Optimal. Leaf size=194 \[ \frac{a^3 (38 A+55 C) \tan (c+d x)}{15 d}+\frac{a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac{3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]
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Rubi [A] time = 0.57228, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {3044, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac{a^3 (38 A+55 C) \tan (c+d x)}{15 d}+\frac{a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (109 A+140 C) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{(11 A+10 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{30 d}+\frac{3 A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{20 a d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{5 d} \]
Antiderivative was successfully verified.
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Rule 3044
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+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 (3 a A+a (A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac{3 A \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+10 C)+a^2 (7 A+20 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{(11 A+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 \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+140 C)+a^3 (43 A+80 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{(11 A+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 \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+140 C)+\left (a^4 (43 A+80 C)+a^4 (109 A+140 C)\right ) \cos (c+d x)+a^4 (43 A+80 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+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 \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+55 C)+15 a^4 (13 A+20 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac{a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+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 \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+20 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{15} \left (a^3 (38 A+55 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+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 \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+55 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a^3 (13 A+20 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (38 A+55 C) \tan (c+d x)}{15 d}+\frac{a^3 (109 A+140 C) \sec (c+d x) \tan (c+d x)}{120 d}+\frac{(11 A+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 \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 [A] time = 1.41525, size = 294, normalized size = 1.52 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (13 A+20 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (3 A+7 C) \sin (2 c+d x)+750 A \sin (c+2 d x)+750 A \sin (3 c+2 d x)+1520 A \sin (2 c+3 d x)+195 A \sin (3 c+4 d x)+195 A \sin (5 c+4 d x)+304 A \sin (4 c+5 d x)+80 (29 A+34 C) \sin (d x)+360 C \sin (c+2 d x)+360 C \sin (3 c+2 d x)+1840 C \sin (2 c+3 d x)-360 C \sin (4 c+3 d x)+180 C \sin (3 c+4 d x)+180 C \sin (5 c+4 d x)+440 C \sin (4 c+5 d x))\right )}{15360 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.08, size = 212, normalized size = 1.1 \begin{align*}{\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}}+{\frac{5\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{38\,A{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{19\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{11\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{3\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.07989, size = 394, normalized size = 2.03 \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} + 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 )} - 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 \, 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 )} + 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 = 1.50459, size = 419, normalized size = 2.16 \begin{align*} \frac{15 \,{\left (13 \, A + 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 + 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 + 55 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \,{\left (13 \, A + 12 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (19 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 90 \, A 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 = 1.27092, size = 332, normalized size = 1.71 \begin{align*} \frac{15 \,{\left (13 \, A 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} + 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} + 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} - 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} + 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} - 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 ) + 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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