Optimal. Leaf size=224 \[ \frac{10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac{10 a^2 b^3 \sec (c+d x)}{d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac{5 a^4 b \sec (c+d x)}{d}+\frac{a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{15 a b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{15 a b^4 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b^5 \sec ^5(c+d x)}{5 d}-\frac{2 b^5 \sec ^3(c+d x)}{3 d}+\frac{b^5 \sec (c+d x)}{d} \]
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Rubi [A] time = 0.233113, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3090, 3770, 2606, 8, 2611, 194} \[ \frac{10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac{10 a^2 b^3 \sec (c+d x)}{d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{d}+\frac{5 a^4 b \sec (c+d x)}{d}+\frac{a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{15 a b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a b^4 \tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac{15 a b^4 \tan (c+d x) \sec (c+d x)}{8 d}+\frac{b^5 \sec ^5(c+d x)}{5 d}-\frac{2 b^5 \sec ^3(c+d x)}{3 d}+\frac{b^5 \sec (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 3090
Rule 3770
Rule 2606
Rule 8
Rule 2611
Rule 194
Rubi steps
\begin{align*} \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \sec (c+d x)+5 a^4 b \sec (c+d x) \tan (c+d x)+10 a^3 b^2 \sec (c+d x) \tan ^2(c+d x)+10 a^2 b^3 \sec (c+d x) \tan ^3(c+d x)+5 a b^4 \sec (c+d x) \tan ^4(c+d x)+b^5 \sec (c+d x) \tan ^5(c+d x)\right ) \, dx\\ &=a^5 \int \sec (c+d x) \, dx+\left (5 a^4 b\right ) \int \sec (c+d x) \tan (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sec (c+d x) \tan ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \sec (c+d x) \tan ^4(c+d x) \, dx+b^5 \int \sec (c+d x) \tan ^5(c+d x) \, dx\\ &=\frac{a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{d}+\frac{5 a b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d}-\left (5 a^3 b^2\right ) \int \sec (c+d x) \, dx-\frac{1}{4} \left (15 a b^4\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\frac{\left (5 a^4 b\right ) \operatorname{Subst}(\int 1 \, dx,x,\sec (c+d x))}{d}+\frac{\left (10 a^2 b^3\right ) \operatorname{Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac{b^5 \operatorname{Subst}\left (\int \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{a^5 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^4 b \sec (c+d x)}{d}-\frac{10 a^2 b^3 \sec (c+d x)}{d}+\frac{10 a^2 b^3 \sec ^3(c+d x)}{3 d}+\frac{5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{d}-\frac{15 a b^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{5 a b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac{1}{8} \left (15 a b^4\right ) \int \sec (c+d x) \, dx+\frac{b^5 \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{a^5 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{15 a b^4 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{5 a^4 b \sec (c+d x)}{d}-\frac{10 a^2 b^3 \sec (c+d x)}{d}+\frac{b^5 \sec (c+d x)}{d}+\frac{10 a^2 b^3 \sec ^3(c+d x)}{3 d}-\frac{2 b^5 \sec ^3(c+d x)}{3 d}+\frac{b^5 \sec ^5(c+d x)}{5 d}+\frac{5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{d}-\frac{15 a b^4 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{5 a b^4 \sec (c+d x) \tan ^3(c+d x)}{4 d}\\ \end{align*}
Mathematica [B] time = 6.32179, size = 1219, normalized size = 5.44 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.26, size = 440, normalized size = 2. \begin{align*}{\frac{{a}^{5}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+5\,{\frac{{a}^{4}b}{d\cos \left ( dx+c \right ) }}+5\,{\frac{{a}^{3}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+5\,{\frac{{a}^{3}{b}^{2}\sin \left ( dx+c \right ) }{d}}-5\,{\frac{{a}^{3}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{10\,{a}^{2}{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{10\,{a}^{2}{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,d\cos \left ( dx+c \right ) }}-{\frac{10\,\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}{b}^{3}}{3\,d}}-{\frac{20\,{a}^{2}{b}^{3}\cos \left ( dx+c \right ) }{3\,d}}+{\frac{5\,a{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{5\,a{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,a{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{15\,a{b}^{4}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,a{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{5\,d\cos \left ( dx+c \right ) }}+{\frac{8\,{b}^{5}\cos \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{5}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{5}}{15\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.13435, size = 311, normalized size = 1.39 \begin{align*} \frac{75 \, a b^{4}{\left (\frac{2 \,{\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \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 )} - 600 \, a^{3} b^{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 )} + 120 \, a^{5}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac{1200 \, a^{4} b}{\cos \left (d x + c\right )} - \frac{800 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2} b^{3}}{\cos \left (d x + c\right )^{3}} + \frac{16 \,{\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} b^{5}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.571347, size = 478, normalized size = 2.13 \begin{align*} \frac{15 \,{\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 48 \, b^{5} + 240 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 160 \,{\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 150 \,{\left (2 \, a b^{4} \cos \left (d x + c\right ) +{\left (8 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\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 [A] time = 1.34456, size = 554, normalized size = 2.47 \begin{align*} \frac{15 \,{\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (600 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 225 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 600 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 1200 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1050 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 2400 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 2400 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 3600 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 5600 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 640 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1200 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1050 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2400 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4000 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 320 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 600 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 225 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 600 \, a^{4} b + 800 \, a^{2} b^{3} - 64 \, b^{5}\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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