Optimal. Leaf size=154 \[ -\frac {a^3 (15 A+13 B) \sin ^3(c+d x)}{60 d}+\frac {a^3 (15 A+13 B) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} a^3 x (15 A+13 B)+\frac {(5 A-B) \sin (c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^4}{5 a d} \]
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Rubi [A] time = 0.23, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2968, 3023, 2751, 2645, 2637, 2635, 8, 2633} \[ -\frac {a^3 (15 A+13 B) \sin ^3(c+d x)}{60 d}+\frac {a^3 (15 A+13 B) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} a^3 x (15 A+13 B)+\frac {(5 A-B) \sin (c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^4}{5 a d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 2637
Rule 2645
Rule 2751
Rule 2968
Rule 3023
Rubi steps
\begin {align*} \int \cos (c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\int (a+a \cos (c+d x))^3 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {\int (a+a \cos (c+d x))^3 (4 a B+a (5 A-B) \cos (c+d x)) \, dx}{5 a}\\ &=\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} (15 A+13 B) \int (a+a \cos (c+d x))^3 \, dx\\ &=\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} (15 A+13 B) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac {1}{20} a^3 (15 A+13 B) x+\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} \left (a^3 (15 A+13 B)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 A+13 B)\right ) \int \cos (c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 A+13 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {1}{20} a^3 (15 A+13 B) x+\frac {3 a^3 (15 A+13 B) \sin (c+d x)}{20 d}+\frac {3 a^3 (15 A+13 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{40} \left (3 a^3 (15 A+13 B)\right ) \int 1 \, dx-\frac {\left (a^3 (15 A+13 B)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac {1}{8} a^3 (15 A+13 B) x+\frac {a^3 (15 A+13 B) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 A+13 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(5 A-B) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}-\frac {a^3 (15 A+13 B) \sin ^3(c+d x)}{60 d}\\ \end {align*}
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Mathematica [A] time = 0.46, size = 108, normalized size = 0.70 \[ \frac {a^3 (60 (26 A+23 B) \sin (c+d x)+480 (A+B) \sin (2 (c+d x))+120 A \sin (3 (c+d x))+15 A \sin (4 (c+d x))+900 A d x+170 B \sin (3 (c+d x))+45 B \sin (4 (c+d x))+6 B \sin (5 (c+d x))+780 B c+780 B d x)}{480 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.87, size = 110, normalized size = 0.71 \[ \frac {15 \, {\left (15 \, A + 13 \, B\right )} a^{3} d x + {\left (24 \, B a^{3} \cos \left (d x + c\right )^{4} + 30 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, A + 19 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (15 \, A + 13 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (45 \, A + 38 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.41, size = 136, normalized size = 0.88 \[ \frac {B a^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (15 \, A a^{3} + 13 \, B a^{3}\right )} x + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (12 \, A a^{3} + 17 \, B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a^{3} + B a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {{\left (26 \, A a^{3} + 23 \, B a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 223, normalized size = 1.45 \[ \frac {A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{3} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,a^{3} \sin \left (d x +c \right )+a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 213, normalized size = 1.38 \[ -\frac {480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} + 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 480 \, A a^{3} \sin \left (d x + c\right )}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.50, size = 277, normalized size = 1.80 \[ \frac {\left (\frac {15\,A\,a^3}{4}+\frac {13\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {35\,A\,a^3}{2}+\frac {91\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (32\,A\,a^3+\frac {416\,B\,a^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {61\,A\,a^3}{2}+\frac {133\,B\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+\frac {51\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^3\,\left (15\,A+13\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+13\,B\right )}{4\,\left (\frac {15\,A\,a^3}{4}+\frac {13\,B\,a^3}{4}\right )}\right )\,\left (15\,A+13\,B\right )}{4\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.81, size = 530, normalized size = 3.44 \[ \begin {cases} \frac {3 A a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {A a^{3} \sin {\left (c + d x \right )}}{d} + \frac {9 B a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {9 B a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {8 B a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 B a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 B a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right )^{3} \cos {\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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