Optimal. Leaf size=211 \[ \frac{2 (13 A-5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-5 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]
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Rubi [A] time = 0.66737, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4265, 4086, 4022, 4013, 3808, 206} \[ \frac{2 (13 A-5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-5 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 4265
Rule 4086
Rule 4022
Rule 4013
Rule 3808
Rule 206
Rubi steps
\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \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+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a (A-5 B)+\frac{1}{2} a (4 A+5 C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=-\frac{2 (A-5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^2 (13 A-5 B+15 C)-\frac{1}{2} a^2 (A-5 B) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 (13 A-5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 (A-5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\left ((A-B+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\\ &=\frac{2 (13 A-5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 (A-5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{\left (2 (A-B+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 )}{d}\\ &=-\frac{\sqrt{2} (A-B+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)}}{\sqrt{a} d}+\frac{2 (13 A-5 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{2 (A-5 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.56791, size = 163, normalized size = 0.77 \[ \frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (\sqrt{1-\sec (c+d x)} \sec ^2(c+d x) (-2 (A-5 B) \cos (c+d x)+3 A \cos (2 (c+d x))+29 A-10 B+30 C)+15 \sqrt{2} (A-B+C) \sec ^{\frac{5}{2}}(c+d x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )\right )}{15 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.404, size = 253, normalized size = 1.2 \begin{align*}{\frac{1}{15\,ad\sin \left ( dx+c \right ) }\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 15\,\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}}A\sin \left ( dx+c \right ) -6\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-15\,\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}}B\sin \left ( dx+c \right ) +15\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sin \left ( dx+c \right ) +8\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-10\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}-28\,A\cos \left ( dx+c \right ) +20\,B\cos \left ( dx+c \right ) -30\,C\cos \left ( dx+c \right ) +26\,A-10\,B+30\,C \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.46897, size = 1045, normalized size = 4.95 \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 = 0.55956, size = 1023, normalized size = 4.85 \begin{align*} \left [\frac{4 \,{\left (3 \, A \cos \left (d x + c\right )^{2} -{\left (A - 5 \, B\right )} \cos \left (d x + c\right ) + 13 \, A - 5 \, B + 15 \, 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 ) + \frac{15 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\right )} a\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}}}{30 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac{15 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\right )} a\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 ) + 2 \,{\left (3 \, A \cos \left (d x + c\right )^{2} -{\left (A - 5 \, B\right )} \cos \left (d x + c\right ) + 13 \, A - 5 \, B + 15 \, 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 )}{15 \,{\left (a d \cos \left (d x + c\right ) + a 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} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{\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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