Optimal. Leaf size=97 \[ \frac{(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d}+\frac{C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
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Rubi [A] time = 0.0843856, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {4047, 3767, 4046, 3768, 3770} \[ \frac{(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d}+\frac{C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 4047
Rule 3767
Rule 4046
Rule 3768
Rule 3770
Rubi steps
\begin{align*} \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=B \int \sec ^4(c+d x) \, dx+\int \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} (4 A+3 C) \int \sec ^3(c+d x) \, dx-\frac{B \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{B \tan (c+d x)}{d}+\frac{(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{B \tan ^3(c+d x)}{3 d}+\frac{1}{8} (4 A+3 C) \int \sec (c+d x) \, dx\\ &=\frac{(4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{B \tan (c+d x)}{d}+\frac{(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{B \tan ^3(c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.238183, size = 71, normalized size = 0.73 \[ \frac{\tan (c+d x) \left (3 (4 A+3 C) \sec (c+d x)+8 B \left (\tan ^2(c+d x)+3\right )+6 C \sec ^3(c+d x)\right )+3 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.032, size = 130, normalized size = 1.3 \begin{align*}{\frac{A\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{C \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,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 = 0.936623, size = 188, normalized size = 1.94 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B - 3 \, C{\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 \, A{\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 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.510543, size = 306, normalized size = 3.15 \begin{align*} \frac{3 \,{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, C\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 + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\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 = 1.22434, size = 311, normalized size = 3.21 \begin{align*} \frac{3 \,{\left (4 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (4 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C \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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