Optimal. Leaf size=211 \[ \frac{(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(75 A-19 B-5 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.639784, 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 = {3041, 2978, 2984, 12, 2782, 205} \[ \frac{(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}-\frac{(75 A-19 B-5 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3041
Rule 2978
Rule 2984
Rule 12
Rule 2782
Rule 205
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (9 A-B+C)-2 a (A-B-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 (49 A-9 B+C)-\frac{1}{2} a^2 (13 A-5 B-3 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int -\frac{a^3 (75 A-19 B-5 C)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a^5}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(75 A-19 B-5 C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{(75 A-19 B-5 C) \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{(75 A-19 B-5 C) \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 )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B+C) \sin (c+d x)}{4 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}}-\frac{(13 A-5 B-3 C) \sin (c+d x)}{16 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}}+\frac{(49 A-9 B+C) \sin (c+d x)}{16 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [C] time = 2.44642, size = 225, normalized size = 1.07 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) (2 (85 A-13 B+5 C) \cos (c+d x)+(49 A-9 B+C) \cos (2 (c+d x))+113 A-9 B+C)}{4 \sqrt{\cos (c+d x)}}-\frac{i (75 A-19 B-5 C) e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{4 d (a (\cos (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.111, size = 675, normalized size = 3.2 \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 = 2.45177, size = 711, normalized size = 3.37 \begin{align*} -\frac{\sqrt{2}{\left ({\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (75 \, A - 19 \, B - 5 \, C\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{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (49 \, A - 9 \, B + C\right )} \cos \left (d x + c\right )^{2} +{\left (85 \, A - 13 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 32 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \,{\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{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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