Optimal. Leaf size=151 \[ -\frac{6 a^2 b^2 \sin (c+d x)}{d}+\frac{6 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{4 a^3 b \cos (c+d x)}{d}+\frac{a^4 \sin (c+d x)}{d}+\frac{4 a b^3 \cos (c+d x)}{d}+\frac{4 a b^3 \sec (c+d x)}{d}+\frac{3 b^4 \sin (c+d x)}{2 d}+\frac{b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{2 d} \]
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Rubi [A] time = 0.164174, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3090, 2637, 2638, 2592, 321, 206, 2590, 14, 288} \[ -\frac{6 a^2 b^2 \sin (c+d x)}{d}+\frac{6 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{4 a^3 b \cos (c+d x)}{d}+\frac{a^4 \sin (c+d x)}{d}+\frac{4 a b^3 \cos (c+d x)}{d}+\frac{4 a b^3 \sec (c+d x)}{d}+\frac{3 b^4 \sin (c+d x)}{2 d}+\frac{b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Rule 3090
Rule 2637
Rule 2638
Rule 2592
Rule 321
Rule 206
Rule 2590
Rule 14
Rule 288
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=\int \left (a^4 \cos (c+d x)+4 a^3 b \sin (c+d x)+6 a^2 b^2 \sin (c+d x) \tan (c+d x)+4 a b^3 \sin (c+d x) \tan ^2(c+d x)+b^4 \sin (c+d x) \tan ^3(c+d x)\right ) \, dx\\ &=a^4 \int \cos (c+d x) \, dx+\left (4 a^3 b\right ) \int \sin (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \sin (c+d x) \tan (c+d x) \, dx+\left (4 a b^3\right ) \int \sin (c+d x) \tan ^2(c+d x) \, dx+b^4 \int \sin (c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac{4 a^3 b \cos (c+d x)}{d}+\frac{a^4 \sin (c+d x)}{d}+\frac{\left (6 a^2 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac{b^4 \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{4 a^3 b \cos (c+d x)}{d}+\frac{a^4 \sin (c+d x)}{d}-\frac{6 a^2 b^2 \sin (c+d x)}{d}+\frac{b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}+\frac{\left (6 a^2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac{6 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{4 a^3 b \cos (c+d x)}{d}+\frac{4 a b^3 \cos (c+d x)}{d}+\frac{4 a b^3 \sec (c+d x)}{d}+\frac{a^4 \sin (c+d x)}{d}-\frac{6 a^2 b^2 \sin (c+d x)}{d}+\frac{3 b^4 \sin (c+d x)}{2 d}+\frac{b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac{6 a^2 b^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{3 b^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{4 a^3 b \cos (c+d x)}{d}+\frac{4 a b^3 \cos (c+d x)}{d}+\frac{4 a b^3 \sec (c+d x)}{d}+\frac{a^4 \sin (c+d x)}{d}-\frac{6 a^2 b^2 \sin (c+d x)}{d}+\frac{3 b^4 \sin (c+d x)}{2 d}+\frac{b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 2.29426, size = 268, normalized size = 1.77 \[ \frac{-24 a^2 b^2 \sin (c+d x)-16 a b \left (a^2-b^2\right ) \cos (c+d x)-24 a^2 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+24 a^2 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 a^4 \sin (c+d x)+32 a b^3 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)+16 a b^3+4 b^4 \sin (c+d x)+\frac{b^4}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^4}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+6 b^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 b^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.136, size = 211, normalized size = 1.4 \begin{align*}{\frac{{a}^{4}\sin \left ( dx+c \right ) }{d}}-4\,{\frac{{a}^{3}b\cos \left ( dx+c \right ) }{d}}-6\,{\frac{{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{a{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}+4\,{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}a{b}^{3}}{d}}+8\,{\frac{a{b}^{3}\cos \left ( dx+c \right ) }{d}}+{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,{b}^{4}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{b}^{4}\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 = 1.23948, size = 192, normalized size = 1.27 \begin{align*} -\frac{b^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\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 ) - 4 \, \sin \left (d x + c\right )\right )} - 16 \, a b^{3}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 12 \, a^{2} b^{2}{\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 )} + 16 \, a^{3} b \cos \left (d x + c\right ) - 4 \, a^{4} \sin \left (d x + c\right )}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.522919, size = 363, normalized size = 2.4 \begin{align*} \frac{16 \, a b^{3} \cos \left (d x + c\right ) - 16 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (b^{4} + 2 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\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 [A] time = 1.27284, size = 278, normalized size = 1.84 \begin{align*} \frac{3 \,{\left (4 \, a^{2} b^{2} - b^{4}\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} - b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{4 \,{\left (a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{3} b + 4 \, a b^{3}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + \frac{2 \,{\left (b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, a b^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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