Optimal. Leaf size=246 \[ \frac{(2 A-5 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(11 A-35 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(43 A-115 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(7 A-15 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.836839, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2977, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(2 A-5 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(11 A-35 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(43 A-115 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{(7 A-15 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2977
Rule 2983
Rule 2982
Rule 2782
Rule 205
Rule 2774
Rule 216
Rubi steps
\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (\frac{5}{2} a (A-B)-a (A-5 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(7 A-15 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{4} a^2 (7 A-15 B)-\frac{1}{2} a^2 (11 A-35 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(7 A-15 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(11 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{4} a^3 (11 A-35 B)+4 a^3 (2 A-5 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(7 A-15 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(11 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(43 A-115 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}+\frac{(2 A-5 B) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{2 a^3}\\ &=\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(7 A-15 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(11 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(43 A-115 B) \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{(2 A-5 B) \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^3 d}\\ &=\frac{(2 A-5 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{5/2} d}-\frac{(43 A-115 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 )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{(7 A-15 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(11 A-35 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [C] time = 3.30646, size = 376, normalized size = 1.53 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((55 B-15 A) \cos (c+d x)-11 A+8 B \cos (2 (c+d x))+43 B)+\frac{\sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (i \sqrt{2} (43 A-115 B) \log \left (1+e^{i (c+d x)}\right )-16 i (2 A-5 B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+32 i A \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-43 i \sqrt{2} A \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+32 A d x-80 i B \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+115 i \sqrt{2} B \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-80 B d x\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{8 d (a (\cos (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.632, size = 647, normalized size = 2.6 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \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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Fricas [A] time = 98.4708, size = 830, normalized size = 3.37 \begin{align*} \frac{\sqrt{2}{\left ({\left (43 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (43 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (43 \, A - 115 \, B\right )} \cos \left (d x + c\right ) + 43 \, A - 115 \, 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 ) + 2 \,{\left (16 \, B \cos \left (d x + c\right )^{2} - 5 \,{\left (3 \, A - 11 \, B\right )} \cos \left (d x + c\right ) - 11 \, A + 35 \, B\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 ({\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 5 \, 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 )}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} 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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \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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