Integrand size = 8, antiderivative size = 105 \[ \int x^2 \sin ^6(x) \, dx=-\frac {245 x}{1152}+\frac {5 x^3}{48}+\frac {245 \cos (x) \sin (x)}{1152}-\frac {5}{16} x^2 \cos (x) \sin (x)+\frac {5}{16} x \sin ^2(x)+\frac {65 \cos (x) \sin ^3(x)}{1728}-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x) \]
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Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3392, 30, 2715, 8} \[ \int x^2 \sin ^6(x) \, dx=\frac {5 x^3}{48}-\frac {1}{6} x^2 \sin ^5(x) \cos (x)-\frac {5}{24} x^2 \sin ^3(x) \cos (x)-\frac {5}{16} x^2 \sin (x) \cos (x)-\frac {245 x}{1152}+\frac {1}{18} x \sin ^6(x)+\frac {5}{48} x \sin ^4(x)+\frac {5}{16} x \sin ^2(x)+\frac {1}{108} \sin ^5(x) \cos (x)+\frac {65 \sin ^3(x) \cos (x)}{1728}+\frac {245 \sin (x) \cos (x)}{1152} \]
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Rule 8
Rule 30
Rule 2715
Rule 3392
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {1}{18} \int \sin ^6(x) \, dx+\frac {5}{6} \int x^2 \sin ^4(x) \, dx \\ & = -\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {5}{108} \int \sin ^4(x) \, dx-\frac {5}{48} \int \sin ^4(x) \, dx+\frac {5}{8} \int x^2 \sin ^2(x) \, dx \\ & = -\frac {5}{16} x^2 \cos (x) \sin (x)+\frac {5}{16} x \sin ^2(x)+\frac {65 \cos (x) \sin ^3(x)}{1728}-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {5}{144} \int \sin ^2(x) \, dx-\frac {5}{64} \int \sin ^2(x) \, dx+\frac {5 \int x^2 \, dx}{16}-\frac {5}{16} \int \sin ^2(x) \, dx \\ & = \frac {5 x^3}{48}+\frac {245 \cos (x) \sin (x)}{1152}-\frac {5}{16} x^2 \cos (x) \sin (x)+\frac {5}{16} x \sin ^2(x)+\frac {65 \cos (x) \sin ^3(x)}{1728}-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {5 \int 1 \, dx}{288}-\frac {5 \int 1 \, dx}{128}-\frac {5 \int 1 \, dx}{32} \\ & = -\frac {245 x}{1152}+\frac {5 x^3}{48}+\frac {245 \cos (x) \sin (x)}{1152}-\frac {5}{16} x^2 \cos (x) \sin (x)+\frac {5}{16} x \sin ^2(x)+\frac {65 \cos (x) \sin ^3(x)}{1728}-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int x^2 \sin ^6(x) \, dx=\frac {1440 x^3-3240 x \cos (2 x)+324 x \cos (4 x)-24 x \cos (6 x)-1620 \left (-1+2 x^2\right ) \sin (2 x)+81 \left (-1+8 x^2\right ) \sin (4 x)-4 \left (-1+18 x^2\right ) \sin (6 x)}{13824} \]
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Time = 0.84 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.64
method | result | size |
risch | \(\frac {5 x^{3}}{48}-\frac {x \cos \left (6 x \right )}{576}-\frac {\left (18 x^{2}-1\right ) \sin \left (6 x \right )}{3456}+\frac {3 x \cos \left (4 x \right )}{128}+\frac {3 \left (8 x^{2}-1\right ) \sin \left (4 x \right )}{512}-\frac {15 x \cos \left (2 x \right )}{64}-\frac {15 \left (2 x^{2}-1\right ) \sin \left (2 x \right )}{128}\) | \(67\) |
default | \(x^{2} \left (-\frac {\left (\sin ^{5}\left (x \right )+\frac {5 \left (\sin ^{3}\left (x \right )\right )}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{6}+\frac {5 x}{16}\right )+\frac {x \left (\sin ^{6}\left (x \right )\right )}{18}+\frac {\left (\sin ^{5}\left (x \right )+\frac {5 \left (\sin ^{3}\left (x \right )\right )}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{108}+\frac {115 x}{1152}+\frac {5 x \left (\sin ^{4}\left (x \right )\right )}{48}+\frac {5 \left (\sin ^{3}\left (x \right )+\frac {3 \sin \left (x \right )}{2}\right ) \cos \left (x \right )}{192}-\frac {5 \left (\cos ^{2}\left (x \right )\right ) x}{16}+\frac {5 \cos \left (x \right ) \sin \left (x \right )}{32}-\frac {5 x^{3}}{24}\) | \(96\) |
