Optimal. Leaf size=160 \[ -\frac {a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x \left (9 a^2+4 b^2\right )+\frac {11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d} \]
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Rubi [A] time = 0.19, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3841, 4047, 2633, 4045, 2635, 8} \[ -\frac {a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x \left (9 a^2+4 b^2\right )+\frac {11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 3841
Rule 4045
Rule 4047
Rubi steps
\begin {align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) \left (11 a^2 b+a \left (4 a^2+15 b^2\right ) \sec (c+d x)+b \left (3 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) \left (11 a^2 b+b \left (3 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (a \left (4 a^2+15 b^2\right )\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac {11 a^2 b \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{4} \left (b \left (9 a^2+4 b^2\right )\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (a \left (4 a^2+15 b^2\right )\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {11 a^2 b \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}-\frac {a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {1}{8} \left (b \left (9 a^2+4 b^2\right )\right ) \int 1 \, dx\\ &=\frac {1}{8} b \left (9 a^2+4 b^2\right ) x+\frac {a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {11 a^2 b \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}-\frac {a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A] time = 0.30, size = 130, normalized size = 0.81 \[ \frac {50 a^3 \sin (3 (c+d x))+6 a^3 \sin (5 (c+d x))+120 \left (3 a^2 b+b^3\right ) \sin (2 (c+d x))+60 a \left (5 a^2+18 b^2\right ) \sin (c+d x)+45 a^2 b \sin (4 (c+d x))+540 a^2 b c+540 a^2 b d x+120 a b^2 \sin (3 (c+d x))+240 b^3 c+240 b^3 d x}{480 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 110, normalized size = 0.69 \[ \frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} d x + {\left (24 \, a^{3} \cos \left (d x + c\right )^{4} + 90 \, a^{2} b \cos \left (d x + c\right )^{3} + 64 \, a^{3} + 240 \, a b^{2} + 8 \, {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 332, normalized size = 2.08 \[ \frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.28, size = 123, normalized size = 0.77 \[ \frac {\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b^{2} a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 119, normalized size = 0.74 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3}}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.93, size = 287, normalized size = 1.79 \[ \frac {\left (2\,a^3-\frac {15\,a^2\,b}{4}+6\,a\,b^2-b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,a^3}{3}-\frac {3\,a^2\,b}{2}+16\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^3}{15}+20\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,a^3}{3}+\frac {3\,a^2\,b}{2}+16\,a\,b^2+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+\frac {15\,a^2\,b}{4}+6\,a\,b^2+b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^2+4\,b^2\right )}{4\,\left (\frac {9\,a^2\,b}{4}+b^3\right )}\right )\,\left (9\,a^2+4\,b^2\right )}{4\,d} \]
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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