Optimal. Leaf size=196 \[ \frac{5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac{(12 A-4 B-7 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}-\frac{(12 A-32 B-35 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac{a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{8} a^4 x (52 A+48 B+35 C)-\frac{a (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^4}{d} \]
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Rubi [A] time = 0.679115, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3043, 2976, 2968, 3023, 2735, 3770} \[ \frac{5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac{(12 A-4 B-7 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 d}-\frac{(12 A-32 B-35 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{24 d}+\frac{a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{8} a^4 x (52 A+48 B+35 C)-\frac{a (4 A-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^4}{d} \]
Antiderivative was successfully verified.
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Rule 3043
Rule 2976
Rule 2968
Rule 3023
Rule 2735
Rule 3770
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x))^4 (a (4 A+B)-a (4 A-C) \cos (c+d x)) \sec (c+d x) \, dx}{a}\\ &=-\frac{a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x))^3 \left (4 a^2 (4 A+B)-a^2 (12 A-4 B-7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{4 a}\\ &=-\frac{a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x))^2 \left (12 a^3 (4 A+B)-a^3 (12 A-32 B-35 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac{a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x)) \left (24 a^4 (4 A+B)+15 a^4 (4 A+8 B+7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=-\frac{a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{\int \left (24 a^5 (4 A+B)+\left (24 a^5 (4 A+B)+15 a^5 (4 A+8 B+7 C)\right ) \cos (c+d x)+15 a^5 (4 A+8 B+7 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac{a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{\int \left (24 a^5 (4 A+B)+3 a^5 (52 A+48 B+35 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac{1}{8} a^4 (52 A+48 B+35 C) x+\frac{5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac{a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}+\left (a^4 (4 A+B)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a^4 (52 A+48 B+35 C) x+\frac{a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^4 (4 A+8 B+7 C) \sin (c+d x)}{8 d}-\frac{a (4 A-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac{A (a+a \cos (c+d x))^4 \tan (c+d x)}{d}\\ \end{align*}
Mathematica [A] time = 1.87439, size = 246, normalized size = 1.26 \[ \frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \left (12 (52 A+48 B+35 C) (c+d x)+24 (16 A+27 B+28 C) \sin (c+d x)+24 (A+4 B+7 C) \sin (2 (c+d x))-96 (4 A+B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+96 (4 A+B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{96 A \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{96 A \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+8 (B+4 C) \sin (3 (c+d x))+3 C \sin (4 (c+d x))\right )}{1536 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.09, size = 289, normalized size = 1.5 \begin{align*}{\frac{13\,A{a}^{4}x}{2}}+{\frac{A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{35\,{a}^{4}Cx}{8}}+{\frac{27\,{a}^{4}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+6\,{a}^{4}Bx+2\,{\frac{{a}^{4}B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+{\frac{B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{3\,d}}+{\frac{20\,{a}^{4}B\sin \left ( dx+c \right ) }{3\,d}}+{\frac{4\,{a}^{4}C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{20\,{a}^{4}C\sin \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{35\,{a}^{4}Cc}{8\,d}}+6\,{\frac{{a}^{4}Bc}{d}}+{\frac{13\,A{a}^{4}c}{2\,d}}+4\,{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.0366, size = 392, normalized size = 2. \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 576 \,{\left (d x + c\right )} A a^{4} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 96 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 384 \,{\left (d x + c\right )} B a^{4} - 128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} + 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 96 \,{\left (d x + c\right )} C a^{4} + 192 \, A a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 384 \, A a^{4} \sin \left (d x + c\right ) + 576 \, B a^{4} \sin \left (d x + c\right ) + 384 \, C a^{4} \sin \left (d x + c\right ) + 96 \, A a^{4} \tan \left (d x + c\right )}{96 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.21651, size = 466, normalized size = 2.38 \begin{align*} \frac{3 \,{\left (52 \, A + 48 \, B + 35 \, C\right )} a^{4} d x \cos \left (d x + c\right ) + 12 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, C a^{4} \cos \left (d x + c\right )^{4} + 8 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + 16 \, B + 27 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \,{\left (3 \, A + 5 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, A a^{4}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \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 = 1.31086, size = 448, normalized size = 2.29 \begin{align*} -\frac{\frac{48 \, A a^{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} - 3 \,{\left (52 \, A a^{4} + 48 \, B a^{4} + 35 \, C a^{4}\right )}{\left (d x + c\right )} - 24 \,{\left (4 \, A a^{4} + B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 24 \,{\left (4 \, A a^{4} + B a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (84 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 105 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 276 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 424 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 385 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 300 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 511 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 108 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 216 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 279 \, C a^{4} \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 )}^{4}}}{24 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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