Optimal. Leaf size=218 \[ \frac {1}{8} b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) x+\frac {a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d} \]
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Rubi [A]
time = 0.45, 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 = {4180, 4179,
4159, 4132, 2717, 4130, 8} \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 2717
Rule 4130
Rule 4132
Rule 4159
Rule 4179
Rule 4180
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 = 0.69, size = 155, normalized size = 0.71 \begin {gather*} \frac {60 b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) (c+d x)+60 a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sin (c+d x)+120 b \left (A b^2+3 a^2 (A+C)\right ) \sin (2 (c+d x))+10 a \left (12 A b^2+a^2 (5 A+4 C)\right ) \sin (3 (c+d x))+45 a^2 A b \sin (4 (c+d x))+6 a^3 A \sin (5 (c+d x))}{480 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 201, normalized size = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {A \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,b^{2} a \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\) | \(201\) |
default | \(\frac {\frac {A \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,b^{2} a \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\) | \(201\) |
risch | \(\frac {9 a^{2} A b x}{8}+\frac {A \,b^{3} x}{2}+\frac {3 C \,a^{2} b x}{2}+C \,b^{3} x +\frac {5 a^{3} A \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) a A \,b^{2}}{4 d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{4 d}+\frac {3 \sin \left (d x +c \right ) C \,b^{2} a}{d}+\frac {A \,a^{3} \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 A \,a^{2} b \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 A \,a^{3} \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a A \,b^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) a^{3} C}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) A \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b C}{4 d}\) | \(241\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 194, normalized size = 0.89 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.00, size = 153, normalized size = 0.70 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 606 vs.
\(2 (206) = 412\).
time = 0.49, size = 606, normalized size = 2.78 \begin {gather*} \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} \end {gather*}
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 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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