Optimal. Leaf size=170 \[ \frac{a \left (a^2 (-C)+12 A b^2+8 b^2 C\right ) \tan (c+d x)}{6 b d}+\frac{\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d} \]
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Rubi [A] time = 0.308335, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4083, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{a \left (a^2 (-C)+12 A b^2+8 b^2 C\right ) \tan (c+d x)}{6 b d}+\frac{\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{C \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}-\frac{a C \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d} \]
Antiderivative was successfully verified.
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Rule 4083
Rule 4002
Rule 3997
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 (b (4 A+3 C)-a C \sec (c+d x)) \, dx}{4 b}\\ &=-\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (a b (12 A+7 C)-\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x)\right ) \, dx}{12 b}\\ &=-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{\int \sec (c+d x) \left (3 b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+4 a \left (12 A b^2-\left (a^2-8 b^2\right ) C\right ) \sec (c+d x)\right ) \, dx}{24 b}\\ &=-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac{\left (a \left (12 A b^2-a^2 C+8 b^2 C\right )\right ) \int \sec ^2(c+d x) \, dx}{6 b}+\frac{1}{8} \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}-\frac{\left (a \left (12 A b^2-a^2 C+8 b^2 C\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 b d}\\ &=\frac{\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (12 A b^2-a^2 C+8 b^2 C\right ) \tan (c+d x)}{6 b d}-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac{C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}\\ \end{align*}
Mathematica [B] time = 6.31943, size = 1123, normalized size = 6.61 \[ \frac{\left (-8 A a^2-4 C a^2-4 A b^2-3 b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \cos ^4(c+d x)}{4 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C)}+\frac{\left (8 A a^2+4 C a^2+4 A b^2+3 b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \cos ^4(c+d x)}{4 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C)}+\frac{2 a b C (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^4(c+d x)}{3 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{4 (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \left (3 a A b \sin \left (\frac{1}{2} (c+d x)\right )+2 a b C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)}{3 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \left (3 a A b \sin \left (\frac{1}{2} (c+d x)\right )+2 a b C \sin \left (\frac{1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)}{3 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\left (12 C a^2+8 b C a+12 A b^2+9 b^2 C\right ) (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \cos ^4(c+d x)}{24 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{\left (-12 C a^2-8 b C a-12 A b^2-9 b^2 C\right ) (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \cos ^4(c+d x)}{24 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 a b C (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^4(c+d x)}{3 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{b^2 C (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \cos ^4(c+d x)}{8 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{b^2 C (a+b \sec (c+d x))^2 \left (C \sec ^2(c+d x)+A\right ) \cos ^4(c+d x)}{8 d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.044, size = 229, normalized size = 1.4 \begin{align*}{\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 ) }{2\,d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{Aab\tan \left ( dx+c \right ) }{d}}+{\frac{4\,abC\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,abC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{A{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.988969, size = 304, normalized size = 1.79 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b - 3 \, C b^{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 )} - 12 \, 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 b^{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 )} + 48 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, A a b \tan \left (d x + c\right )}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.537459, size = 424, normalized size = 2.49 \begin{align*} \frac{3 \,{\left (4 \,{\left (2 \, A + C\right )} a^{2} +{\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \,{\left (2 \, A + C\right )} a^{2} +{\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (3 \, A + 2 \, C\right )} a b \cos \left (d x + c\right )^{3} + 16 \, C a b \cos \left (d x + c\right ) + 6 \, C b^{2} + 3 \,{\left (4 \, C a^{2} +{\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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*} \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{2} \sec{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.21912, size = 575, normalized size = 3.38 \begin{align*} \frac{3 \,{\left (8 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (8 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 144 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 80 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 144 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C b^{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 )}^{4}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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