Optimal. Leaf size=169 \[ -\frac{(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(19 A-75 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac{(5 A-13 B) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.420275, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2977, 2968, 3019, 2751, 2649, 206} \[ -\frac{(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(19 A-75 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac{(5 A-13 B) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2977
Rule 2968
Rule 3019
Rule 2751
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\cos (c+d x) \left (2 a (A-B)-\frac{1}{2} a (A-9 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{2 a (A-B) \cos (c+d x)-\frac{1}{2} a (A-9 B) \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{\int \frac{-\frac{3}{4} a^2 (5 A-13 B)+a^2 (A-9 B) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(19 A-75 B) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(19 A-75 B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{(19 A-75 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac{(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.685333, size = 100, normalized size = 0.59 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) ((85 B-13 A) \cos (c+d x)-9 A+16 B \cos (2 (c+d x))+65 B)+2 (19 A-75 B) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{16 a d (a (\cos (c+d x)+1))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 2.579, size = 327, normalized size = 1.9 \begin{align*}{\frac{1}{32\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 19\,A\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-75\,B\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+64\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-13\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+21\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+2\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-2\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{a}^{-{\frac{7}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \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 = 1.69405, size = 628, normalized size = 3.72 \begin{align*} -\frac{\sqrt{2}{\left ({\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right ) + 19 \, A - 75 \, B\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left (32 \, B \cos \left (d x + c\right )^{2} -{\left (13 \, A - 85 \, B\right )} \cos \left (d x + c\right ) - 9 \, A + 49 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \,{\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 [A] time = 2.18537, size = 244, normalized size = 1.44 \begin{align*} \frac{\frac{{\left ({\left (\frac{2 \,{\left (\sqrt{2} A a^{6} - \sqrt{2} B a^{6}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8}} - \frac{9 \, \sqrt{2} A a^{6} - 17 \, \sqrt{2} B a^{6}}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{11 \, \sqrt{2} A a^{6} - 83 \, \sqrt{2} B a^{6}}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{{\left (19 \, \sqrt{2} A - 75 \, \sqrt{2} B\right )} \log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac{5}{2}}}}{32 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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