Integrand size = 40, antiderivative size = 124 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {5}{8} a^3 (3 B+4 C) x+\frac {a^3 (3 B+4 C) \sin (c+d x)}{d}+\frac {3 a^3 (3 B+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {B \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {a^3 (3 B+4 C) \sin ^3(c+d x)}{12 d} \]
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Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4098, 3876, 2717, 2715, 8, 2713} \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a^3 (3 B+4 C) \sin ^3(c+d x)}{12 d}+\frac {a^3 (3 B+4 C) \sin (c+d x)}{d}+\frac {3 a^3 (3 B+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5}{8} a^3 x (3 B+4 C)+\frac {B \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
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Rule 8
Rule 2713
Rule 2715
Rule 2717
Rule 3876
Rule 4098
Rule 4157
Rubi steps \begin{align*} \text {integral}& = \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx \\ & = \frac {B \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} (3 B+4 C) \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx \\ & = \frac {B \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} (3 B+4 C) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx \\ & = \frac {1}{4} a^3 (3 B+4 C) x+\frac {B \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \left (a^3 (3 B+4 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{4} \left (3 a^3 (3 B+4 C)\right ) \int \cos (c+d x) \, dx+\frac {1}{4} \left (3 a^3 (3 B+4 C)\right ) \int \cos ^2(c+d x) \, dx \\ & = \frac {1}{4} a^3 (3 B+4 C) x+\frac {3 a^3 (3 B+4 C) \sin (c+d x)}{4 d}+\frac {3 a^3 (3 B+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {B \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{8} \left (3 a^3 (3 B+4 C)\right ) \int 1 \, dx-\frac {\left (a^3 (3 B+4 C)\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d} \\ & = \frac {5}{8} a^3 (3 B+4 C) x+\frac {a^3 (3 B+4 C) \sin (c+d x)}{d}+\frac {3 a^3 (3 B+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {B \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {a^3 (3 B+4 C) \sin ^3(c+d x)}{12 d} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \sin (c+d x) \left (72 B+88 C+9 (5 B+4 C) \cos (c+d x)+8 (3 B+C) \cos ^2(c+d x)+6 B \cos ^3(c+d x)+\frac {30 (3 B+4 C) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{\sqrt {\sin ^2(c+d x)}}\right )}{24 d} \]
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Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(\frac {a^{3} \left (\left (32 B +24 C \right ) \sin \left (2 d x +2 c \right )+\left (8 B +\frac {8 C}{3}\right ) \sin \left (3 d x +3 c \right )+B \sin \left (4 d x +4 c \right )+\left (104 B +120 C \right ) \sin \left (d x +c \right )+60 x d \left (B +\frac {4 C}{3}\right )\right )}{32 d}\) | \(78\) |
risch | \(\frac {15 a^{3} B x}{8}+\frac {5 a^{3} x C}{2}+\frac {13 a^{3} B \sin \left (d x +c \right )}{4 d}+\frac {15 a^{3} C \sin \left (d x +c \right )}{4 d}+\frac {B \,a^{3} \sin \left (4 d x +4 c \right )}{32 d}+\frac {B \,a^{3} \sin \left (3 d x +3 c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) a^{3} C}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{3}}{d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{3} C}{4 d}\) | \(135\) |
derivativedivides | \(\frac {B \,a^{3} \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \sin \left (d x +c \right )+B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(176\) |
default | \(\frac {B \,a^{3} \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \sin \left (d x +c \right )+B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(176\) |
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Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, B + 4 \, C\right )} a^{3} d x + {\left (6 \, B a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{2} + 9 \, {\left (5 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (9 \, B + 11 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \]
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Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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Time = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.35 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 96 \, {\left (d x + c\right )} C a^{3} - 96 \, B a^{3} \sin \left (d x + c\right ) - 288 \, C a^{3} \sin \left (d x + c\right )}{96 \, d} \]
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Time = 0.33 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.42 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 4 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 165 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 220 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 219 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 147 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 132 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
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Time = 16.74 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15\,B\,a^3\,x}{8}+\frac {5\,C\,a^3\,x}{2}+\frac {13\,B\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {15\,C\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]
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