Optimal. Leaf size=197 \[ \frac{(2 A-3 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(5 A-9 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{2 a d \sqrt{a \cos (c+d x)+a}} \]
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Rubi [A] time = 0.636338, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{(2 A-3 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(5 A-9 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{2 a d \sqrt{a \cos (c+d x)+a}} \]
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{3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} a (A-B)-a (A-3 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{2} a^2 (A-3 B)+a^2 (2 A-3 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^3}\\ &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}-\frac{(5 A-9 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a}+\frac{(2 A-3 B) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{(5 A-9 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 )}{2 d}-\frac{(2 A-3 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^2 d}\\ &=\frac{(2 A-3 B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{3/2} d}-\frac{(5 A-9 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 )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [C] time = 2.10118, size = 362, normalized size = 1.84 \[ \frac{\cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (\frac{2 \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) (-A+2 B \cos (c+d x)+3 B)}{d}+\frac{\sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (i \sqrt{2} (5 A-9 B) \log \left (1+e^{i (c+d x)}\right )-2 i (2 A-3 B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+4 i A \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-5 i \sqrt{2} A \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+4 A d x-6 i B \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+9 i \sqrt{2} B \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-6 B d x\right )}{d \sqrt{1+e^{2 i (c+d x)}}}\right )}{2 (a (\cos (c+d x)+1))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.602, size = 379, normalized size = 1.9 \begin{align*} -{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{7}{a}^{2}} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 2\,A \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+5\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \sqrt{2}\cos \left ( dx+c \right ) -4\,B\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-9\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \sqrt{2}\cos \left ( dx+c \right ) -2\,A \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}+8\,A\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -2\,B\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-12\,B\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +6\,B\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \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{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 40.7271, size = 644, normalized size = 3.27 \begin{align*} \frac{\sqrt{2}{\left ({\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 9 \, 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 (2 \, B \cos \left (d x + c\right ) - A + 3 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 4 \,{\left ({\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 3 \, 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 )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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