Optimal. Leaf size=212 \[ \frac{a^3 (38 A+45 B+55 C) \tan (c+d x)}{15 d}+\frac{a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (109 A+135 B+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(11 A+15 B+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac{(3 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]
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Rubi [A] time = 0.637081, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {3043, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac{a^3 (38 A+45 B+55 C) \tan (c+d x)}{15 d}+\frac{a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (109 A+135 B+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(11 A+15 B+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac{(3 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]
Antiderivative was successfully verified.
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Rule 3043
Rule 2975
Rule 2968
Rule 3021
Rule 2748
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^3 \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))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^3 (a (3 A+5 B)+a (A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac{(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^2 \left (2 a^2 (11 A+15 B+10 C)+a^2 (7 A+5 B+20 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x)) \left (a^3 (109 A+135 B+140 C)+a^3 (43 A+45 B+80 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (a^4 (109 A+135 B+140 C)+\left (a^4 (43 A+45 B+80 C)+a^4 (109 A+135 B+140 C)\right ) \cos (c+d x)+a^4 (43 A+45 B+80 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (8 a^4 (38 A+45 B+55 C)+15 a^4 (13 A+15 B+20 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac{a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{8} \left (a^3 (13 A+15 B+20 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{15} \left (a^3 (38 A+45 B+55 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{\left (a^3 (38 A+45 B+55 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a^3 (13 A+15 B+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (38 A+45 B+55 C) \tan (c+d x)}{15 d}+\frac{a^3 (109 A+135 B+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+15 B+10 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{(3 A+5 B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{20 a d}+\frac{A (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [B] time = 6.19492, size = 931, normalized size = 4.39 \[ \frac{(-13 A-15 B-20 C) (\cos (c+d x) a+a)^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{64 d}+\frac{(13 A+15 B+20 C) (\cos (c+d x) a+a)^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{64 d}+\frac{A (\cos (c+d x) a+a)^3 \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{160 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}+\frac{(\cos (c+d x) a+a)^3 \left (79 A \sin \left (\frac{1}{2} (c+d x)\right )+60 B \sin \left (\frac{1}{2} (c+d x)\right )+20 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{960 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{(\cos (c+d x) a+a)^3 \left (79 A \sin \left (\frac{1}{2} (c+d x)\right )+60 B \sin \left (\frac{1}{2} (c+d x)\right )+20 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{960 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{(\cos (c+d x) a+a)^3 \left (38 A \sin \left (\frac{1}{2} (c+d x)\right )+45 B \sin \left (\frac{1}{2} (c+d x)\right )+55 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{120 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{(\cos (c+d x) a+a)^3 \left (38 A \sin \left (\frac{1}{2} (c+d x)\right )+45 B \sin \left (\frac{1}{2} (c+d x)\right )+55 C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{120 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{(274 A+285 B+200 C) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{1920 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(-274 A-285 B-200 C) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{1920 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{(17 A+5 B) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{640 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{(-17 A-5 B) (\cos (c+d x) a+a)^3 \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{640 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{A (\cos (c+d x) a+a)^3 \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{160 d \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^5} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.095, size = 316, normalized size = 1.5 \begin{align*}{\frac{38\,A{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{19\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,{a}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{11\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{3\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{13\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{13\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+3\,{\frac{{a}^{3}B\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{5\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.06863, size = 602, normalized size = 2.84 \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^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 45 \, A a^{3}{\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^{3}{\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 \, A a^{3}{\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 )} - 180 \, B a^{3}{\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 )} - 180 \, C a^{3}{\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^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B a^{3} \tan \left (d x + c\right ) + 720 \, C a^{3} \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 = 2.04784, size = 477, normalized size = 2.25 \begin{align*} \frac{15 \,{\left (13 \, A + 15 \, B + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (13 \, A + 15 \, B + 20 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (38 \, A + 45 \, B + 55 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \,{\left (13 \, A + 15 \, B + 12 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (19 \, A + 15 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 24 \, A a^{3}\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 = 1.2301, size = 460, normalized size = 2.17 \begin{align*} \frac{15 \,{\left (13 \, A a^{3} + 15 \, B a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (13 \, A a^{3} + 15 \, B a^{3} + 20 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (195 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 225 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 300 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 910 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1050 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1400 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1920 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2560 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1330 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1830 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2120 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 735 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 660 \, C a^{3} \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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