Optimal. Leaf size=170 \[ \frac{4 (2 A-3 B) \sin ^3(c+d x)}{3 a^2 d}-\frac{4 (2 A-3 B) \sin (c+d x)}{a^2 d}+\frac{(7 A-10 B) \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{(7 A-10 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{x (7 A-10 B)}{2 a^2}+\frac{(A-B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.322422, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2977, 2748, 2635, 8, 2633} \[ \frac{4 (2 A-3 B) \sin ^3(c+d x)}{3 a^2 d}-\frac{4 (2 A-3 B) \sin (c+d x)}{a^2 d}+\frac{(7 A-10 B) \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{(7 A-10 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{x (7 A-10 B)}{2 a^2}+\frac{(A-B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2977
Rule 2748
Rule 2635
Rule 8
Rule 2633
Rubi steps
\begin{align*} \int \frac{\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^3(c+d x) (4 a (A-B)-3 a (A-2 B) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac{(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \cos ^2(c+d x) \left (3 a^2 (7 A-10 B)-12 a^2 (2 A-3 B) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac{(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(7 A-10 B) \int \cos ^2(c+d x) \, dx}{a^2}-\frac{(4 (2 A-3 B)) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=\frac{(7 A-10 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{(7 A-10 B) \int 1 \, dx}{2 a^2}+\frac{(4 (2 A-3 B)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=\frac{(7 A-10 B) x}{2 a^2}-\frac{4 (2 A-3 B) \sin (c+d x)}{a^2 d}+\frac{(7 A-10 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{(7 A-10 B) \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{4 (2 A-3 B) \sin ^3(c+d x)}{3 a^2 d}\\ \end{align*}
Mathematica [B] time = 0.617073, size = 369, normalized size = 2.17 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (36 d x (7 A-10 B) \cos \left (c+\frac{d x}{2}\right )+36 d x (7 A-10 B) \cos \left (\frac{d x}{2}\right )+147 A \sin \left (c+\frac{d x}{2}\right )-239 A \sin \left (c+\frac{3 d x}{2}\right )-63 A \sin \left (2 c+\frac{3 d x}{2}\right )-15 A \sin \left (2 c+\frac{5 d x}{2}\right )-15 A \sin \left (3 c+\frac{5 d x}{2}\right )+3 A \sin \left (3 c+\frac{7 d x}{2}\right )+3 A \sin \left (4 c+\frac{7 d x}{2}\right )+84 A d x \cos \left (c+\frac{3 d x}{2}\right )+84 A d x \cos \left (2 c+\frac{3 d x}{2}\right )-381 A \sin \left (\frac{d x}{2}\right )-156 B \sin \left (c+\frac{d x}{2}\right )+342 B \sin \left (c+\frac{3 d x}{2}\right )+118 B \sin \left (2 c+\frac{3 d x}{2}\right )+30 B \sin \left (2 c+\frac{5 d x}{2}\right )+30 B \sin \left (3 c+\frac{5 d x}{2}\right )-3 B \sin \left (3 c+\frac{7 d x}{2}\right )-3 B \sin \left (4 c+\frac{7 d x}{2}\right )+B \sin \left (4 c+\frac{9 d x}{2}\right )+B \sin \left (5 c+\frac{9 d x}{2}\right )-120 B d x \cos \left (c+\frac{3 d x}{2}\right )-120 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+516 B \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.066, size = 322, normalized size = 1.9 \begin{align*}{\frac{A}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7\,A}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{9\,B}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}A}{{a}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+10\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}B}{{a}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-8\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}A}{{a}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{40\,B}{3\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-3\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{{a}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+6\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{{a}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{{a}^{2}d}}-10\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{{a}^{2}d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.01015, size = 502, normalized size = 2.95 \begin{align*} \frac{B{\left (\frac{4 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{60 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - A{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.38919, size = 387, normalized size = 2.28 \begin{align*} \frac{3 \,{\left (7 \, A - 10 \, B\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (7 \, A - 10 \, B\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (7 \, A - 10 \, B\right )} d x +{\left (2 \, B \cos \left (d x + c\right )^{4} +{\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} - 6 \,{\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} -{\left (43 \, A - 66 \, B\right )} \cos \left (d x + c\right ) - 32 \, A + 48 \, B\right )} \sin \left (d x + c\right )}{6 \,{\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 [A] time = 21.0575, size = 1425, normalized size = 8.38 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19495, size = 259, normalized size = 1.52 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}{\left (7 \, A - 10 \, B\right )}}{a^{2}} - \frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 30 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 18 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 21 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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