Optimal. Leaf size=198 \[ -\frac{b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (2 a^2 A+9 a b B+9 A b^2\right ) \tan (c+d x)}{3 d}+\frac{a \left (4 a^2 A b+a^3 B+12 a b^2 B+8 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+b^3 x (4 a B+A b)+\frac{a (a B+2 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d} \]
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Rubi [A] time = 0.580128, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2989, 3047, 3031, 3023, 2735, 3770} \[ -\frac{b^2 \left (3 a^2 B+8 a A b-6 b^2 B\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (2 a^2 A+9 a b B+9 A b^2\right ) \tan (c+d x)}{3 d}+\frac{a \left (4 a^2 A b+a^3 B+12 a b^2 B+8 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+b^3 x (4 a B+A b)+\frac{a (a B+2 A b) \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^3}{3 d} \]
Antiderivative was successfully verified.
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Rule 2989
Rule 3047
Rule 3031
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac{a A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x))^2 \left (3 a (2 A b+a B)+\left (2 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)-b (a A-3 b B) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a (2 A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (a+b \cos (c+d x)) \left (2 a \left (2 a^2 A+9 A b^2+9 a b B\right )+\left (8 a^2 A b+6 A b^3+3 a^3 B+18 a b^2 B\right ) \cos (c+d x)-b \left (8 a A b+3 a^2 B-6 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a^2 \left (2 a^2 A+9 A b^2+9 a b B\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right )-6 b^3 (A b+4 a B) \cos (c+d x)+b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (2 a^2 A+9 A b^2+9 a b B\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right )-6 b^3 (A b+4 a B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 (A b+4 a B) x-\frac{b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (2 a^2 A+9 A b^2+9 a b B\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} \left (a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right )\right ) \int \sec (c+d x) \, dx\\ &=b^3 (A b+4 a B) x+\frac{a \left (4 a^2 A b+8 A b^3+a^3 B+12 a b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (8 a A b+3 a^2 B-6 b^2 B\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (2 a^2 A+9 A b^2+9 a b B\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+a B) (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}
Mathematica [B] time = 5.9631, size = 415, normalized size = 2.1 \[ \frac{\frac{8 a^2 \left (a^2 A+6 a b B+9 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{8 a^2 \left (a^2 A+6 a b B+9 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-6 a \left (4 a^2 A b+a^3 B+12 a b^2 B+8 A b^3\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 a \left (4 a^2 A b+a^3 B+12 a b^2 B+8 A b^3\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^3 (a (A+3 B)+12 A b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^3 (a (A+3 B)+12 A b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+12 b^3 (c+d x) (4 a B+A b)+12 b^4 B \sin (c+d x)}{12 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.102, size = 262, normalized size = 1.3 \begin{align*}{\frac{2\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{A{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{B{a}^{3}b\tan \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+6\,{\frac{B{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,Ba{b}^{3}x+4\,{\frac{Ba{b}^{3}c}{d}}+A{b}^{4}x+{\frac{A{b}^{4}c}{d}}+{\frac{B{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 = 1.16211, size = 331, normalized size = 1.67 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 48 \,{\left (d x + c\right )} B a b^{3} + 12 \,{\left (d x + c\right )} A b^{4} - 3 \, B 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 a^{3} b{\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 )} + 36 \, B 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 )} + 24 \, A a b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B b^{4} \sin \left (d x + c\right ) + 48 \, B a^{3} b \tan \left (d x + c\right ) + 72 \, A a^{2} b^{2} \tan \left (d x + c\right )}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.54464, size = 524, normalized size = 2.65 \begin{align*} \frac{12 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, B b^{4} \cos \left (d x + c\right )^{3} + 2 \, A a^{4} + 4 \,{\left (A a^{4} + 6 \, B a^{3} b + 9 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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 [B] time = 1.61798, size = 522, normalized size = 2.64 \begin{align*} \frac{\frac{12 \, B 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} + 6 \,{\left (4 \, B a b^{3} + A b^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (B a^{4} + 4 \, A a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, A a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 48 \, B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, A a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, B a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \, A a^{2} b^{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 )}^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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