Optimal. Leaf size=213 \[ \frac {7 a^3 b \sin (c+d x) \cos ^4(c+d x)}{15 d}-\frac {4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d}+\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (5 a^4+36 a^2 b^2+8 b^4\right ) \]
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Rubi [A] time = 0.38, antiderivative size = 213, normalized size of antiderivative = 1.00, 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 = {3841, 4074, 4047, 2633, 4045, 2635, 8} \[ -\frac {4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {\left (36 a^2 b^2+5 a^4+8 b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (36 a^2 b^2+5 a^4+8 b^4\right )+\frac {7 a^3 b \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 3841
Rule 4045
Rule 4047
Rule 4074
Rubi steps
\begin {align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac {a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (14 a^2 b+a \left (5 a^2+18 b^2\right ) \sec (c+d x)+3 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac {1}{30} \int \cos ^4(c+d x) \left (-5 a^2 \left (5 a^2+32 b^2\right )-24 a b \left (4 a^2+5 b^2\right ) \sec (c+d x)-15 b^2 \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac {1}{30} \int \cos ^4(c+d x) \left (-5 a^2 \left (5 a^2+32 b^2\right )-15 b^2 \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (4 a b \left (4 a^2+5 b^2\right )\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac {a^2 \left (5 a^2+32 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac {1}{8} \left (-5 a^4-36 a^2 b^2-8 b^4\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (4 a b \left (4 a^2+5 b^2\right )\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac {4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}-\frac {1}{16} \left (-5 a^4-36 a^2 b^2-8 b^4\right ) \int 1 \, dx\\ &=\frac {1}{16} \left (5 a^4+36 a^2 b^2+8 b^4\right ) x+\frac {4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac {a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac {4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 156, normalized size = 0.73 \[ \frac {5 a^4 \sin (6 (c+d x))+48 a^3 b \sin (5 (c+d x))+45 a^2 \left (a^2+4 b^2\right ) \sin (4 (c+d x))+480 a b \left (5 a^2+6 b^2\right ) \sin (c+d x)+80 a b \left (5 a^2+4 b^2\right ) \sin (3 (c+d x))+60 \left (5 a^4+36 a^2 b^2+8 b^4\right ) (c+d x)+15 \left (15 a^4+96 a^2 b^2+16 b^4\right ) \sin (2 (c+d x))}{960 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.54, size = 150, normalized size = 0.70 \[ \frac {15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} d x + {\left (40 \, a^{4} \cos \left (d x + c\right )^{5} + 192 \, a^{3} b \cos \left (d x + c\right )^{4} + 512 \, a^{3} b + 640 \, a b^{3} + 10 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 64 \, {\left (4 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.25, size = 550, normalized size = 2.58 \[ \frac {15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (165 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 120 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2240 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 3520 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 450 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4992 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5760 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 450 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4992 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5760 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3520 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 360 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 165 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, b^{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 )}^{6}}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.60, size = 174, normalized size = 0.82 \[ \frac {a^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {4 a^{3} b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+6 a^{2} b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.37, size = 170, normalized size = 0.80 \[ -\frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} b - 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b^{2} + 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4}}{960 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.09, size = 214, normalized size = 1.00 \[ \frac {5\,a^4\,x}{16}+\frac {b^4\,x}{2}+\frac {9\,a^2\,b^2\,x}{4}+\frac {15\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {a^4\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {5\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a^3\,b\,\sin \left (5\,c+5\,d\,x\right )}{20\,d}+\frac {3\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,a^2\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{16\,d}+\frac {3\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,a^3\,b\,\sin \left (c+d\,x\right )}{2\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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