Optimal. Leaf size=218 \[ \frac {a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \sin (c+d x)}{15 d}+\frac {a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{30 d}+\frac {3 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} b x \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d} \]
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Rubi [A] time = 0.65, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4095, 4094, 4074, 4047, 2637, 4045, 8} \[ \frac {a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \sin (c+d x)}{15 d}+\frac {a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{30 d}+\frac {3 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} b x \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2637
Rule 4045
Rule 4047
Rule 4074
Rule 4094
Rule 4095
Rubi steps
\begin {align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (4 A+5 C) \sec (c+d x)+b (A+5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (2 \left (3 A b^2+2 a^2 (4 A+5 C)\right )+a b (29 A+40 C) \sec (c+d x)+b^2 (7 A+20 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac {1}{60} \int \cos ^2(c+d x) \left (-9 b \left (2 A b^2+5 a^2 (3 A+4 C)\right )-4 a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sec (c+d x)-3 b^3 (7 A+20 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac {1}{60} \int \cos ^2(c+d x) \left (-9 b \left (2 A b^2+5 a^2 (3 A+4 C)\right )-3 b^3 (7 A+20 C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{15} \left (a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \int \cos (c+d x) \, dx\\ &=\frac {a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{8} \left (b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right )\right ) \int 1 \, dx\\ &=\frac {1}{8} b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) x+\frac {a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.81, size = 155, normalized size = 0.71 \[ \frac {6 a^3 A \sin (5 (c+d x))+60 b (c+d x) \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+60 a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \sin (c+d x)+10 a \left (a^2 (5 A+4 C)+12 A b^2\right ) \sin (3 (c+d x))+120 b \left (3 a^2 (A+C)+A b^2\right ) \sin (2 (c+d x))+45 a^2 A b \sin (4 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 153, normalized size = 0.70 \[ \frac {15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 90 \, A a^{2} b \cos \left (d x + c\right )^{3} + 16 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 120 \, {\left (2 \, A + 3 \, C\right )} a b^{2} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.30, size = 606, normalized size = 2.78 \[ \frac {15 \, {\left (9 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 8 \, C b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1440 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2160 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1440 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{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 )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.56, size = 201, normalized size = 0.92 \[ \frac {\frac {A \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C a \,b^{2} \sin \left (d x +c \right )+b^{3} C \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 194, normalized size = 0.89 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \, {\left (d x + c\right )} C b^{3} + 1440 \, C a b^{2} \sin \left (d x + c\right )}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.90, size = 442, normalized size = 2.03 \[ \frac {\left (2\,A\,a^3-A\,b^3+2\,C\,a^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,A\,a^3}{3}-2\,A\,b^3+\frac {16\,C\,a^3}{3}+16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}+24\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+\frac {20\,C\,a^3}{3}+20\,A\,a\,b^2+36\,C\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,A\,a^3}{3}+2\,A\,b^3+\frac {16\,C\,a^3}{3}+16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}+24\,C\,a\,b^2+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2+3\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{4\,\left (A\,b^3+2\,C\,b^3+\frac {9\,A\,a^2\,b}{4}+3\,C\,a^2\,b\right )}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{4\,d} \]
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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