Optimal. Leaf size=197 \[ \frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{75 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.458366, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4264, 3817, 4020, 4013, 3808, 206} \[ \frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{75 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 4264
Rule 3817
Rule 4020
Rule 4013
Rule 3808
Rule 206
Rubi steps
\begin{align*} \int \frac{\sqrt{\cos (c+d x)}}{(a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{\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{9 a}{2}+2 a \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{13 \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{49 a^2}{4}+\frac{13}{2} a^2 \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{\left (75 \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{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{49 \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{\left (75 \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{75 \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{\sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{13 \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{49 \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 = 0.930474, size = 141, normalized size = 0.72 \[ \frac{\sqrt{1-\sec (c+d x)} (32 \sin (c+d x)+\tan (c+d x) (49 \sec (c+d x)+85))+150 \sqrt{2} \sin (c+d x) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )}{16 d \sqrt{\cos (c+d x)-1} (a (\sec (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.18, size = 234, normalized size = 1.2 \begin{align*}{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{32\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 75\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\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 ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+150\,\cos \left ( dx+c \right ) \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 ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+75\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) -64\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-106\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+72\,\cos \left ( dx+c \right ) +98 \right ) } \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 = 1.843, size = 1148, normalized size = 5.83 \begin{align*} \left [\frac{75 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\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 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 49\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{75 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\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 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 49\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{\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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