Optimal. Leaf size=109 \[ \frac {d \cos (a+b x)}{8 b^2}-\frac {d \cos (3 a+3 b x)}{144 b^2}-\frac {d \cos (5 a+5 b x)}{400 b^2}+\frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b} \]
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Rubi [A] time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4406, 3296, 2638} \[ \frac {d \cos (a+b x)}{8 b^2}-\frac {d \cos (3 a+3 b x)}{144 b^2}-\frac {d \cos (5 a+5 b x)}{400 b^2}+\frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b} \]
Antiderivative was successfully verified.
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Rule 2638
Rule 3296
Rule 4406
Rubi steps
\begin {align*} \int (c+d x) \cos ^3(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac {1}{8} (c+d x) \cos (a+b x)-\frac {1}{16} (c+d x) \cos (3 a+3 b x)-\frac {1}{16} (c+d x) \cos (5 a+5 b x)\right ) \, dx\\ &=-\left (\frac {1}{16} \int (c+d x) \cos (3 a+3 b x) \, dx\right )-\frac {1}{16} \int (c+d x) \cos (5 a+5 b x) \, dx+\frac {1}{8} \int (c+d x) \cos (a+b x) \, dx\\ &=\frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b}+\frac {d \int \sin (5 a+5 b x) \, dx}{80 b}+\frac {d \int \sin (3 a+3 b x) \, dx}{48 b}-\frac {d \int \sin (a+b x) \, dx}{8 b}\\ &=\frac {d \cos (a+b x)}{8 b^2}-\frac {d \cos (3 a+3 b x)}{144 b^2}-\frac {d \cos (5 a+5 b x)}{400 b^2}+\frac {(c+d x) \sin (a+b x)}{8 b}-\frac {(c+d x) \sin (3 a+3 b x)}{48 b}-\frac {(c+d x) \sin (5 a+5 b x)}{80 b}\\ \end {align*}
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Mathematica [A] time = 0.36, size = 110, normalized size = 1.01 \[ -\frac {-450 b c \sin (a+b x)+75 b c \sin (3 (a+b x))+45 b c \sin (5 (a+b x))-450 b d x \sin (a+b x)+75 b d x \sin (3 (a+b x))+45 b d x \sin (5 (a+b x))-450 d \cos (a+b x)+25 d \cos (3 (a+b x))+9 d \cos (5 (a+b x))}{3600 b^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.76, size = 91, normalized size = 0.83 \[ -\frac {9 \, d \cos \left (b x + a\right )^{5} - 5 \, d \cos \left (b x + a\right )^{3} - 30 \, d \cos \left (b x + a\right ) + 15 \, {\left (3 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{4} - 2 \, b d x - {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} - 2 \, b c\right )} \sin \left (b x + a\right )}{225 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.97, size = 106, normalized size = 0.97 \[ -\frac {d \cos \left (5 \, b x + 5 \, a\right )}{400 \, b^{2}} - \frac {d \cos \left (3 \, b x + 3 \, a\right )}{144 \, b^{2}} + \frac {d \cos \left (b x + a\right )}{8 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (5 \, b x + 5 \, a\right )}{80 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (3 \, b x + 3 \, a\right )}{48 \, b^{2}} + \frac {{\left (b d x + b c\right )} \sin \left (b x + a\right )}{8 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 175, normalized size = 1.61 \[ \frac {\frac {d \left (\frac {\left (b x +a \right ) \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}+\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{45}+\frac {2 \cos \left (b x +a \right )}{15}-\frac {\left (b x +a \right ) \left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{5}-\frac {\left (\cos ^{5}\left (b x +a \right )\right )}{25}\right )}{b}-\frac {d a \left (-\frac {\left (\cos ^{4}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{5}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{15}\right )}{b}+c \left (-\frac {\left (\cos ^{4}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{5}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{15}\right )}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.95, size = 139, normalized size = 1.28 \[ -\frac {240 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )} c - \frac {240 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )} a d}{b} + \frac {{\left (45 \, {\left (b x + a\right )} \sin \left (5 \, b x + 5 \, a\right ) + 75 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + 9 \, \cos \left (5 \, b x + 5 \, a\right ) + 25 \, \cos \left (3 \, b x + 3 \, a\right ) - 450 \, \cos \left (b x + a\right )\right )} d}{b}}{3600 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.47, size = 119, normalized size = 1.09 \[ \frac {26\,d\,{\cos \left (a+b\,x\right )}^5+65\,d\,{\cos \left (a+b\,x\right )}^3\,{\sin \left (a+b\,x\right )}^2+30\,d\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^4+30\,b\,c\,{\sin \left (a+b\,x\right )}^5+30\,b\,d\,x\,{\sin \left (a+b\,x\right )}^5+75\,b\,c\,{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3+75\,b\,d\,x\,{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3}{225\,b^2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.10, size = 163, normalized size = 1.50 \[ \begin {cases} \frac {2 c \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {c \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {2 d x \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {d x \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {2 d \sin ^{4}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{15 b^{2}} + \frac {13 d \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{45 b^{2}} + \frac {26 d \cos ^{5}{\left (a + b x \right )}}{225 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{2}{\relax (a )} \cos ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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