Integrand size = 17, antiderivative size = 97 \[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b^2 n^2 x^3}{3 \left (9+4 b^2 n^2\right )}-\frac {2 b n x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}+\frac {3 x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2} \]
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Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4575, 30} \[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3 x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}-\frac {2 b n x^3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}+\frac {2 b^2 n^2 x^3}{3 \left (4 b^2 n^2+9\right )} \]
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Rule 30
Rule 4575
Rubi steps \begin{align*} \text {integral}& = -\frac {2 b n x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}+\frac {3 x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}+\frac {\left (2 b^2 n^2\right ) \int x^2 \, dx}{9+4 b^2 n^2} \\ & = \frac {2 b^2 n^2 x^3}{3 \left (9+4 b^2 n^2\right )}-\frac {2 b n x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}+\frac {3 x^3 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.63 \[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^3 \left (9+4 b^2 n^2-9 \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-6 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right )}{6 \left (9+4 b^2 n^2\right )} \]
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\[\int x^{2} {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}d x\]
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none
Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {6 \, b n x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + 9 \, x^{3} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - {\left (2 \, b^{2} n^{2} + 9\right )} x^{3}}{3 \, {\left (4 \, b^{2} n^{2} + 9\right )}} \]
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\[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \int x^{2} \sin ^{2}{\left (a - \frac {3 i \log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {3 i}{2 n} \\\int x^{2} \sin ^{2}{\left (a + \frac {3 i \log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {3 i}{2 n} \\\frac {2 b^{2} n^{2} x^{3} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{12 b^{2} n^{2} + 27} + \frac {2 b^{2} n^{2} x^{3} \cos ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{12 b^{2} n^{2} + 27} - \frac {6 b n x^{3} \sin {\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{12 b^{2} n^{2} + 27} + \frac {9 x^{3} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{12 b^{2} n^{2} + 27} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (95) = 190\).
Time = 0.21 (sec) , antiderivative size = 301, normalized size of antiderivative = 3.10 \[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {3 \, {\left (2 \, {\left (b \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) - b \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \sin \left (2 \, b \log \left (c\right )\right )\right )} n + 3 \, \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) + 3 \, \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + 3 \, \cos \left (2 \, b \log \left (c\right )\right )\right )} x^{3} \cos \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) + 3 \, {\left (2 \, {\left (b \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) + b \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \cos \left (2 \, b \log \left (c\right )\right )\right )} n - 3 \, \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) + 3 \, \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - 3 \, \sin \left (2 \, b \log \left (c\right )\right )\right )} x^{3} \sin \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) - 2 \, {\left (4 \, {\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + 9 \, \cos \left (2 \, b \log \left (c\right )\right )^{2} + 9 \, \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} x^{3}}{12 \, {\left (4 \, {\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + 9 \, \cos \left (2 \, b \log \left (c\right )\right )^{2} + 9 \, \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 833 vs. \(2 (95) = 190\).
Time = 0.46 (sec) , antiderivative size = 833, normalized size of antiderivative = 8.59 \[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
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Time = 28.00 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.69 \[ \int x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^3}{6}-\frac {x^3\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}\,1{}\mathrm {i}}{8\,b\,n+12{}\mathrm {i}}-\frac {x^3\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}{12+b\,n\,8{}\mathrm {i}} \]
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