Optimal. Leaf size=270 \[ \frac{(95 A-39 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{48 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(299 A-147 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{48 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(163 A-75 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.928702, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2961, 2978, 2984, 12, 2782, 205} \[ \frac{(95 A-39 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{48 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(299 A-147 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{48 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(163 A-75 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2961
Rule 2978
Rule 2984
Rule 12
Rule 2782
Rule 205
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (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) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (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) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (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) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(299 A-147 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{3 a^4 (163 A-75 B)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{6 a^6}\\ &=-\frac{(299 A-147 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((163 A-75 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(299 A-147 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (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^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac{(163 A-75 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{16 \sqrt{2} a^{5/2} d}-\frac{(299 A-147 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [C] time = 3.58877, size = 243, normalized size = 0.9 \[ \frac{i \cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (3 (163 A-75 B) e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\frac{1}{8} i \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((1537 A-825 B) \cos (c+d x)+2 (503 A-255 B) \cos (2 (c+d x))+299 A \cos (3 (c+d x))+878 A-147 B \cos (3 (c+d x))-510 B)\right )}{12 d (a (\cos (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.638, size = 585, normalized size = 2.2 \begin{align*} \text{result too large to display} \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 = 1.83528, size = 672, normalized size = 2.49 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) + \frac{2 \,{\left ({\left (299 \, A - 147 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (503 \, A - 255 \, B\right )} \cos \left (d x + c\right )^{2} + 32 \,{\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 32 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{96 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\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 (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \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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