Optimal. Leaf size=207 \[ \frac{a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac{a^4 (7 A+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(7 A+5 C) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 d}+\frac{(7 A+8 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+a^4 C x+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{5 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^4}{5 d} \]
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Rubi [A] time = 0.665237, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3044, 2975, 2968, 3021, 2735, 3770} \[ \frac{a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac{a^4 (7 A+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(7 A+5 C) \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 d}+\frac{(7 A+8 C) \tan (c+d x) \sec (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+a^4 C x+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3}{5 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^4}{5 d} \]
Antiderivative was successfully verified.
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Rule 3044
Rule 2975
Rule 2968
Rule 3021
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^4 (4 a A+5 a C \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac{a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^3 \left (4 a^2 (7 A+5 C)+20 a^2 C \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^2 \left (20 a^3 (7 A+8 C)+60 a^3 C \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x)) \left (60 a^4 (7 A+10 C)+120 a^4 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac{(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (60 a^5 (7 A+10 C)+\left (120 a^5 C+60 a^5 (7 A+10 C)\right ) \cos (c+d x)+120 a^5 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{120 a}\\ &=\frac{a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac{(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (60 a^5 (7 A+12 C)+120 a^5 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{120 a}\\ &=a^4 C x+\frac{a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac{(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{2} \left (a^4 (7 A+12 C)\right ) \int \sec (c+d x) \, dx\\ &=a^4 C x+\frac{a^4 (7 A+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^4 (7 A+10 C) \tan (c+d x)}{2 d}+\frac{(7 A+8 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{(7 A+5 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A (a+a \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 1.79188, size = 389, normalized size = 1.88 \[ \frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (\sec (c) (-480 A \sin (2 c+d x)+330 A \sin (c+2 d x)+330 A \sin (3 c+2 d x)+800 A \sin (2 c+3 d x)-30 A \sin (4 c+3 d x)+105 A \sin (3 c+4 d x)+105 A \sin (5 c+4 d x)+166 A \sin (4 c+5 d x)+1180 A \sin (d x)-780 C \sin (2 c+d x)+120 C \sin (c+2 d x)+120 C \sin (3 c+2 d x)+820 C \sin (2 c+3 d x)-180 C \sin (4 c+3 d x)+60 C \sin (3 c+4 d x)+60 C \sin (5 c+4 d x)+200 C \sin (4 c+5 d x)+150 C d x \cos (2 c+d x)+75 C d x \cos (2 c+3 d x)+75 C d x \cos (4 c+3 d x)+15 C d x \cos (4 c+5 d x)+15 C d x \cos (6 c+5 d x)+1220 C \sin (d x)+150 C d x \cos (d x))-240 (7 A+12 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{7680 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.089, size = 226, normalized size = 1.1 \begin{align*}{\frac{83\,A{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{a}^{4}Cx+{\frac{C{a}^{4}c}{d}}+{\frac{7\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{7\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+6\,{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{34\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{20\,{a}^{4}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+2\,{\frac{{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{4}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 = 1.02836, size = 425, normalized size = 2.05 \begin{align*} \frac{4 \,{\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^{4} + 120 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 20 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 60 \,{\left (d x + c\right )} C a^{4} - 15 \, A a^{4}{\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^{4}{\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 )} - 60 \, C a^{4}{\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^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 60 \, A a^{4} \tan \left (d x + c\right ) + 360 \, C a^{4} \tan \left (d x + c\right )}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.58336, size = 452, normalized size = 2.18 \begin{align*} \frac{60 \, C a^{4} d x \cos \left (d x + c\right )^{5} + 15 \,{\left (7 \, A + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (7 \, A + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (83 \, A + 100 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \,{\left (7 \, A + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (34 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, A a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{60 \, 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 = 1.26489, size = 347, normalized size = 1.68 \begin{align*} \frac{30 \,{\left (d x + c\right )} C a^{4} + 15 \,{\left (7 \, A a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (7 \, A a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (105 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 150 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 490 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 680 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 896 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1180 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 790 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 920 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 375 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 270 \, C a^{4} \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}}}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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