Optimal. Leaf size=125 \[ -\frac{\left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}-\frac{2 a b \tan (c+d x)}{3 d}+\frac{a b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d} \]
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Rubi [A] time = 0.439664, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {4397, 2889, 3048, 3031, 3021, 2748, 3767, 8, 3770} \[ -\frac{\left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}-\frac{2 a b \tan (c+d x)}{3 d}+\frac{a b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 2889
Rule 3048
Rule 3031
Rule 3021
Rule 2748
Rule 3767
Rule 8
Rule 3770
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx &=\int (b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=\int (b+a \cos (c+d x))^2 \left (1-\cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (b+a \cos (c+d x)) \left (2 a-b \cos (c+d x)-3 a \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{12} \int \left (-3 \left (2 a^2-b^2\right )+8 a b \cos (c+d x)+9 a^2 \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (16 a b+3 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{3} (2 a b) \int \sec ^2(c+d x) \, dx-\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{\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{(2 a b) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=-\frac{\left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{2 a b \tan (c+d x)}{3 d}+\frac{\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [B] time = 0.635513, size = 336, normalized size = 2.69 \[ \frac{\sec ^4(c+d x) \left (12 \left (4 a^2+b^2\right ) \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+3 \left (4 a^2+b^2\right ) \cos (4 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+24 a^2 \sin (c+d x)+24 a^2 \sin (3 (c+d x))+36 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-36 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+32 a b \sin (2 (c+d x))-16 a b \sin (4 (c+d x))+42 b^2 \sin (c+d x)-6 b^2 \sin (3 (c+d x))+9 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-9 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{192 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.08, size = 169, normalized size = 1.4 \begin{align*}{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}\sin \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{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\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{{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{{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 = 1.12989, size = 174, normalized size = 1.39 \begin{align*} \frac{32 \, a b \tan \left (d x + c\right )^{3} + 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 )}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.524828, size = 325, normalized size = 2.6 \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 ) + 2 \,{\left (16 \, a b \cos \left (d x + c\right )^{3} - 16 \, a b \cos \left (d x + c\right ) - 3 \,{\left (4 \, a^{2} - 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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.90036, size = 305, normalized size = 2.44 \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} - 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 64 \, a b \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} - 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 64 \, a b \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} + 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 )\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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