Optimal. Leaf size=236 \[ \frac{\left (30 a^2 b C+8 a^3 B+30 a b^2 B+15 b^3 C\right ) \tan (c+d x)}{15 d}+\frac{\left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (4 a^2 B+15 a b C+12 b^2 B\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{\left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^2 (5 a C+7 b B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{a B \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
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Rubi [A] time = 0.561617, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.225, Rules used = {3029, 2989, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\left (30 a^2 b C+8 a^3 B+30 a b^2 B+15 b^3 C\right ) \tan (c+d x)}{15 d}+\frac{\left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (4 a^2 B+15 a b C+12 b^2 B\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{\left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a^2 (5 a C+7 b B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{a B \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 3029
Rule 2989
Rule 3031
Rule 3021
Rule 2748
Rule 3768
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\int (a+b \cos (c+d x))^3 (B+C \cos (c+d x)) \sec ^6(c+d x) \, dx\\ &=\frac{a B (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x)) \left (a (7 b B+5 a C)+\left (4 a^2 B+5 b^2 B+10 a b C\right ) \cos (c+d x)+b (2 a B+5 b C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a^2 (7 b B+5 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{20} \int \left (-4 a \left (4 a^2 B+12 b^2 B+15 a b C\right )-5 \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x)-4 b^2 (2 a B+5 b C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 (7 b B+5 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{60} \int \left (-15 \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right )-4 \left (8 a^3 B+30 a b^2 B+30 a^2 b C+15 b^3 C\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 (7 b B+5 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{4} \left (-9 a^2 b B-4 b^3 B-3 a^3 C-12 a b^2 C\right ) \int \sec ^3(c+d x) \, dx-\frac{1}{15} \left (-8 a^3 B-30 a b^2 B-30 a^2 b C-15 b^3 C\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 (7 b B+5 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{8} \left (-9 a^2 b B-4 b^3 B-3 a^3 C-12 a b^2 C\right ) \int \sec (c+d x) \, dx-\frac{\left (8 a^3 B+30 a b^2 B+30 a^2 b C+15 b^3 C\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (8 a^3 B+30 a b^2 B+30 a^2 b C+15 b^3 C\right ) \tan (c+d x)}{15 d}+\frac{\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a^2 (7 b B+5 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 3.41093, size = 181, normalized size = 0.77 \[ \frac{15 \left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 a \left (2 a^2 B+3 a b C+3 b^2 B\right ) \tan ^2(c+d x)+15 \left (3 a^2 b C+a^3 B+3 a b^2 B+b^3 C\right )+3 a^3 B \tan ^4(c+d x)\right )+15 \left (9 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \sec (c+d x)+30 a^2 (a C+3 b B) \sec ^3(c+d x)\right )}{120 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.059, size = 382, normalized size = 1.6 \begin{align*}{\frac{C{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{3\,Ca{b}^{2}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{a{b}^{2}B\tan \left ( dx+c \right ) }{d}}+{\frac{a{b}^{2}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}bC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{2}bB\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,{a}^{2}bB\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,{a}^{2}bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,{a}^{3}B\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.06425, size = 460, normalized size = 1.95 \begin{align*} \frac{16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b^{2} - 15 \, C a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \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 )\right )} - 45 \, B a^{2} b{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \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 )\right )} - 180 \, C a b^{2}{\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 )} - 60 \, B b^{3}{\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 )} + 240 \, C b^{3} \tan \left (d x + c\right )}{240 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.58003, size = 612, normalized size = 2.59 \begin{align*} \frac{15 \,{\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (8 \, B a^{3} + 30 \, C a^{2} b + 30 \, B a b^{2} + 15 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, B a^{3} + 15 \,{\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (4 \, B a^{3} + 15 \, C a^{2} b + 15 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \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 = 1.86497, size = 975, normalized size = 4.13 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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