Optimal. Leaf size=99 \[ \frac{\left (4 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{5 a b \sec ^3(c+d x)}{12 d}+\frac{b \sec ^3(c+d x) (a+b \tan (c+d x))}{4 d} \]
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Rubi [A] time = 0.0988038, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3508, 3486, 3768, 3770} \[ \frac{\left (4 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{5 a b \sec ^3(c+d x)}{12 d}+\frac{b \sec ^3(c+d x) (a+b \tan (c+d x))}{4 d} \]
Antiderivative was successfully verified.
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Rule 3508
Rule 3486
Rule 3768
Rule 3770
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac{b \sec ^3(c+d x) (a+b \tan (c+d x))}{4 d}+\frac{1}{4} \int \sec ^3(c+d x) \left (4 a^2-b^2+5 a b \tan (c+d x)\right ) \, dx\\ &=\frac{5 a b \sec ^3(c+d x)}{12 d}+\frac{b \sec ^3(c+d x) (a+b \tan (c+d x))}{4 d}+\frac{1}{4} \left (4 a^2-b^2\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{5 a b \sec ^3(c+d x)}{12 d}+\frac{\left (4 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) (a+b \tan (c+d x))}{4 d}+\frac{1}{8} \left (4 a^2-b^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (4 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a b \sec ^3(c+d x)}{12 d}+\frac{\left (4 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) (a+b \tan (c+d x))}{4 d}\\ \end{align*}
Mathematica [A] time = 0.0594228, size = 120, normalized size = 1.21 \[ \frac{a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 \tan (c+d x) \sec (c+d x)}{2 d}+\frac{2 a b \sec ^3(c+d x)}{3 d}-\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{b^2 \tan (c+d x) \sec (c+d x)}{8 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.054, size = 143, normalized size = 1.4 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{\sin \left ( dx+c \right ){b}^{2}}{8\,d}}-{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,ab}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.42239, size = 174, normalized size = 1.76 \begin{align*} \frac{3 \, b^{2}{\left (\frac{2 \,{\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \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^{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 )} + \frac{32 \, a b}{\cos \left (d x + c\right )^{3}}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.04285, size = 289, normalized size = 2.92 \begin{align*} \frac{3 \,{\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 32 \, a b \cos \left (d x + c\right ) + 6 \,{\left ({\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, b^{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 + b \tan{\left (c + d x \right )}\right )^{2} \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 = 1.49328, size = 336, normalized size = 3.39 \begin{align*} \frac{3 \,{\left (4 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (4 \, a^{2} - 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 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 21 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 16 \, a b\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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