Optimal. Leaf size=178 \[ \frac{a^2 (18 A+25 C) \tan (c+d x)}{15 d}+\frac{a^2 (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac{a^2 (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{10 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
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Rubi [A] time = 0.4736, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3044, 2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{a^2 (18 A+25 C) \tan (c+d x)}{15 d}+\frac{a^2 (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac{a^2 (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{10 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 3044
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+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 (2 a A+a (2 A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 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 (2 a^2 (9 A+10 C)+4 a^2 (3 A+5 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (2 a^3 (9 A+10 C)+\left (4 a^3 (3 A+5 C)+2 a^3 (9 A+10 C)\right ) \cos (c+d x)+4 a^3 (3 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac{a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{\int \left (30 a^3 (3 A+4 C)+4 a^3 (18 A+25 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac{a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{2} \left (a^2 (3 A+4 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{15} \left (a^2 (18 A+25 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac{a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 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 (3 A+4 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (18 A+25 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a^2 (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 (18 A+25 C) \tan (c+d x)}{15 d}+\frac{a^2 (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac{a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 1.39847, size = 292, normalized size = 1.64 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (3 A+4 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-120 (A+3 C) \sin (2 c+d x)+210 A \sin (c+2 d x)+210 A \sin (3 c+2 d x)+360 A \sin (2 c+3 d x)+45 A \sin (3 c+4 d x)+45 A \sin (5 c+4 d x)+72 A \sin (4 c+5 d x)+40 (15 A+16 C) \sin (d x)+120 C \sin (c+2 d x)+120 C \sin (3 c+2 d x)+440 C \sin (2 c+3 d x)-60 C \sin (4 c+3 d x)+60 C \sin (3 c+4 d x)+60 C \sin (5 c+4 d x)+100 C \sin (4 c+5 d x))\right )}{3840 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.095, size = 210, normalized size = 1.2 \begin{align*}{\frac{6\,A{a}^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{5\,{a}^{2}C\tan \left ( dx+c \right ) }{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{{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}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{2}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 = 1.09912, size = 294, normalized size = 1.65 \begin{align*} \frac{8 \,{\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} + 40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, 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 )} - 60 \, 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 )} + 120 \, C a^{2} \tan \left (d x + c\right )}{120 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.52733, size = 409, normalized size = 2.3 \begin{align*} \frac{15 \,{\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (4 \,{\left (18 \, A + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \,{\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \,{\left (9 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, A a^{2} \cos \left (d x + c\right ) + 12 \, A a^{2}\right )} \sin \left (d x + c\right )}{120 \, 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 = 1.20525, size = 332, normalized size = 1.87 \begin{align*} \frac{15 \,{\left (3 \, A a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (3 \, A a^{2} + 4 \, 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 (45 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 210 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 280 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 432 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 560 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 270 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 520 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 195 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 180 \, 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}}}{60 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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