Optimal. Leaf size=110 \[ \frac{\left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2+3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
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Rubi [A] time = 0.0941703, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3788, 3767, 4046, 3768, 3770} \[ \frac{\left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (4 a^2+3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{b^2 \tan (c+d x) \sec ^3(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 3788
Rule 3767
Rule 4046
Rule 3768
Rule 3770
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^2 \, dx &=(2 a b) \int \sec ^4(c+d x) \, dx+\int \sec ^3(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \left (4 a^2+3 b^2\right ) \int \sec ^3(c+d x) \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{2 a b \tan (c+d x)}{d}+\frac{\left (4 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{1}{8} \left (4 a^2+3 b^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{2 a b \tan (c+d x)}{d}+\frac{\left (4 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{2 a b \tan ^3(c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.277427, size = 82, normalized size = 0.75 \[ \frac{3 \left (4 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 a^2+3 b^2\right ) \sec (c+d x)+16 a b \left (\tan ^2(c+d x)+3\right )+6 b^2 \sec ^3(c+d x)\right )}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.028, size = 142, normalized size = 1.3 \begin{align*}{\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}}+{\frac{4\,ab\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,ab\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{2}\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 = 1.10035, size = 194, normalized size = 1.76 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b - 3 \, 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 \, 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 )}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.76404, size = 332, normalized size = 3.02 \begin{align*} \frac{3 \,{\left (4 \, a^{2} + 3 \, 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} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (32 \, a b \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right ) + 3 \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, 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 \sec{\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 = 1.35778, size = 348, normalized size = 3.16 \begin{align*} \frac{3 \,{\left (4 \, a^{2} + 3 \, 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} + 3 \, 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} - 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 80 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, 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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