Optimal. Leaf size=281 \[ \frac{(5 A-49 B) \sin (c+d x)}{64 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac{(5 A-177 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}+\frac{(5 A-17 B) \sin (c+d x)}{48 a d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}+\frac{(A-B) \sin (c+d x)}{6 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]
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Rubi [A] time = 0.919124, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2961, 2977, 2982, 2782, 205, 2774, 216} \[ \frac{(5 A-49 B) \sin (c+d x)}{64 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac{(5 A-177 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{7/2} d}+\frac{(5 A-17 B) \sin (c+d x)}{48 a d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}+\frac{(A-B) \sin (c+d x)}{6 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]
Antiderivative was successfully verified.
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Rule 2961
Rule 2977
Rule 2982
Rule 2782
Rule 205
Rule 2774
Rule 216
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{5}{2} a (A-B)+6 a B \cos (c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(5 A-17 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{4} a^2 (5 A-17 B)+24 a^2 B \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(5 A-17 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(5 A-49 B) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{8} a^3 (5 A-49 B)+48 a^3 B \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(5 A-17 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(5 A-49 B) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left ((5 A-177 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}{128 a^3}+\frac{\left (B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{a^4}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(5 A-17 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(5 A-49 B) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}-\frac{\left ((5 A-177 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 )}{64 a^2 d}-\frac{\left (2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^4 d}\\ &=\frac{2 B \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{a^{7/2} d}+\frac{(5 A-177 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)}}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{5}{2}}(c+d x)}+\frac{(5 A-17 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(5 A-49 B) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 3.91197, size = 281, normalized size = 1. \[ \frac{\cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{8} \left (\sin \left (\frac{3}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} (4 (25 A-181 B) \cos (c+d x)+(67 A-247 B) \cos (2 (c+d x))+97 A-541 B)-3 i \sqrt{2} e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (-\sqrt{2} (5 A-177 B) \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+128 B \sinh ^{-1}\left (e^{i (c+d x)}\right )-128 B \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )\right )}{48 d (a (\cos (c+d x)+1))^{7/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.675, size = 667, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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 = 98.8766, size = 927, normalized size = 3.3 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (5 \, A - 177 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 177 \, B\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 ) + 768 \,{\left (B \cos \left (d x + c\right )^{4} + 4 \, B \cos \left (d x + c\right )^{3} + 6 \, B \cos \left (d x + c\right )^{2} + 4 \, B \cos \left (d x + c\right ) + B\right )} \sqrt{a} \arctan \left (\frac{\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 (67 \, A - 247 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (25 \, A - 181 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (5 \, A - 49 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\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{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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