Optimal. Leaf size=270 \[ \frac{(95 A-39 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{48 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{(163 A-75 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.909944, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2955, 4020, 4022, 4013, 3808, 206} \[ \frac{(95 A-39 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{48 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{(163 A-75 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2955
Rule 4020
Rule 4022
Rule 4013
Rule 3808
Rule 206
Rubi steps
\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (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{A+B \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (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 (11 A-3 B)-3 a (A-B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (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 (95 A-39 B)-a^2 (17 A-9 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(95 A-39 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a^3 (299 A-147 B)+\frac{1}{4} a^3 (95 A-39 B) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(95 A-39 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\left ((163 A-75 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}{32 a^2}\\ &=-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(95 A-39 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\left ((163 A-75 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 )}{16 a^2 d}\\ &=\frac{(163 A-75 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)}}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{(95 A-39 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.66488, size = 183, normalized size = 0.68 \[ \frac{2 \tan (c+d x) \sqrt{1-\sec (c+d x)} \sec (c+d x) ((255 B-479 A) \cos (c+d x)+(48 B-80 A) \cos (2 (c+d x))+8 A \cos (3 (c+d x))-379 A+195 B)-12 \sqrt{2} (163 A-75 B) \sin (c+d x) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )}{96 d \sqrt{\cos (c+d x)-1} (a (\sec (c+d x)+1))^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.325, size = 439, normalized size = 1.6 \begin{align*} -{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{96\,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 ( 489\,A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\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}}-225\,B\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\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}}+978\,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 ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+64\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}-450\,B\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 ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+489\,\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 ) -384\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-225\,\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 ) +192\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}-686\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+318\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+408\,A\cos \left ( dx+c \right ) -216\,B\cos \left ( dx+c \right ) +598\,A-294\,B \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 = 0.558409, size = 1457, normalized size = 5.4 \begin{align*} \text{result too large to display} \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 (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\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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