Optimal. Leaf size=219 \[ \frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{5 (15 A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.687886, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {4265, 4085, 4020, 4013, 3808, 206} \[ \frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{5 (15 A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 4265
Rule 4085
Rule 4020
Rule 4013
Rule 3808
Rule 206
Rubi steps
\begin{align*} \int \frac{\sqrt{\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a (9 A+C)+2 a (A-C) \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{4} a^2 (49 A+C)+\frac{1}{2} a^2 (13 A-3 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{\left (5 (15 A-C) \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}{32 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{\left (5 (15 A-C) \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 )}{16 a^2 d}\\ &=-\frac{5 (15 A-C) \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)}}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(13 A-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 2.67953, size = 118, normalized size = 0.54 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) (5 (17 A+C) \cos (c+d x)+16 A \cos (2 (c+d x))+65 A+C)-40 (15 A-C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{64 a d \cos ^{\frac{3}{2}}(c+d x) (a (\sec (c+d x)+1))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.354, size = 365, normalized size = 1.7 \begin{align*} -{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{16\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 32\,A\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}+75\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) +53\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-5\,C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) +5\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+75\,A\sin \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) -36\,A\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-5\,C\sin \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) -4\,C\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-49\,A\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \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 [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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Fricas [A] time = 0.553944, size = 1303, normalized size = 5.95 \begin{align*} \left [-\frac{5 \, \sqrt{2}{\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - C\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{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 ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left (32 \, A \cos \left (d x + c\right )^{2} + 5 \,{\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 49 \, A + C\right )} \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 )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac{5 \, \sqrt{2}{\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - C\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{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 )}}{a \sin \left (d x + c\right )}\right ) + 2 \,{\left (32 \, A \cos \left (d x + c\right )^{2} + 5 \,{\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 49 \, A + C\right )} \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 )}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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