Optimal. Leaf size=236 \[ -\frac{b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}+\frac{3 b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{4 b^2 d}-\frac{\left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{4 b d}-\frac{a \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a x \left (a^2-12 b^2\right )+\frac{a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d} \]
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Rubi [A] time = 0.74808, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2893, 3049, 3033, 3023, 2735, 3770} \[ -\frac{b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}+\frac{3 b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{\left (4 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^3}{4 b^2 d}-\frac{\left (6 a^2-b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{4 b d}-\frac{a \left (21 a^2-2 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a x \left (a^2-12 b^2\right )+\frac{a \tan (c+d x) (a \cos (c+d x)+b)^4}{b^2 d}+\frac{\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^4}{2 b d} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2893
Rule 3049
Rule 3033
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^3 \sin ^4(c+d x) \, dx &=-\int (-b-a \cos (c+d x))^3 \sin (c+d x) \tan ^3(c+d x) \, dx\\ &=\frac{a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac{(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\int (-b-a \cos (c+d x))^3 \left (-3 \left (2 a^2-b^2\right )+3 a b \cos (c+d x)+2 \left (4 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{2 b^2}\\ &=-\frac{\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac{a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac{(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\int (-b-a \cos (c+d x))^2 \left (12 b \left (2 a^2-b^2\right )-18 a b^2 \cos (c+d x)-6 b \left (6 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{8 b^2}\\ &=-\frac{\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac{\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac{a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac{(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\int (-b-a \cos (c+d x)) \left (-36 b^2 \left (2 a^2-b^2\right )+78 a b^3 \cos (c+d x)+6 b^2 \left (21 a^2-2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 b^2}\\ &=-\frac{a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac{\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac{a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac{(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\int \left (72 b^3 \left (2 a^2-b^2\right )+18 a b^2 \left (a^2-12 b^2\right ) \cos (c+d x)-24 b^3 \left (17 a^2-b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{48 b^2}\\ &=-\frac{b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}-\frac{a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac{\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac{a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac{(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\int \left (72 b^3 \left (2 a^2-b^2\right )+18 a b^2 \left (a^2-12 b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx}{48 b^2}\\ &=\frac{3}{8} a \left (a^2-12 b^2\right ) x-\frac{b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}-\frac{a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac{\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac{a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac{(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{1}{2} \left (3 b \left (2 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac{3}{8} a \left (a^2-12 b^2\right ) x+\frac{3 b \left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b \left (17 a^2-b^2\right ) \sin (c+d x)}{2 d}-\frac{a \left (21 a^2-2 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{\left (6 a^2-b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{4 b d}-\frac{\left (4 a^2-b^2\right ) (b+a \cos (c+d x))^3 \sin (c+d x)}{4 b^2 d}+\frac{a (b+a \cos (c+d x))^4 \tan (c+d x)}{b^2 d}+\frac{(b+a \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 b d}\\ \end{align*}
Mathematica [B] time = 6.16191, size = 696, normalized size = 2.95 \[ \frac{3 a \left (a^2-12 b^2\right ) (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{8 d (a \cos (c+d x)+b)^3}+\frac{b \left (4 b^2-15 a^2\right ) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (a \cos (c+d x)+b)^3}-\frac{a \left (a^2-3 b^2\right ) \sin (2 (c+d x)) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (a \cos (c+d x)+b)^3}+\frac{3 \left (b^3-2 a^2 b\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b)^3}-\frac{3 \left (b^3-2 a^2 b\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b)^3}+\frac{a^2 b \sin (3 (c+d x)) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (a \cos (c+d x)+b)^3}+\frac{a^3 \sin (4 (c+d x)) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{32 d (a \cos (c+d x)+b)^3}+\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}+\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^3}+\frac{b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b)^3}-\frac{b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.047, size = 276, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}x}{8}}+{\frac{3\,{a}^{3}c}{8\,d}}-{\frac{{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{\cos \left ( dx+c \right ) a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{9\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) a{b}^{2}}{2\,d}}-{\frac{9\,a{b}^{2}x}{2}}-{\frac{9\,a{b}^{2}c}{2\,d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,{b}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{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 = 1.48542, size = 247, normalized size = 1.05 \begin{align*} \frac{{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 16 \,{\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{2} b - 48 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{2} - 8 \, b^{3}{\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 )}}{32 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.9365, size = 466, normalized size = 1.97 \begin{align*} \frac{3 \,{\left (a^{3} - 12 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \,{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{2} b \cos \left (d x + c\right )^{4} + 24 \, a b^{2} \cos \left (d x + c\right ) -{\left (5 \, a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 4 \, b^{3} - 8 \,{\left (4 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, 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 = 1.53695, size = 582, normalized size = 2.47 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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