Optimal. Leaf size=180 \[ -\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \sin (c+d x)}{5 d}+\frac {\left (3 a^2 C+6 a b B+4 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2 C+6 a b B+4 b^2 C\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {a (a C+2 b B) \sin (c+d x) \cos ^3(c+d x)}{4 d} \]
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Rubi [A] time = 0.34, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4072, 4024, 4047, 2633, 4045, 2635, 8} \[ -\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \sin (c+d x)}{5 d}+\frac {\left (3 a^2 C+6 a b B+4 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2 C+6 a b B+4 b^2 C\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {a (a C+2 b B) \sin (c+d x) \cos ^3(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 4024
Rule 4045
Rule 4047
Rule 4072
Rubi steps
\begin {align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^5(c+d x) (a+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac {a^2 B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 a (2 b B+a C)+\left (\left (-4 a^2-5 b^2\right ) B-10 a b C\right ) \sec (c+d x)-5 b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 a (2 b B+a C)-5 b^2 C \sec ^2(c+d x)\right ) \, dx-\frac {1}{5} \left (-4 a^2 B-5 b^2 B-10 a b C\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac {a (2 b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^2 B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{4} \left (-6 a b B-3 a^2 C-4 b^2 C\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (4 a^2 B+5 b^2 B+10 a b C\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a b B+3 a^2 C+4 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^2 B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sin ^3(c+d x)}{15 d}-\frac {1}{8} \left (-6 a b B-3 a^2 C-4 b^2 C\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (6 a b B+3 a^2 C+4 b^2 C\right ) x+\frac {\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a b B+3 a^2 C+4 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^2 B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 0.60, size = 146, normalized size = 0.81 \[ \frac {60 (c+d x) \left (3 a^2 C+6 a b B+4 b^2 C\right )+60 \left (5 a^2 B+12 a b C+6 b^2 B\right ) \sin (c+d x)+120 \left (a^2 C+2 a b B+b^2 C\right ) \sin (2 (c+d x))+10 \left (5 a^2 B+8 a b C+4 b^2 B\right ) \sin (3 (c+d x))+6 a^2 B \sin (5 (c+d x))+15 a (a C+2 b B) \sin (4 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 142, normalized size = 0.79 \[ \frac {15 \, {\left (3 \, C a^{2} + 6 \, B a b + 4 \, C b^{2}\right )} d x + {\left (24 \, B a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )^{3} + 64 \, B a^{2} + 160 \, C a b + 80 \, B b^{2} + 8 \, {\left (4 \, B a^{2} + 10 \, C a b + 5 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, C a^{2} + 6 \, B a b + 4 \, C b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 487, normalized size = 2.71 \[ \frac {15 \, {\left (3 \, C a^{2} + 6 \, B a b + 4 \, C b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 150 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 640 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 640 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 150 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C 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 )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.73, size = 184, normalized size = 1.02 \[ \frac {\frac {a^{2} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 B a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 C a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {b^{2} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 176, normalized size = 0.98 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2}}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.63, size = 307, normalized size = 1.71 \[ \frac {x\,\left (\frac {3\,C\,a^2}{4}+\frac {3\,B\,a\,b}{2}+C\,b^2\right )}{2}+\frac {\left (2\,B\,a^2+2\,B\,b^2-\frac {5\,C\,a^2}{4}-C\,b^2-\frac {5\,B\,a\,b}{2}+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,B\,a^2}{3}+\frac {16\,B\,b^2}{3}-\frac {C\,a^2}{2}-2\,C\,b^2-B\,a\,b+\frac {32\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,B\,a^2}{15}+\frac {40\,C\,a\,b}{3}+\frac {20\,B\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,B\,a^2}{3}+\frac {16\,B\,b^2}{3}+\frac {C\,a^2}{2}+2\,C\,b^2+B\,a\,b+\frac {32\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a^2+2\,B\,b^2+\frac {5\,C\,a^2}{4}+C\,b^2+\frac {5\,B\,a\,b}{2}+4\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
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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