Optimal. Leaf size=243 \[ \frac {\left (-3 a^2 C+15 a b B+16 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}+\frac {\left (-6 a^3 C+30 a^2 b B+71 a b^2 C+45 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac {1}{8} x \left (4 a^3 C+12 a^2 b B+9 a b^2 C+3 b^3 B\right )+\frac {\left (-3 a^4 C+15 a^3 b B+52 a^2 b^2 C+60 a b^3 B+16 b^4 C\right ) \sin (c+d x)}{30 b d}+\frac {(5 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d} \]
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Rubi [A] time = 0.29, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3023, 2753, 2734} \[ \frac {\left (52 a^2 b^2 C+15 a^3 b B-3 a^4 C+60 a b^3 B+16 b^4 C\right ) \sin (c+d x)}{30 b d}+\frac {\left (-3 a^2 C+15 a b B+16 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}+\frac {\left (30 a^2 b B-6 a^3 C+71 a b^2 C+45 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac {1}{8} x \left (12 a^2 b B+4 a^3 C+9 a b^2 C+3 b^3 B\right )+\frac {(5 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d} \]
Antiderivative was successfully verified.
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Rule 2734
Rule 2753
Rule 3023
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x))^3 (4 b C+(5 b B-a C) \cos (c+d x)) \, dx}{5 b}\\ &=\frac {(5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x))^2 \left (b (15 b B+13 a C)+\left (15 a b B-3 a^2 C+16 b^2 C\right ) \cos (c+d x)\right ) \, dx}{20 b}\\ &=\frac {\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x)) \left (b \left (75 a b B+33 a^2 C+32 b^2 C\right )+\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \cos (c+d x)\right ) \, dx}{60 b}\\ &=\frac {1}{8} \left (12 a^2 b B+3 b^3 B+4 a^3 C+9 a b^2 C\right ) x+\frac {\left (15 a^3 b B+60 a b^3 B-3 a^4 C+52 a^2 b^2 C+16 b^4 C\right ) \sin (c+d x)}{30 b d}+\frac {\left (30 a^2 b B+45 b^3 B-6 a^3 C+71 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{120 d}+\frac {\left (15 a b B-3 a^2 C+16 b^2 C\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}+\frac {(5 b B-a C) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}\\ \end {align*}
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Mathematica [A] time = 0.69, size = 176, normalized size = 0.72 \[ \frac {10 b \left (12 a^2 C+12 a b B+5 b^2 C\right ) \sin (3 (c+d x))+60 (c+d x) \left (4 a^3 C+12 a^2 b B+9 a b^2 C+3 b^3 B\right )+60 \left (8 a^3 B+18 a^2 b C+18 a b^2 B+5 b^3 C\right ) \sin (c+d x)+120 \left (a^3 C+3 a^2 b B+3 a b^2 C+b^3 B\right ) \sin (2 (c+d x))+15 b^2 (3 a C+b B) \sin (4 (c+d x))+6 b^3 C \sin (5 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 174, normalized size = 0.72 \[ \frac {15 \, {\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} d x + {\left (24 \, C b^{3} \cos \left (d x + c\right )^{4} + 120 \, B a^{3} + 240 \, C a^{2} b + 240 \, B a b^{2} + 64 \, C b^{3} + 30 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, C a^{2} b + 15 \, B a b^{2} + 4 \, C b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.40, size = 188, normalized size = 0.77 \[ \frac {C b^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (4 \, C a^{3} + 12 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} x + \frac {{\left (3 \, C a b^{2} + B b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (12 \, C a^{2} b + 12 \, B a b^{2} + 5 \, C b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (C a^{3} + 3 \, B a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, B a^{3} + 18 \, C a^{2} b + 18 \, B a b^{2} + 5 \, C b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 227, normalized size = 0.93 \[ \frac {C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} B \sin \left (d x +c \right )+C \,a^{2} b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{2} b B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C a \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {b^{3} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+b^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 217, normalized size = 0.89 \[ \frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{2} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{3} + 480 \, B a^{3} \sin \left (d x + c\right )}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.03, size = 277, normalized size = 1.14 \[ \frac {3\,B\,b^3\,x}{8}+\frac {C\,a^3\,x}{2}+\frac {3\,B\,a^2\,b\,x}{2}+\frac {9\,C\,a\,b^2\,x}{8}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,C\,b^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {5\,C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {C\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,C\,a\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {9\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,C\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.09, size = 552, normalized size = 2.27 \[ \begin {cases} \frac {B a^{3} \sin {\left (c + d x \right )}}{d} + \frac {3 B a^{2} b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 B a b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {C a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C a^{2} b \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 C a^{2} b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 C a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 C a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {9 C a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {9 C a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {15 C a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 C b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{3} \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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