Optimal. Leaf size=49 \[ -\frac {\sin ^3(2 a+2 b x)}{12 b}-\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{8 b}+\frac {x}{4} \]
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Rubi [A] time = 0.06, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4286, 2635, 8, 2564, 30} \[ -\frac {\sin ^3(2 a+2 b x)}{12 b}-\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{8 b}+\frac {x}{4} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2564
Rule 2635
Rule 4286
Rubi steps
\begin {align*} \int \sin ^2(a+b x) \sin ^2(2 a+2 b x) \, dx &=\frac {1}{2} \int \sin ^2(2 a+2 b x) \, dx-\frac {1}{2} \int \cos (2 a+2 b x) \sin ^2(2 a+2 b x) \, dx\\ &=-\frac {\cos (2 a+2 b x) \sin (2 a+2 b x)}{8 b}+\frac {\int 1 \, dx}{4}-\frac {\operatorname {Subst}\left (\int x^2 \, dx,x,\sin (2 a+2 b x)\right )}{4 b}\\ &=\frac {x}{4}-\frac {\cos (2 a+2 b x) \sin (2 a+2 b x)}{8 b}-\frac {\sin ^3(2 a+2 b x)}{12 b}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 40, normalized size = 0.82 \[ \frac {-3 \sin (2 (a+b x))-3 \sin (4 (a+b x))+\sin (6 (a+b x))+12 b x}{48 b} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 46, normalized size = 0.94 \[ \frac {3 \, b x + {\left (8 \, \cos \left (b x + a\right )^{5} - 14 \, \cos \left (b x + a\right )^{3} + 3 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{12 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 46, normalized size = 0.94 \[ \frac {1}{4} \, x + \frac {\sin \left (6 \, b x + 6 \, a\right )}{48 \, b} - \frac {\sin \left (4 \, b x + 4 \, a\right )}{16 \, b} - \frac {\sin \left (2 \, b x + 2 \, a\right )}{16 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.67, size = 47, normalized size = 0.96 \[ \frac {x}{4}-\frac {\sin \left (2 b x +2 a \right )}{16 b}-\frac {\sin \left (4 b x +4 a \right )}{16 b}+\frac {\sin \left (6 b x +6 a \right )}{48 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 41, normalized size = 0.84 \[ \frac {12 \, b x + \sin \left (6 \, b x + 6 \, a\right ) - 3 \, \sin \left (4 \, b x + 4 \, a\right ) - 3 \, \sin \left (2 \, b x + 2 \, a\right )}{48 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.30, size = 43, normalized size = 0.88 \[ \frac {x}{4}-\frac {\frac {\sin \left (2\,a+2\,b\,x\right )}{16}+\frac {\sin \left (4\,a+4\,b\,x\right )}{16}-\frac {\sin \left (6\,a+6\,b\,x\right )}{48}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 11.96, size = 231, normalized size = 4.71 \[ \begin {cases} \frac {x \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac {x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{4} + \frac {x \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {x \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{4} - \frac {7 \sin ^{2}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{24 b} - \frac {\sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{6 b} - \frac {\sin {\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{3 b} + \frac {\sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{24 b} & \text {for}\: b \neq 0 \\x \sin ^{2}{\relax (a )} \sin ^{2}{\left (2 a \right )} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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