Optimal. Leaf size=196 \[ \frac{a^2 (18 A+20 B+25 C) \tan (c+d x)}{15 d}+\frac{a^2 (6 A+7 B+8 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (18 A+25 B+20 C) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{a^2 (6 A+7 B+8 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{(2 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
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Rubi [A] time = 0.512652, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.22, Rules used = {3043, 2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{a^2 (18 A+20 B+25 C) \tan (c+d x)}{15 d}+\frac{a^2 (6 A+7 B+8 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (18 A+25 B+20 C) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac{a^2 (6 A+7 B+8 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{(2 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{20 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 3043
Rule 2975
Rule 2968
Rule 3021
Rule 2748
Rule 3768
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^2 (a (2 A+5 B)+a (2 A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac{(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x)) \left (a^2 (18 A+25 B+20 C)+2 a^2 (6 A+5 B+10 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (a^3 (18 A+25 B+20 C)+\left (2 a^3 (6 A+5 B+10 C)+a^3 (18 A+25 B+20 C)\right ) \cos (c+d x)+2 a^3 (6 A+5 B+10 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (15 a^3 (6 A+7 B+8 C)+4 a^3 (18 A+20 B+25 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} \left (a^2 (6 A+7 B+8 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{15} \left (a^2 (18 A+20 B+25 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (6 A+7 B+8 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{8} \left (a^2 (6 A+7 B+8 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (18 A+20 B+25 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a^2 (6 A+7 B+8 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (18 A+20 B+25 C) \tan (c+d x)}{15 d}+\frac{a^2 (6 A+7 B+8 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [B] time = 5.30991, size = 502, normalized size = 2.56 \[ \frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (\frac{16 (18 A+20 B+25 C) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{2 (39 A+20 (2 B+C)) \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{16 (18 A+20 B+25 C) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{129 A+145 B+140 C}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{-129 A-5 (29 B+28 C)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 (39 A+20 (2 B+C)) \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}-30 (6 A+7 B+8 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+30 (6 A+7 B+8 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{3 (12 A+5 B)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{3 (12 A+5 B)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{12 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}+\frac{12 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}\right )}{960 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.088, size = 315, normalized size = 1.6 \begin{align*}{\frac{6\,A{a}^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{3\,A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{{a}^{2}B \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{7\,{a}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{5\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{4\,{a}^{2}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,{a}^{2}B \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.01797, size = 486, normalized size = 2.48 \begin{align*} \frac{16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 30 \, A a^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, C a^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, C a^{2} \tan \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.09134, size = 463, normalized size = 2.36 \begin{align*} \frac{15 \,{\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (18 \, A + 20 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \,{\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (9 \, A + 10 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \,{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 24 \, A a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \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 = 1.2344, size = 460, normalized size = 2.35 \begin{align*} \frac{15 \,{\left (6 \, A a^{2} + 7 \, B a^{2} + 8 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (6 \, A a^{2} + 7 \, B a^{2} + 8 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (90 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 105 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 120 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 420 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 490 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 560 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 864 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 800 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1120 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 540 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 790 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1040 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 390 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 375 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 360 \, C a^{2} \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 )}^{5}}}{120 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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