Optimal. Leaf size=89 \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d} \]
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Rubi [A] time = 0.146448, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3904, 3887, 481, 203} \[ \frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 \sqrt{2} A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 3904
Rule 3887
Rule 481
Rule 203
Rubi steps
\begin{align*} \int \frac{A+A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx &=-\left ((a A) \int \frac{\tan ^2(c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx\right )\\ &=\frac{(2 a A) \operatorname{Subst}\left (\int \frac{x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{d}\\ &=-\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\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{\tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 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}\\ \end{align*}
Mathematica [C] time = 0.5378, size = 140, normalized size = 1.57 \[ -\frac{i A \left (-1+e^{i (c+d x)}\right ) \left (\sqrt{2} \sinh ^{-1}\left (e^{i (c+d x)}\right )-4 \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\sqrt{2} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{\sqrt{2} d \sqrt{1+e^{2 i (c+d x)}} \sqrt{a-a \sec (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.243, size = 120, normalized size = 1.4 \begin{align*}{\frac{A\sqrt{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }{d\sin \left ( dx+c \right ) a} \left ( \sqrt{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \right ) \sqrt{{\frac{a \left ( -1+\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \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 = 0.505888, size = 780, normalized size = 8.76 \begin{align*} \left [\frac{\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 ) - 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 )}{a d}, \frac{2 \,{\left (\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 ) - 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 )\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*} A \left (\int \frac{\sec{\left (c + d x \right )}}{\sqrt{- a \sec{\left (c + d x \right )} + a}}\, dx + \int \frac{1}{\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 = 1.88414, size = 225, normalized size = 2.53 \begin{align*} -\frac{2 \,{\left (A a{\left (\frac{\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{\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 )}\right )} + \frac{{\left (\sqrt{2} A \sqrt{a} \arctan \left (-i\right ) - A \sqrt{a} \arctan \left (-\frac{1}{2} i \, \sqrt{2}\right )\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{a}\right )}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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