Optimal. Leaf size=169 \[ \frac{a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac{a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac{b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (9 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{11 a^2 b \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))}{5 d} \]
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Rubi [A] time = 0.216605, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2792, 3021, 2748, 3767, 3768, 3770} \[ \frac{a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac{a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac{b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (9 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{11 a^2 b \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))}{5 d} \]
Antiderivative was successfully verified.
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Rule 2792
Rule 3021
Rule 2748
Rule 3767
Rule 3768
Rule 3770
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx &=\frac{a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int \left (11 a^2 b+a \left (4 a^2+15 b^2\right ) \cos (c+d x)+b \left (3 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int \left (4 a \left (4 a^2+15 b^2\right )+5 b \left (9 a^2+4 b^2\right ) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} \left (b \left (9 a^2+4 b^2\right )\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{5} \left (a \left (4 a^2+15 b^2\right )\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{b \left (9 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{8} \left (b \left (9 a^2+4 b^2\right )\right ) \int \sec (c+d x) \, dx-\frac{\left (a \left (4 a^2+15 b^2\right )\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac{b \left (9 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d}\\ \end{align*}
Mathematica [A] time = 0.839601, size = 120, normalized size = 0.71 \[ \frac{15 b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 a \left (5 \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)+15 \left (a^2+3 b^2\right )+3 a^2 \tan ^4(c+d x)\right )+15 b \left (9 a^2+4 b^2\right ) \sec (c+d x)+90 a^2 b \sec ^3(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.064, size = 206, normalized size = 1.2 \begin{align*}{\frac{8\,{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{3\,{a}^{2}b \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{9\,{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+2\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{a{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) \sec \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 = 0.976783, size = 244, normalized size = 1.44 \begin{align*} \frac{16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{2} - 45 \, a^{2} b{\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 )} - 60 \, 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 )}}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.00369, size = 421, normalized size = 2.49 \begin{align*} \frac{15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 90 \, a^{2} b \cos \left (d x + c\right ) + 15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 24 \, a^{3} + 8 \,{\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \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 = 1.38761, size = 495, normalized size = 2.93 \begin{align*} \frac{15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 225 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 360 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 60 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 160 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 90 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 960 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 120 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 464 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1200 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 160 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 90 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 960 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 120 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 225 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 360 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 60 \, 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 )^{2} - 1\right )}^{5}}}{120 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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