Optimal. Leaf size=96 \[ \frac{a^2 (A+C) \tan (c+d x)}{d}+\frac{a^2 (2 A+C) \tanh ^{-1}(\sin (c+d x))}{d}+a^2 A x+\frac{C \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]
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Rubi [A] time = 0.140093, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4055, 3917, 3914, 3767, 8, 3770} \[ \frac{a^2 (A+C) \tan (c+d x)}{d}+\frac{a^2 (2 A+C) \tanh ^{-1}(\sin (c+d x))}{d}+a^2 A x+\frac{C \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 4055
Rule 3917
Rule 3914
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^2 \left (A+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+a \sec (c+d x))^2 (3 a A+2 a C \sec (c+d x)) \, dx}{3 a}\\ &=\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{3 d}+\frac{\int (a+a \sec (c+d x)) \left (6 a^2 A+6 a^2 (A+C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=a^2 A x+\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{3 d}+\left (a^2 (A+C)\right ) \int \sec ^2(c+d x) \, dx+\left (a^2 (2 A+C)\right ) \int \sec (c+d x) \, dx\\ &=a^2 A x+\frac{a^2 (2 A+C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{3 d}-\frac{\left (a^2 (A+C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=a^2 A x+\frac{a^2 (2 A+C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 (A+C) \tan (c+d x)}{d}+\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{C \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{3 d}\\ \end{align*}
Mathematica [B] time = 6.51247, size = 1090, normalized size = 11.35 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.051, size = 134, normalized size = 1.4 \begin{align*}{a}^{2}Ax+{\frac{A{a}^{2}c}{d}}+{\frac{5\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+2\,{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \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{{a}^{2}A\tan \left ( dx+c \right ) }{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 = 0.931493, size = 177, normalized size = 1.84 \begin{align*} \frac{6 \,{\left (d x + c\right )} A a^{2} + 2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 3 \, 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 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 6 \, A a^{2} \tan \left (d x + c\right ) + 6 \, C a^{2} \tan \left (d x + c\right )}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.517296, size = 331, normalized size = 3.45 \begin{align*} \frac{6 \, A a^{2} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, A + C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left ({\left (3 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, C a^{2} \cos \left (d x + c\right ) + C a^{2}\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] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int A\, dx + \int 2 A \sec{\left (c + d x \right )}\, dx + \int A \sec ^{2}{\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 [B] time = 1.20263, size = 252, normalized size = 2.62 \begin{align*} \frac{3 \,{\left (d x + c\right )} A a^{2} + 3 \,{\left (2 \, A a^{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 (2 \, A a^{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 (3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, 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}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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