Optimal. Leaf size=120 \[ -\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac{3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac{3 d^3 \sin (a+b x) \cos (a+b x)}{8 b^4}+\frac{(c+d x)^3 \sin ^2(a+b x)}{2 b}+\frac{3 d^3 x}{8 b^3}-\frac{(c+d x)^3}{4 b} \]
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Rubi [A] time = 0.0834477, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4404, 3311, 32, 2635, 8} \[ -\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac{3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{4 b^2}-\frac{3 d^3 \sin (a+b x) \cos (a+b x)}{8 b^4}+\frac{(c+d x)^3 \sin ^2(a+b x)}{2 b}+\frac{3 d^3 x}{8 b^3}-\frac{(c+d x)^3}{4 b} \]
Antiderivative was successfully verified.
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Rule 4404
Rule 3311
Rule 32
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int (c+d x)^3 \cos (a+b x) \sin (a+b x) \, dx &=\frac{(c+d x)^3 \sin ^2(a+b x)}{2 b}-\frac{(3 d) \int (c+d x)^2 \sin ^2(a+b x) \, dx}{2 b}\\ &=\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac{(c+d x)^3 \sin ^2(a+b x)}{2 b}-\frac{(3 d) \int (c+d x)^2 \, dx}{4 b}+\frac{\left (3 d^3\right ) \int \sin ^2(a+b x) \, dx}{4 b^3}\\ &=-\frac{(c+d x)^3}{4 b}-\frac{3 d^3 \cos (a+b x) \sin (a+b x)}{8 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac{(c+d x)^3 \sin ^2(a+b x)}{2 b}+\frac{\left (3 d^3\right ) \int 1 \, dx}{8 b^3}\\ &=\frac{3 d^3 x}{8 b^3}-\frac{(c+d x)^3}{4 b}-\frac{3 d^3 \cos (a+b x) \sin (a+b x)}{8 b^4}+\frac{3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{4 b^2}-\frac{3 d^2 (c+d x) \sin ^2(a+b x)}{4 b^3}+\frac{(c+d x)^3 \sin ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 0.300051, size = 71, normalized size = 0.59 \[ \frac{3 d \sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )-2 b (c+d x) \cos (2 (a+b x)) \left (2 b^2 (c+d x)^2-3 d^2\right )}{16 b^4} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.015, size = 466, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.13191, size = 462, normalized size = 3.85 \begin{align*} -\frac{8 \, c^{3} \cos \left (b x + a\right )^{2} - \frac{24 \, a c^{2} d \cos \left (b x + a\right )^{2}}{b} + \frac{24 \, a^{2} c d^{2} \cos \left (b x + a\right )^{2}}{b^{2}} - \frac{8 \, a^{3} d^{3} \cos \left (b x + a\right )^{2}}{b^{3}} + \frac{6 \,{\left (2 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2} d}{b} - \frac{12 \,{\left (2 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} a c d^{2}}{b^{2}} + \frac{6 \,{\left (2 \,{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} d^{3}}{b^{3}} + \frac{6 \,{\left ({\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} c d^{2}}{b^{2}} - \frac{6 \,{\left ({\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \,{\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} a d^{3}}{b^{3}} + \frac{{\left (2 \,{\left (2 \,{\left (b x + a\right )}^{3} - 3 \, b x - 3 \, a\right )} \cos \left (2 \, b x + 2 \, a\right ) - 3 \,{\left (2 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{3}}{b^{3}}}{16 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.483148, size = 348, normalized size = 2.9 \begin{align*} \frac{2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} - 2 \,{\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 2 \, b^{3} c^{3} - 3 \, b c d^{2} + 3 \,{\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (b x + a\right )^{2} + 3 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 \,{\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x}{8 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 3.10669, size = 342, normalized size = 2.85 \begin{align*} \begin{cases} \frac{c^{3} \sin ^{2}{\left (a + b x \right )}}{2 b} + \frac{3 c^{2} d x \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac{3 c^{2} d x \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac{3 c d^{2} x^{2} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac{3 c d^{2} x^{2} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac{d^{3} x^{3} \sin ^{2}{\left (a + b x \right )}}{4 b} - \frac{d^{3} x^{3} \cos ^{2}{\left (a + b x \right )}}{4 b} + \frac{3 c^{2} d \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 b^{2}} + \frac{3 c d^{2} x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b^{2}} + \frac{3 d^{3} x^{2} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 b^{2}} - \frac{3 c d^{2} \sin ^{2}{\left (a + b x \right )}}{4 b^{3}} - \frac{3 d^{3} x \sin ^{2}{\left (a + b x \right )}}{8 b^{3}} + \frac{3 d^{3} x \cos ^{2}{\left (a + b x \right )}}{8 b^{3}} - \frac{3 d^{3} \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b^{4}} & \text{for}\: b \neq 0 \\\left (c^{3} x + \frac{3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac{d^{3} x^{4}}{4}\right ) \sin{\left (a \right )} \cos{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.14494, size = 163, normalized size = 1.36 \begin{align*} -\frac{{\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{2} d x + 2 \, b^{3} c^{3} - 3 \, b d^{3} x - 3 \, b c d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{4}} + \frac{3 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \sin \left (2 \, b x + 2 \, a\right )}{16 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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