Optimal. Leaf size=79 \[ -\frac{d (c+d x) \cos (4 a+4 b x)}{64 b^2}+\frac{d^2 \sin (4 a+4 b x)}{256 b^3}-\frac{(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^3}{24 d} \]
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Rubi [A] time = 0.122533, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4406, 3296, 2637} \[ -\frac{d (c+d x) \cos (4 a+4 b x)}{64 b^2}+\frac{d^2 \sin (4 a+4 b x)}{256 b^3}-\frac{(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^3}{24 d} \]
Antiderivative was successfully verified.
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Rule 4406
Rule 3296
Rule 2637
Rubi steps
\begin{align*} \int (c+d x)^2 \cos ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac{1}{8} (c+d x)^2-\frac{1}{8} (c+d x)^2 \cos (4 a+4 b x)\right ) \, dx\\ &=\frac{(c+d x)^3}{24 d}-\frac{1}{8} \int (c+d x)^2 \cos (4 a+4 b x) \, dx\\ &=\frac{(c+d x)^3}{24 d}-\frac{(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac{d \int (c+d x) \sin (4 a+4 b x) \, dx}{16 b}\\ &=\frac{(c+d x)^3}{24 d}-\frac{d (c+d x) \cos (4 a+4 b x)}{64 b^2}-\frac{(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac{d^2 \int \cos (4 a+4 b x) \, dx}{64 b^2}\\ &=\frac{(c+d x)^3}{24 d}-\frac{d (c+d x) \cos (4 a+4 b x)}{64 b^2}+\frac{d^2 \sin (4 a+4 b x)}{256 b^3}-\frac{(c+d x)^2 \sin (4 a+4 b x)}{32 b}\\ \end{align*}
Mathematica [A] time = 0.43819, size = 77, normalized size = 0.97 \[ \frac{-3 \sin (4 (a+b x)) \left (8 b^2 (c+d x)^2-d^2\right )-12 b d (c+d x) \cos (4 (a+b x))+32 b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )}{768 b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.02, size = 519, normalized size = 6.6 \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 = 1.22758, size = 313, normalized size = 3.96 \begin{align*} \frac{24 \,{\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} c^{2} - \frac{48 \,{\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} a c d}{b} + \frac{24 \,{\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} a^{2} d^{2}}{b^{2}} + \frac{12 \,{\left (8 \,{\left (b x + a\right )}^{2} - 4 \,{\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) - \cos \left (4 \, b x + 4 \, a\right )\right )} c d}{b} - \frac{12 \,{\left (8 \,{\left (b x + a\right )}^{2} - 4 \,{\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) - \cos \left (4 \, b x + 4 \, a\right )\right )} a d^{2}}{b^{2}} + \frac{{\left (32 \,{\left (b x + a\right )}^{3} - 12 \,{\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) - 3 \,{\left (8 \,{\left (b x + a\right )}^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} d^{2}}{b^{2}}}{768 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.500507, size = 398, normalized size = 5.04 \begin{align*} \frac{8 \, b^{3} d^{2} x^{3} + 24 \, b^{3} c d x^{2} - 24 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} + 24 \,{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 3 \,{\left (8 \, b^{3} c^{2} - b d^{2}\right )} x - 3 \,{\left (2 \,{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{3} -{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{192 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 5.12042, size = 484, normalized size = 6.13 \begin{align*} \begin{cases} \frac{c^{2} x \sin ^{4}{\left (a + b x \right )}}{8} + \frac{c^{2} x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac{c^{2} x \cos ^{4}{\left (a + b x \right )}}{8} + \frac{c d x^{2} \sin ^{4}{\left (a + b x \right )}}{8} + \frac{c d x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac{c d x^{2} \cos ^{4}{\left (a + b x \right )}}{8} + \frac{d^{2} x^{3} \sin ^{4}{\left (a + b x \right )}}{24} + \frac{d^{2} x^{3} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{12} + \frac{d^{2} x^{3} \cos ^{4}{\left (a + b x \right )}}{24} + \frac{c^{2} \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b} - \frac{c^{2} \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} + \frac{c d x \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{4 b} - \frac{c d x \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{4 b} + \frac{d^{2} x^{2} \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{8 b} - \frac{d^{2} x^{2} \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} + \frac{c d \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{8 b^{2}} - \frac{d^{2} x \sin ^{4}{\left (a + b x \right )}}{64 b^{2}} + \frac{3 d^{2} x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{32 b^{2}} - \frac{d^{2} x \cos ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac{d^{2} \sin ^{3}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{64 b^{3}} + \frac{d^{2} \sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text{for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) \sin ^{2}{\left (a \right )} \cos ^{2}{\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 = 1.08375, size = 127, normalized size = 1.61 \begin{align*} \frac{1}{24} \, d^{2} x^{3} + \frac{1}{8} \, c d x^{2} + \frac{1}{8} \, c^{2} x - \frac{{\left (b d^{2} x + b c d\right )} \cos \left (4 \, b x + 4 \, a\right )}{64 \, b^{3}} - \frac{{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \sin \left (4 \, b x + 4 \, a\right )}{256 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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