Integrand size = 41, antiderivative size = 215 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac {\left (10 a b B+a^2 (4 A+5 C)+b^2 (4 A+5 C)\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sin ^3(c+d x)}{15 d} \]
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Time = 0.57 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4179, 4159, 4132, 2715, 8, 4129, 3092} \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {\sin ^3(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac {\sin (c+d x) \left (a^2 (4 A+5 C)+10 a b B+b^2 (4 A+5 C)\right )}{5 d}+\frac {\sin (c+d x) \cos (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac {1}{8} x \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]
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Rule 8
Rule 2715
Rule 3092
Rule 4129
Rule 4132
Rule 4159
Rule 4179
Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (2 A b+5 a B+(4 a A+5 b B+5 a C) \sec (c+d x)+b (2 A+5 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{20} \int \cos ^3(c+d x) \left (-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )-5 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x)-4 b^2 (2 A+5 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{20} \int \cos ^3(c+d x) \left (-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )-4 b^2 (2 A+5 C) \sec ^2(c+d x)\right ) \, dx-\frac {1}{4} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \cos ^2(c+d x) \, dx \\ & = \frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{20} \int \cos (c+d x) \left (-4 b^2 (2 A+5 C)-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \cos ^2(c+d x)\right ) \, dx-\frac {1}{8} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int 1 \, dx \\ & = \frac {1}{8} \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {\text {Subst}\left (\int \left (-4 b^2 (2 A+5 C)-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )+4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d} \\ & = \frac {1}{8} \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac {\left (10 a b B+a^2 (4 A+5 C)+b^2 (4 A+5 C)\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sin ^3(c+d x)}{15 d} \\ \end{align*}
Time = 1.81 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.79 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {60 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) (c+d x)+60 \left (12 a b B+2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sin (c+d x)+120 \left (a^2 B+b^2 B+2 a b (A+C)\right ) \sin (2 (c+d x))+10 \left (4 A b^2+8 a b B+a^2 (5 A+4 C)\right ) \sin (3 (c+d x))+15 a (2 A b+a B) \sin (4 (c+d x))+6 a^2 A \sin (5 (c+d x))}{480 d} \]
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Time = 0.85 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.78
method | result | size |
parallelrisch | \(\frac {120 \left (B \,a^{2}+2 a b \left (A +C \right )+B \,b^{2}\right ) \sin \left (2 d x +2 c \right )+10 \left (\left (5 A +4 C \right ) a^{2}+8 B a b +4 A \,b^{2}\right ) \sin \left (3 d x +3 c \right )+15 \left (2 a A b +B \,a^{2}\right ) \sin \left (4 d x +4 c \right )+6 a^{2} A \sin \left (5 d x +5 c \right )+60 \left (a^{2} \left (5 A +6 C \right )+12 B a b +6 b^{2} \left (A +\frac {4 C}{3}\right )\right ) \sin \left (d x +c \right )+360 d x \left (\frac {B \,a^{2}}{2}+a \left (A +\frac {4 C}{3}\right ) b +\frac {2 B \,b^{2}}{3}\right )}{480 d}\) | \(167\) |
derivativedivides | \(\frac {\frac {a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a A b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {2 B a b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b^{2}}{d}\) | \(244\) |
default | \(\frac {\frac {a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a A b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {2 B a b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b^{2}}{d}\) | \(244\) |
risch | \(\frac {3 a A b x}{4}+\frac {3 a^{2} B x}{8}+\frac {x B \,b^{2}}{2}+x C a b +\frac {5 \sin \left (d x +c \right ) a^{2} A}{8 d}+\frac {3 \sin \left (d x +c \right ) A \,b^{2}}{4 d}+\frac {3 \sin \left (d x +c \right ) B a b}{2 d}+\frac {3 \sin \left (d x +c \right ) C \,a^{2}}{4 d}+\frac {\sin \left (d x +c \right ) C \,b^{2}}{d}+\frac {a^{2} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a A b}{16 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{2}}{32 d}+\frac {5 a^{2} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) A \,b^{2}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) B a b}{6 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a A b}{2 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,b^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a b}{2 d}\) | \(294\) |
