Optimal. Leaf size=68 \[ \frac{\sqrt{a} (B+2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{a B \sin (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]
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Rubi [A] time = 0.209768, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4072, 4015, 3774, 203} \[ \frac{\sqrt{a} (B+2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{a B \sin (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4015
Rule 3774
Rule 203
Rubi steps
\begin{align*} \int \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos (c+d x) \sqrt{a+a \sec (c+d x)} (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+\frac{1}{2} (B+2 C) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a B \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}-\frac{(a (B+2 C)) \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{\sqrt{a} (B+2 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}+\frac{a B \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.241721, size = 93, normalized size = 1.37 \[ \frac{\sqrt{\cos (c+d x)} \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (\sqrt{2} (B+2 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 B \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)}\right )}{2 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.316, size = 198, normalized size = 2.9 \begin{align*} -{\frac{1}{2\,d\sin \left ( dx+c \right ) } \left ( B\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\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}}}\sin \left ( dx+c \right ) +2\,C\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) +2\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,B\cos \left ( dx+c \right ) \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.97053, size = 1268, normalized size = 18.65 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.636405, size = 694, normalized size = 10.21 \begin{align*} \left [\frac{2 \, B \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left ({\left (B + 2 \, C\right )} \cos \left (d x + c\right ) + B + 2 \, C\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right )}{2 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, \frac{B \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left ({\left (B + 2 \, C\right )} \cos \left (d x + c\right ) + B + 2 \, C\right )} \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 )}{d \cos \left (d x + c\right ) + 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 [B] time = 6.32136, size = 450, normalized size = 6.62 \begin{align*} -\frac{{\left (B \sqrt{-a} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, C \sqrt{-a} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right ) -{\left (B \sqrt{-a} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, C \sqrt{-a} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right ) + \frac{4 \, \sqrt{2}{\left (3 \,{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} B \sqrt{-a} a \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - B \sqrt{-a} a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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