Optimal. Leaf size=108 \[ -\frac{b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac{a^2 \left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a^3 b \tan (c+d x)}{d}+\frac{a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+4 a b^3 x \]
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Rubi [A] time = 0.252111, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2792, 3031, 3023, 2735, 3770} \[ -\frac{b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac{a^2 \left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{3 a^3 b \tan (c+d x)}{d}+\frac{a^2 \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+4 a b^3 x \]
Antiderivative was successfully verified.
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Rule 2792
Rule 3031
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^4 \sec ^3(c+d x) \, dx &=\frac{a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+b \cos (c+d x)) \left (6 a^2 b+a \left (a^2+6 b^2\right ) \cos (c+d x)-b \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{3 a^3 b \tan (c+d x)}{d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-a^2 \left (a^2+12 b^2\right )-8 a b^3 \cos (c+d x)+b^2 \left (a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a^3 b \tan (c+d x)}{d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-a^2 \left (a^2+12 b^2\right )-8 a b^3 \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=4 a b^3 x-\frac{b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a^3 b \tan (c+d x)}{d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^2 \left (a^2+12 b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=4 a b^3 x+\frac{a^2 \left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (a^2-2 b^2\right ) \sin (c+d x)}{2 d}+\frac{3 a^3 b \tan (c+d x)}{d}+\frac{a^2 (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 2.4354, size = 174, normalized size = 1.61 \[ \frac{a \left (-2 a \left (a^2+12 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a \left (a^2+12 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^3}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^3}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+16 b^3 c+16 b^3 d x\right )+16 a^3 b \tan (c+d x)+4 b^4 \sin (c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.067, size = 114, normalized size = 1.1 \begin{align*}{\frac{{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+4\,{\frac{{a}^{3}b\tan \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\,a{b}^{3}x+4\,{\frac{a{b}^{3}c}{d}}+{\frac{{b}^{4}\sin \left ( dx+c \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.975915, size = 155, normalized size = 1.44 \begin{align*} \frac{16 \,{\left (d x + c\right )} a b^{3} - a^{4}{\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 )} + 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 )\right )} + 4 \, b^{4} \sin \left (d x + c\right ) + 16 \, a^{3} b \tan \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 = 2.02003, size = 324, normalized size = 3. \begin{align*} \frac{16 \, a b^{3} d x \cos \left (d x + c\right )^{2} +{\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{4} + 12 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, b^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{3} b \cos \left (d x + c\right ) + a^{4}\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 = 1.47515, size = 239, normalized size = 2.21 \begin{align*} \frac{8 \,{\left (d x + c\right )} a b^{3} + \frac{4 \, b^{4} \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} +{\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (a^{4} + 12 \, a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, a^{3} b \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 )}^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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