Optimal. Leaf size=103 \[ \frac{2 a^2 (3 B+2 C) \tan (c+d x)}{3 d}+\frac{a^2 (3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (3 B+2 C) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]
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Rubi [A] time = 0.106857, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {4054, 12, 3788, 3767, 8, 4046, 3770} \[ \frac{2 a^2 (3 B+2 C) \tan (c+d x)}{3 d}+\frac{a^2 (3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (3 B+2 C) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 4054
Rule 12
Rule 3788
Rule 3767
Rule 8
Rule 4046
Rule 3770
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{\int a (3 B+2 C) \sec (c+d x) (a+a \sec (c+d x))^2 \, dx}{3 a}\\ &=\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} (3 B+2 C) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} (3 B+2 C) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} \left (2 a^2 (3 B+2 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (3 B+2 C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{2} \left (a^2 (3 B+2 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (2 a^2 (3 B+2 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{a^2 (3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{2 a^2 (3 B+2 C) \tan (c+d x)}{3 d}+\frac{a^2 (3 B+2 C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.334829, size = 63, normalized size = 0.61 \[ \frac{a^2 \left ((9 B+6 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (B+2 C) \sec (c+d x)+12 (B+C)+2 C \tan ^2(c+d x)\right )\right )}{6 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.041, size = 141, normalized size = 1.4 \begin{align*}{\frac{3\,B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{5\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+2\,{\frac{B{a}^{2}\tan \left ( dx+c \right ) }{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{B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,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 = 0.941812, size = 225, normalized size = 2.18 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 3 \, B 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 )} - 6 \, 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 )} + 12 \, B a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 24 \, B a^{2} \tan \left (d x + c\right ) + 12 \, C a^{2} \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 = 0.508882, size = 315, normalized size = 3.06 \begin{align*} \frac{3 \,{\left (3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (6 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sin \left (d x + c\right )}{12 \, 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] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int B \sec{\left (c + d x \right )}\, dx + \int 2 B \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.18855, size = 240, normalized size = 2.33 \begin{align*} \frac{3 \,{\left (3 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (3 \, B a^{2} + 2 \, 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 (9 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 24 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, 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 )}^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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