Optimal. Leaf size=184 \[ -\frac{\sqrt{2} (a-b) (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 b B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]
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Rubi [A] time = 0.602296, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 54, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4265, 4086, 4023, 3808, 206, 3801, 215} \[ -\frac{\sqrt{2} (a-b) (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 b B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 4265
Rule 4086
Rule 4023
Rule 3808
Rule 206
Rule 3801
Rule 215
Rubi steps
\begin{align*} \int \frac{\sqrt{\cos (c+d x)} \left (a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{2} a (A b-a (A-B))+\frac{1}{2} a b B \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{a}\\ &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\left ((a-b) (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx+\frac{\left (b B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{a}\\ &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{\left (2 (a-b) (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}-\frac{\left (2 b 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 \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a d}\\ &=\frac{2 b B \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{\sqrt{a} d}-\frac{\sqrt{2} (a-b) (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{\sqrt{a} d}+\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.416894, size = 137, normalized size = 0.74 \[ \frac{\sin (c+d x) \left (\sqrt{2} (a-b) (A-B) \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )+2 a A \sqrt{1-\sec (c+d x)}-2 b B \sqrt{\sec (c+d x)} \sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right )}{d \sqrt{\cos (c+d x)-1} \sqrt{a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.389, size = 282, normalized size = 1.5 \begin{align*}{\frac{-1+\cos \left ( dx+c \right ) }{ad \left ( \sin \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( -2\,A\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}a-B\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) b+B\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) b+2\,A\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) a-2\,A\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) b-2\,B\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) a+2\,B\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) b \right ) \sqrt{\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.69847, size = 1202, normalized size = 6.53 \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 = 0.804709, size = 1439, normalized size = 7.82 \begin{align*} \left [\frac{4 \, A a \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left (B b \cos \left (d x + c\right ) + B b\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 4 \, \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}{\left (\cos \left (d x + c\right ) - 2\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac{\sqrt{2}{\left ({\left (A - B\right )} a^{2} -{\left (A - B\right )} a b +{\left ({\left (A - B\right )} a^{2} -{\left (A - B\right )} a b\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} + \frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}}}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac{2 \, A a \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt{2}{\left ({\left (A - B\right )} a^{2} -{\left (A - B\right )} a b +{\left ({\left (A - B\right )} a^{2} -{\left (A - B\right )} a b\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{1}{a}} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \sqrt{\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) +{\left (B b \cos \left (d x + c\right ) + B b\right )} \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right ) \sqrt{\cos{\left (c + d x \right )}}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\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{{\left (B b \sec \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \sec \left (d x + c\right )\right )} \sqrt{\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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