\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-\cos \left ( dx+c \right ) \sqrt{2} \right ) \sqrt{{\frac{a \left ( -1+\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \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 (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\sqrt{-a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.527598, size = 1112, normalized size = 9.67 \begin{align*} \left [\frac{2 \, \sqrt{2} A a \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} -{\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, A \sqrt{-a} \log \left (\frac{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} -{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \,{\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{2 \, a d \sin \left (d x + c\right )}, \frac{2 \, \sqrt{2} A \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, A \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) -{\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{a d \sin \left (d x + c\right )}\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*} A \left (\int \frac{\cos{\left (c + d x \right )}}{\sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx + \int \frac{\cos{\left (c + d x \right )} \sec{\left (c + d x \right )}}{\sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [C] time = 1.92026, size = 346, normalized size = 3.01 \begin{align*} -\frac{A a{\left (\frac{2 \, \sqrt{2} \arctan \left (\frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{\sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{3 \, \arctan \left (\frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{2 \, \sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}\right )} + \frac{{\left (2 i \, \sqrt{2} A \sqrt{-a} \arctan \left (-i\right ) - 3 i \, A \sqrt{-a} \arctan \left (-\frac{1}{2} i \, \sqrt{2}\right ) - \sqrt{2} A \sqrt{-a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{a}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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