Optimal. Leaf size=224 \[ \frac {\sin (c+d x) \left (5 a^2 (3 A+2 C)+20 a b B+2 b^2 (5 A+4 C)\right )}{15 d}+\frac {\sin (c+d x) \cos ^2(c+d x) \left (2 a^2 C+10 a b B+5 A b^2+4 b^2 C\right )}{15 d}+\frac {\sin (c+d x) \cos (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )}{8 d}+\frac {1}{8} x \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )+\frac {b (2 a C+5 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
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Rubi [A] time = 0.33, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3049, 3033, 3023, 2734} \[ \frac {\sin (c+d x) \left (5 a^2 (3 A+2 C)+20 a b B+2 b^2 (5 A+4 C)\right )}{15 d}+\frac {\sin (c+d x) \cos ^2(c+d x) \left (2 a^2 C+10 a b B+5 A b^2+4 b^2 C\right )}{15 d}+\frac {\sin (c+d x) \cos (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )}{8 d}+\frac {1}{8} x \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )+\frac {b (2 a C+5 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d} \]
Antiderivative was successfully verified.
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Rule 2734
Rule 3023
Rule 3033
Rule 3049
Rubi steps
\begin {align*} \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos (c+d x) (a+b \cos (c+d x)) \left (a (5 A+2 C)+(5 A b+5 a B+4 b C) \cos (c+d x)+(5 b B+2 a C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b (5 b B+2 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos (c+d x) \left (4 a^2 (5 A+2 C)+5 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)+4 \left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {\left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac {b (5 b B+2 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{60} \int \cos (c+d x) \left (4 \left (20 a b B+5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right )+15 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {1}{8} \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) x+\frac {\left (20 a b B+5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {\left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac {b (5 b B+2 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.84, size = 169, normalized size = 0.75 \[ \frac {60 (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )+60 \sin (c+d x) \left (a^2 (8 A+6 C)+12 a b B+b^2 (6 A+5 C)\right )+120 \sin (2 (c+d x)) \left (a^2 B+2 a b (A+C)+b^2 B\right )+10 \sin (3 (c+d x)) \left (4 a^2 C+8 a b B+4 A b^2+5 b^2 C\right )+15 b (2 a C+b B) \sin (4 (c+d x))+6 b^2 C \sin (5 (c+d x))}{480 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 171, normalized size = 0.76 \[ \frac {15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} d x + {\left (24 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + 40 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 160 \, B a b + 16 \, {\left (5 \, A + 4 \, C\right )} b^{2} + 8 \, {\left (5 \, C a^{2} + 10 \, B a b + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 184, normalized size = 0.82 \[ \frac {C b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (4 \, B a^{2} + 8 \, A a b + 6 \, C a b + 3 \, B b^{2}\right )} x + \frac {{\left (2 \, C a b + B b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, C a^{2} + 8 \, B a b + 4 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{2} + 2 \, A a b + 2 \, C a b + B b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, A a^{2} + 6 \, C a^{2} + 12 \, B a b + 6 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.29, size = 244, normalized size = 1.09 \[ \frac {a^{2} A \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 C a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{2} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 233, normalized size = 1.04 \[ \frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{2} + 480 \, A a^{2} \sin \left (d x + c\right )}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.44, size = 256, normalized size = 1.14 \[ \frac {30\,B\,a^2\,\sin \left (2\,c+2\,d\,x\right )+10\,A\,b^2\,\sin \left (3\,c+3\,d\,x\right )+30\,B\,b^2\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,a^2\,\sin \left (3\,c+3\,d\,x\right )+\frac {15\,B\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{4}+\frac {25\,C\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,C\,b^2\,\sin \left (5\,c+5\,d\,x\right )}{2}+120\,A\,a^2\,\sin \left (c+d\,x\right )+90\,A\,b^2\,\sin \left (c+d\,x\right )+90\,C\,a^2\,\sin \left (c+d\,x\right )+75\,C\,b^2\,\sin \left (c+d\,x\right )+60\,A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+20\,B\,a\,b\,\sin \left (3\,c+3\,d\,x\right )+60\,C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+\frac {15\,C\,a\,b\,\sin \left (4\,c+4\,d\,x\right )}{2}+60\,B\,a^2\,d\,x+45\,B\,b^2\,d\,x+180\,B\,a\,b\,\sin \left (c+d\,x\right )+120\,A\,a\,b\,d\,x+90\,C\,a\,b\,d\,x}{120\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.15, size = 570, normalized size = 2.54 \[ \begin {cases} \frac {A a^{2} \sin {\left (c + d x \right )}}{d} + A a b x \sin ^{2}{\left (c + d x \right )} + A a b x \cos ^{2}{\left (c + d x \right )} + \frac {A a b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {2 A b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 B a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 B a b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {5 C a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {8 C b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b^{2} \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 )^{2} \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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