Optimal. Leaf size=106 \[ -\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}+\frac {a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a b x}{4} \]
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Rubi [A] time = 0.16, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2862, 2669, 2635, 8} \[ -\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}+\frac {a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a b x}{4} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2635
Rule 2669
Rule 2862
Rubi steps
\begin {align*} \int \cos ^2(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx &=-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{5} \int \cos ^2(c+d x) (2 b+2 a \sin (c+d x)) (a+b \sin (c+d x)) \, dx\\ &=-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{20} \int \cos ^2(c+d x) \left (10 a b+2 \left (a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{2} (a b) \int \cos ^2(c+d x) \, dx\\ &=-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac {1}{4} (a b) \int 1 \, dx\\ &=\frac {a b x}{4}-\frac {\left (a^2+4 b^2\right ) \cos ^3(c+d x)}{30 d}+\frac {a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac {a \cos ^3(c+d x) (a+b \sin (c+d x))}{10 d}-\frac {\cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\\ \end {align*}
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Mathematica [A] time = 0.37, size = 77, normalized size = 0.73 \[ \frac {-30 \left (2 a^2+b^2\right ) \cos (c+d x)-5 \left (4 a^2+b^2\right ) \cos (3 (c+d x))+3 b (20 a (c+d x)-5 a \sin (4 (c+d x))+b \cos (5 (c+d x)))}{240 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 73, normalized size = 0.69 \[ \frac {12 \, b^{2} \cos \left (d x + c\right )^{5} + 15 \, a b d x - 20 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (2 \, a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 82, normalized size = 0.77 \[ \frac {1}{4} \, a b x + \frac {b^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a b \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac {{\left (4 \, a^{2} + b^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.19, size = 94, normalized size = 0.89 \[ \frac {-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.38, size = 68, normalized size = 0.64 \[ -\frac {80 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b - 16 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{2}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.77, size = 180, normalized size = 1.70 \[ \frac {a\,b\,x}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {4\,a^2}{3}+\frac {4\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {8\,a^2}{3}-\frac {4\,b^2}{3}\right )+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {2\,a^2}{3}+\frac {4\,b^2}{15}-3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.96, size = 172, normalized size = 1.62 \[ \begin {cases} - \frac {a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 b^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\relax (c )}\right )^{2} \sin {\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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