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Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.69 \[ \int x^2 \sin ^6(x) \, dx=-\frac {1}{18} \, x \cos \left (x\right )^{6} + \frac {13}{48} \, x \cos \left (x\right )^{4} + \frac {5}{48} \, x^{3} - \frac {11}{16} \, x \cos \left (x\right )^{2} - \frac {1}{3456} \, {\left (32 \, {\left (18 \, x^{2} - 1\right )} \cos \left (x\right )^{5} - 2 \, {\left (936 \, x^{2} - 97\right )} \cos \left (x\right )^{3} + 3 \, {\left (792 \, x^{2} - 299\right )} \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {299}{1152} \, x \]
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Time = 0.55 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.83 \[ \int x^2 \sin ^6(x) \, dx=\frac {5 x^{3} \sin ^{6}{\left (x \right )}}{48} + \frac {5 x^{3} \sin ^{4}{\left (x \right )} \cos ^{2}{\left (x \right )}}{16} + \frac {5 x^{3} \sin ^{2}{\left (x \right )} \cos ^{4}{\left (x \right )}}{16} + \frac {5 x^{3} \cos ^{6}{\left (x \right )}}{48} - \frac {11 x^{2} \sin ^{5}{\left (x \right )} \cos {\left (x \right )}}{16} - \frac {5 x^{2} \sin ^{3}{\left (x \right )} \cos ^{3}{\left (x \right )}}{6} - \frac {5 x^{2} \sin {\left (x \right )} \cos ^{5}{\left (x \right )}}{16} + \frac {299 x \sin ^{6}{\left (x \right )}}{1152} + \frac {35 x \sin ^{4}{\left (x \right )} \cos ^{2}{\left (x \right )}}{384} - \frac {125 x \sin ^{2}{\left (x \right )} \cos ^{4}{\left (x \right )}}{384} - \frac {245 x \cos ^{6}{\left (x \right )}}{1152} + \frac {299 \sin ^{5}{\left (x \right )} \cos {\left (x \right )}}{1152} + \frac {25 \sin ^{3}{\left (x \right )} \cos ^{3}{\left (x \right )}}{54} + \frac {245 \sin {\left (x \right )} \cos ^{5}{\left (x \right )}}{1152} \]
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Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63 \[ \int x^2 \sin ^6(x) \, dx=\frac {5}{48} \, x^{3} - \frac {1}{576} \, x \cos \left (6 \, x\right ) + \frac {3}{128} \, x \cos \left (4 \, x\right ) - \frac {15}{64} \, x \cos \left (2 \, x\right ) - \frac {1}{3456} \, {\left (18 \, x^{2} - 1\right )} \sin \left (6 \, x\right ) + \frac {3}{512} \, {\left (8 \, x^{2} - 1\right )} \sin \left (4 \, x\right ) - \frac {15}{128} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63 \[ \int x^2 \sin ^6(x) \, dx=\frac {5}{48} \, x^{3} - \frac {1}{576} \, x \cos \left (6 \, x\right ) + \frac {3}{128} \, x \cos \left (4 \, x\right ) - \frac {15}{64} \, x \cos \left (2 \, x\right ) - \frac {1}{3456} \, {\left (18 \, x^{2} - 1\right )} \sin \left (6 \, x\right ) + \frac {3}{512} \, {\left (8 \, x^{2} - 1\right )} \sin \left (4 \, x\right ) - \frac {15}{128} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]
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Time = 0.42 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.84 \[ \int x^2 \sin ^6(x) \, dx=\frac {15\,\sin \left (2\,x\right )}{128}-\frac {3\,\sin \left (4\,x\right )}{512}+\frac {\sin \left (6\,x\right )}{3456}-\frac {3\,x\,\left (2\,{\sin \left (2\,x\right )}^2-1\right )}{128}+\frac {x\,\left (2\,{\sin \left (3\,x\right )}^2-1\right )}{576}-\frac {15\,x^2\,\sin \left (2\,x\right )}{64}+\frac {3\,x^2\,\sin \left (4\,x\right )}{64}-\frac {x^2\,\sin \left (6\,x\right )}{192}+\frac {5\,x^3}{48}+\frac {15\,x\,\left (2\,{\sin \left (x\right )}^2-1\right )}{64} \]
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