Optimal. Leaf size=192 \[ \frac{5 a^4 (A-2 C) \sin (c+d x)}{2 d}+\frac{2 a^4 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(3 A+22 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (13 A+2 C)-\frac{a (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d} \]
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Rubi [A] time = 0.527648, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4087, 4018, 3996, 3770} \[ \frac{5 a^4 (A-2 C) \sin (c+d x)}{2 d}+\frac{2 a^4 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(3 A+22 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (13 A+2 C)-\frac{a (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d} \]
Antiderivative was successfully verified.
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Rule 4087
Rule 4018
Rule 3996
Rule 3770
Rubi steps
\begin{align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^4 (4 a A-a (3 A-2 C) \sec (c+d x)) \, dx}{2 a}\\ &=-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^3 \left (a^2 (15 A-2 C)-6 a^2 (A-2 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (4 a^3 (9 A-4 C)+2 a^3 (3 A+22 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (30 a^4 (A-2 C)+24 a^4 (2 A+3 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac{\int \left (-6 a^5 (13 A+2 C)-24 a^5 (2 A+3 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{1}{2} a^4 (13 A+2 C) x+\frac{5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (2 a^4 (2 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^4 (13 A+2 C) x+\frac{2 a^4 (2 A+3 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end{align*}
Mathematica [B] time = 6.41151, size = 1420, normalized size = 7.4 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.102, size = 189, normalized size = 1. \begin{align*}{\frac{A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{13\,{a}^{4}Ax}{2}}+{\frac{13\,A{a}^{4}c}{2\,d}}+{a}^{4}Cx+{\frac{C{a}^{4}c}{d}}+4\,{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{20\,{a}^{4}C\tan \left ( dx+c \right ) }{3\,d}}+4\,{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{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 ) }{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 = 0.953046, size = 285, normalized size = 1.48 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 72 \,{\left (d x + c\right )} A a^{4} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 12 \,{\left (d x + c\right )} C a^{4} - 12 \, 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 )} + 24 \, A a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, 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 )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 12 \, A a^{4} \tan \left (d x + c\right ) + 72 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.543052, size = 425, normalized size = 2.21 \begin{align*} \frac{3 \,{\left (13 \, A + 2 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \,{\left (2 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 24 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (3 \, A + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 12 \, C a^{4} \cos \left (d x + c\right ) + 2 \, C a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \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 = 1.26086, size = 335, normalized size = 1.74 \begin{align*} \frac{3 \,{\left (13 \, A a^{4} + 2 \, C a^{4}\right )}{\left (d x + c\right )} + 12 \,{\left (2 \, A a^{4} + 3 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 12 \,{\left (2 \, A a^{4} + 3 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{6 \,{\left (7 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A 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 )}^{2}} - \frac{4 \,{\left (3 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 38 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, 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 )}^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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