norman | \(\frac {\left (-\frac {3}{4} a A b -\frac {3}{8} B \,a^{2}-\frac {1}{2} B \,b^{2}-C a b \right ) x +\left (-\frac {9}{2} a A b -\frac {9}{4} B \,a^{2}-3 B \,b^{2}-6 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-\frac {3}{2} a A b -\frac {3}{4} B \,a^{2}-B \,b^{2}-2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {3}{2} a A b -\frac {3}{4} B \,a^{2}-B \,b^{2}-2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {3}{2} a A b +\frac {3}{4} B \,a^{2}+B \,b^{2}+2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3}{2} a A b +\frac {3}{4} B \,a^{2}+B \,b^{2}+2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {3}{4} a A b +\frac {3}{8} B \,a^{2}+\frac {1}{2} B \,b^{2}+C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (\frac {9}{2} a A b +\frac {9}{4} B \,a^{2}+3 B \,b^{2}+6 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (8 a^{2} A -10 a A b +8 A \,b^{2}-5 B \,a^{2}+16 B a b -4 B \,b^{2}+8 C \,a^{2}-8 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {\left (8 a^{2} A +10 a A b +8 A \,b^{2}+5 B \,a^{2}+16 B a b +4 B \,b^{2}+8 C \,a^{2}+8 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (40 a^{2} A -78 a A b +8 A \,b^{2}-39 B \,a^{2}+16 B a b -12 B \,b^{2}+8 C \,a^{2}-24 C a b -24 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}+\frac {\left (40 a^{2} A +78 a A b +8 A \,b^{2}+39 B \,a^{2}+16 B a b +12 B \,b^{2}+8 C \,a^{2}+24 C a b -24 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (344 a^{2} A -270 a A b -200 A \,b^{2}-135 B \,a^{2}-400 B a b +180 B \,b^{2}-200 C \,a^{2}+360 C a b -360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{60 d}-\frac {\left (344 a^{2} A +270 a A b -200 A \,b^{2}+135 B \,a^{2}-400 B a b -180 B \,b^{2}-200 C \,a^{2}-360 C a b -360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d}-\frac {\left (872 a^{2} A -30 a A b +40 A \,b^{2}-15 B \,a^{2}+80 B a b +180 B \,b^{2}+40 C \,a^{2}+360 C a b +360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d}+\frac {\left (872 a^{2} A +30 a A b +40 A \,b^{2}+15 B \,a^{2}+80 B a b -180 B \,b^{2}+40 C \,a^{2}-360 C a b +360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{60 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\) | \(849\) |
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Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} d x + {\left (24 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 160 \, B a b + 40 \, {\left (2 \, A + 3 \, C\right )} b^{2} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
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Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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Time = 0.25 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.08 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 480 \, C b^{2} \sin \left (d x + c\right )}{480 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 720 vs. \(2 (203) = 406\).
Time = 0.33 (sec) , antiderivative size = 720, normalized size of antiderivative = 3.35 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
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Time = 17.08 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.19 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {25\,A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,A\,a^2\,\sin \left (5\,c+5\,d\,x\right )}{2}+30\,B\,a^2\,\sin \left (2\,c+2\,d\,x\right )+10\,A\,b^2\,\sin \left (3\,c+3\,d\,x\right )+\frac {15\,B\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{4}+30\,B\,b^2\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,a^2\,\sin \left (3\,c+3\,d\,x\right )+75\,A\,a^2\,\sin \left (c+d\,x\right )+90\,A\,b^2\,\sin \left (c+d\,x\right )+90\,C\,a^2\,\sin \left (c+d\,x\right )+120\,C\,b^2\,\sin \left (c+d\,x\right )+60\,A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+\frac {15\,A\,a\,b\,\sin \left (4\,c+4\,d\,x\right )}{2}+20\,B\,a\,b\,\sin \left (3\,c+3\,d\,x\right )+60\,C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+45\,B\,a^2\,d\,x+60\,B\,b^2\,d\,x+180\,B\,a\,b\,\sin \left (c+d\,x\right )+90\,A\,a\,b\,d\,x+120\,C\,a\,b\,d\,x}{120\,d} \]
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