Optimal. Leaf size=157 \[ -\frac{2 a^2 \sin ^5(c+d x)}{5 d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \tan (c+d x)}{d}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{25 a^2 x}{16} \]
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Rubi [A] time = 0.269959, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3872, 2872, 2637, 2633, 2635, 8, 3770, 3767} \[ -\frac{2 a^2 \sin ^5(c+d x)}{5 d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \tan (c+d x)}{d}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{25 a^2 x}{16} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2872
Rule 2637
Rule 2633
Rule 2635
Rule 8
Rule 3770
Rule 3767
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{\int \left (-2 a^8-6 a^8 \cos (c+d x)+6 a^8 \cos ^3(c+d x)+2 a^8 \cos ^4(c+d x)-2 a^8 \cos ^5(c+d x)-a^8 \cos ^6(c+d x)+2 a^8 \sec (c+d x)+a^8 \sec ^2(c+d x)\right ) \, dx}{a^6}\\ &=-2 a^2 x-a^2 \int \cos ^6(c+d x) \, dx+a^2 \int \sec ^2(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^4(c+d x) \, dx-\left (2 a^2\right ) \int \cos ^5(c+d x) \, dx+\left (2 a^2\right ) \int \sec (c+d x) \, dx-\left (6 a^2\right ) \int \cos (c+d x) \, dx+\left (6 a^2\right ) \int \cos ^3(c+d x) \, dx\\ &=-2 a^2 x+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{6 a^2 \sin (c+d x)}{d}+\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{1}{6} \left (5 a^2\right ) \int \cos ^4(c+d x) \, dx+\frac{1}{2} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac{a^2 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{\left (6 a^2\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-2 a^2 x+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a^2 \sin (c+d x)}{d}+\frac{3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \tan (c+d x)}{d}-\frac{1}{8} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac{5 a^2 x}{4}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a^2 \sin (c+d x)}{d}+\frac{7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \tan (c+d x)}{d}-\frac{1}{16} \left (5 a^2\right ) \int 1 \, dx\\ &=-\frac{25 a^2 x}{16}+\frac{2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a^2 \sin (c+d x)}{d}+\frac{7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a^2 \sin ^3(c+d x)}{3 d}-\frac{2 a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \tan (c+d x)}{d}\\ \end{align*}
Mathematica [A] time = 0.547296, size = 124, normalized size = 0.79 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (384 \sin ^5(c+d x)+640 \sin ^3(c+d x)+1920 \sin (c+d x)-255 \sin (2 (c+d x))-15 \sin (4 (c+d x))+5 \sin (6 (c+d x))+420 \tan ^{-1}(\tan (c+d x))-960 \tan (c+d x)-1920 \tanh ^{-1}(\sin (c+d x))+1080 c+1080 d x\right )}{3840 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.043, size = 172, normalized size = 1.1 \begin{align*}{\frac{5\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{25\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{25\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}-{\frac{25\,{a}^{2}x}{16}}-{\frac{25\,{a}^{2}c}{16\,d}}-{\frac{2\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{2\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-2\,{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{d\cos \left ( dx+c \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.54821, size = 235, normalized size = 1.5 \begin{align*} -\frac{64 \,{\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a^{2} - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 120 \,{\left (15 \, d x + 15 \, c - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{2}}{960 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.97077, size = 423, normalized size = 2.69 \begin{align*} -\frac{375 \, a^{2} d x \cos \left (d x + c\right ) - 240 \, a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 240 \, a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (40 \, a^{2} \cos \left (d x + c\right )^{6} + 96 \, a^{2} \cos \left (d x + c\right )^{5} - 70 \, a^{2} \cos \left (d x + c\right )^{4} - 352 \, a^{2} \cos \left (d x + c\right )^{3} - 105 \, a^{2} \cos \left (d x + c\right )^{2} + 736 \, a^{2} \cos \left (d x + c\right ) - 240 \, a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )} \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 = 1.42387, size = 261, normalized size = 1.66 \begin{align*} -\frac{375 \,{\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 480 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{480 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (615 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 3485 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 7926 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 8586 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2595 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 345 \, a^{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 )}^{6}}}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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