Optimal. Leaf size=134 \[ -\frac {\sin (a+b x) \cos ^3(a+b x)}{32 b^3}-\frac {15 \sin (a+b x) \cos (a+b x)}{64 b^3}+\frac {x \cos ^4(a+b x)}{8 b^2}+\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x^2 \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {3 x^2 \sin (a+b x) \cos (a+b x)}{8 b}-\frac {15 x}{64 b^2}+\frac {x^3}{8} \]
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Rubi [A] time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3311, 30, 2635, 8} \[ \frac {x \cos ^4(a+b x)}{8 b^2}+\frac {3 x \cos ^2(a+b x)}{8 b^2}-\frac {\sin (a+b x) \cos ^3(a+b x)}{32 b^3}-\frac {15 \sin (a+b x) \cos (a+b x)}{64 b^3}+\frac {x^2 \sin (a+b x) \cos ^3(a+b x)}{4 b}+\frac {3 x^2 \sin (a+b x) \cos (a+b x)}{8 b}-\frac {15 x}{64 b^2}+\frac {x^3}{8} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2635
Rule 3311
Rubi steps
\begin {align*} \int x^2 \cos ^4(a+b x) \, dx &=\frac {x \cos ^4(a+b x)}{8 b^2}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac {3}{4} \int x^2 \cos ^2(a+b x) \, dx-\frac {\int \cos ^4(a+b x) \, dx}{8 b^2}\\ &=\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x \cos ^4(a+b x)}{8 b^2}+\frac {3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}+\frac {3 \int x^2 \, dx}{8}-\frac {3 \int \cos ^2(a+b x) \, dx}{32 b^2}-\frac {3 \int \cos ^2(a+b x) \, dx}{8 b^2}\\ &=\frac {x^3}{8}+\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x \cos ^4(a+b x)}{8 b^2}-\frac {15 \cos (a+b x) \sin (a+b x)}{64 b^3}+\frac {3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}-\frac {3 \int 1 \, dx}{64 b^2}-\frac {3 \int 1 \, dx}{16 b^2}\\ &=-\frac {15 x}{64 b^2}+\frac {x^3}{8}+\frac {3 x \cos ^2(a+b x)}{8 b^2}+\frac {x \cos ^4(a+b x)}{8 b^2}-\frac {15 \cos (a+b x) \sin (a+b x)}{64 b^3}+\frac {3 x^2 \cos (a+b x) \sin (a+b x)}{8 b}-\frac {\cos ^3(a+b x) \sin (a+b x)}{32 b^3}+\frac {x^2 \cos ^3(a+b x) \sin (a+b x)}{4 b}\\ \end {align*}
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Mathematica [A] time = 0.18, size = 92, normalized size = 0.69 \[ \frac {64 b^2 x^2 \sin (2 (a+b x))+8 b^2 x^2 \sin (4 (a+b x))-32 \sin (2 (a+b x))-\sin (4 (a+b x))+64 b x \cos (2 (a+b x))+4 b x \cos (4 (a+b x))+32 b^3 x^3}{256 b^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 88, normalized size = 0.66 \[ \frac {8 \, b^{3} x^{3} + 8 \, b x \cos \left (b x + a\right )^{4} + 24 \, b x \cos \left (b x + a\right )^{2} - 15 \, b x + {\left (2 \, {\left (8 \, b^{2} x^{2} - 1\right )} \cos \left (b x + a\right )^{3} + 3 \, {\left (8 \, b^{2} x^{2} - 5\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{64 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.36, size = 84, normalized size = 0.63 \[ \frac {1}{8} \, x^{3} + \frac {x \cos \left (4 \, b x + 4 \, a\right )}{64 \, b^{2}} + \frac {x \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} + \frac {{\left (8 \, b^{2} x^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right )}{256 \, b^{3}} + \frac {{\left (2 \, b^{2} x^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.03, size = 241, normalized size = 1.80 \[ \frac {\left (b x +a \right )^{2} \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )+\frac {\left (b x +a \right ) \left (\cos ^{4}\left (b x +a \right )\right )}{8}-\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{32}-\frac {15 b x}{64}-\frac {15 a}{64}+\frac {3 \left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{8}-\frac {3 \cos \left (b x +a \right ) \sin \left (b x +a \right )}{16}-\frac {\left (b x +a \right )^{3}}{4}-2 a \left (\left (b x +a \right ) \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )-\frac {3 \left (b x +a \right )^{2}}{16}+\frac {\left (\cos ^{4}\left (b x +a \right )\right )}{16}+\frac {3 \left (\cos ^{2}\left (b x +a \right )\right )}{16}\right )+a^{2} \left (\frac {\left (\cos ^{3}\left (b x +a \right )+\frac {3 \cos \left (b x +a \right )}{2}\right ) \sin \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )}{b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 188, normalized size = 1.40 \[ \frac {32 \, {\left (b x + a\right )}^{3} + 8 \, {\left (12 \, b x + 12 \, a + \sin \left (4 \, b x + 4 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} - 4 \, {\left (24 \, {\left (b x + a\right )}^{2} + 4 \, {\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right ) + 16 \, \cos \left (2 \, b x + 2 \, a\right )\right )} a + 4 \, {\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) + 64 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (8 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right ) + 32 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )}{256 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.54, size = 104, normalized size = 0.78 \[ \frac {x^3}{8}-\frac {\frac {\sin \left (2\,a+2\,b\,x\right )}{8}+\frac {\sin \left (4\,a+4\,b\,x\right )}{256}+b\,\left (\frac {x\,\left (2\,{\sin \left (a+b\,x\right )}^2-1\right )}{4}+\frac {x\,\left (2\,{\sin \left (2\,a+2\,b\,x\right )}^2-1\right )}{64}\right )-b^2\,\left (\frac {x^2\,\sin \left (2\,a+2\,b\,x\right )}{4}+\frac {x^2\,\sin \left (4\,a+4\,b\,x\right )}{32}\right )}{b^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.22, size = 209, normalized size = 1.56 \[ \begin {cases} \frac {x^{3} \sin ^{4}{\left (a + b x \right )}}{8} + \frac {x^{3} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {x^{3} \cos ^{4}{\left (a + b x \right )}}{8} + \frac {3 x^{2} \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} + \frac {5 x^{2} \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac {15 x \sin ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac {3 x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{32 b^{2}} + \frac {17 x \cos ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac {15 \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{64 b^{3}} - \frac {17 \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \cos ^{4}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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