Optimal. Leaf size=138 \[ \frac{b \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{1}{2} a x \left (a^2-6 b^2\right )-\frac{15 a^2 b \sin (c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a \tan (c+d x) (a \cos (c+d x)+b)^2}{2 d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^3}{2 d} \]
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Rubi [A] time = 0.503821, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2889, 3048, 3047, 3033, 3023, 2735, 3770} \[ \frac{b \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{1}{2} a x \left (a^2-6 b^2\right )-\frac{15 a^2 b \sin (c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a \tan (c+d x) (a \cos (c+d x)+b)^2}{2 d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^3}{2 d} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2889
Rule 3048
Rule 3047
Rule 3033
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^3 \sin ^2(c+d x) \, dx &=-\int (-b-a \cos (c+d x))^3 \sec (c+d x) \tan ^2(c+d x) \, dx\\ &=-\int (-b-a \cos (c+d x))^3 \left (1-\cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int (-b-a \cos (c+d x))^2 \left (-3 a+b \cos (c+d x)+4 a \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int (-b-a \cos (c+d x)) \left (6 a^2-b^2-5 a b \cos (c+d x)-10 a^2 \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{4} \int \left (-2 b \left (6 a^2-b^2\right )-2 a \left (a^2-6 b^2\right ) \cos (c+d x)+30 a^2 b \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{15 a^2 b \sin (c+d x)}{2 d}-\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{4} \int \left (-2 b \left (6 a^2-b^2\right )-2 a \left (a^2-6 b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} a \left (a^2-6 b^2\right ) x-\frac{15 a^2 b \sin (c+d x)}{2 d}-\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (b \left (6 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a \left (a^2-6 b^2\right ) x+\frac{b \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{15 a^2 b \sin (c+d x)}{2 d}-\frac{5 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a (b+a \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac{(b+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [B] time = 0.865642, size = 327, normalized size = 2.37 \[ \frac{\sec ^2(c+d x) \left (\left (2 b^3-3 a^2 b\right ) \sin (c+d x)+\cos (2 (c+d x)) \left (a \left (a^2-6 b^2\right ) (c+d x)+\left (b^3-6 a^2 b\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-b \left (b^2-6 a^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-3 a^2 b \sin (3 (c+d x))-6 a^2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 a^2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\frac{1}{2} a^3 \sin (2 (c+d x))-\frac{1}{4} a^3 \sin (4 (c+d x))+a^3 c+a^3 d x+6 a b^2 \sin (2 (c+d x))-6 a b^2 c-6 a b^2 d x+b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.042, size = 167, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}x}{2}}+{\frac{{a}^{3}c}{2\,d}}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) }{d}}-3\,a{b}^{2}x+3\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}-3\,{\frac{a{b}^{2}c}{d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{{b}^{3}\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 = 1.49323, size = 174, normalized size = 1.26 \begin{align*} \frac{{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2} - b^{3}{\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 )} + 6 \, a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.91372, size = 359, normalized size = 2.6 \begin{align*} \frac{2 \,{\left (a^{3} - 6 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} +{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{2} b \cos \left (d x + c\right )^{2} - 6 \, a b^{2} \cos \left (d x + c\right ) - b^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \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 [B] time = 1.58889, size = 467, normalized size = 3.38 \begin{align*} \frac{{\left (a^{3} - 6 \, a b^{2}\right )}{\left (d x + c\right )} +{\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \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 )^{4} - 1\right )}^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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