Integrand size = 10, antiderivative size = 44 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=\frac {4}{9} x \cos (x)-\frac {4 \sin (x)}{9}+\frac {2}{9} x \cos (x) \sin ^2(x)-\frac {2 \sin ^3(x)}{27}+\frac {1}{3} x^2 \sin ^3(x) \] Output:
4/9*x*cos(x)-4/9*sin(x)+2/9*x*cos(x)*sin(x)^2-2/27*sin(x)^3+1/3*x^2*sin(x) ^3
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=\frac {1}{54} \left (27 x \cos (x)-3 x \cos (3 x)+\left (-26+9 x^2+\left (2-9 x^2\right ) \cos (2 x)\right ) \sin (x)\right ) \] Input:
Integrate[x^2*Cos[x]*Sin[x]^2,x]
Output:
(27*x*Cos[x] - 3*x*Cos[3*x] + (-26 + 9*x^2 + (2 - 9*x^2)*Cos[2*x])*Sin[x]) /54
Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3924, 3042, 3791, 3042, 3777, 3042, 3117}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \sin ^2(x) \cos (x) \, dx\) |
\(\Big \downarrow \) 3924 |
\(\displaystyle \frac {1}{3} x^2 \sin ^3(x)-\frac {2}{3} \int x \sin ^3(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^2 \sin ^3(x)-\frac {2}{3} \int x \sin (x)^3dx\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle \frac {1}{3} x^2 \sin ^3(x)-\frac {2}{3} \left (\frac {2}{3} \int x \sin (x)dx+\frac {\sin ^3(x)}{9}-\frac {1}{3} x \sin ^2(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^2 \sin ^3(x)-\frac {2}{3} \left (\frac {2}{3} \int x \sin (x)dx+\frac {\sin ^3(x)}{9}-\frac {1}{3} x \sin ^2(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{3} x^2 \sin ^3(x)-\frac {2}{3} \left (\frac {2}{3} (\int \cos (x)dx-x \cos (x))+\frac {\sin ^3(x)}{9}-\frac {1}{3} x \sin ^2(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} x^2 \sin ^3(x)-\frac {2}{3} \left (\frac {2}{3} \left (\int \sin \left (x+\frac {\pi }{2}\right )dx-x \cos (x)\right )+\frac {\sin ^3(x)}{9}-\frac {1}{3} x \sin ^2(x) \cos (x)\right )\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{3} x^2 \sin ^3(x)-\frac {2}{3} \left (\frac {\sin ^3(x)}{9}-\frac {1}{3} x \sin ^2(x) \cos (x)+\frac {2}{3} (\sin (x)-x \cos (x))\right )\) |
Input:
Int[x^2*Cos[x]*Sin[x]^2,x]
Output:
(x^2*Sin[x]^3)/3 - (2*(-1/3*(x*Cos[x]*Sin[x]^2) + Sin[x]^3/9 + (2*(-(x*Cos [x]) + Sin[x]))/3))/3
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^ (p_.), x_Symbol] :> Simp[x^(m - n + 1)*(Sin[a + b*x^n]^(p + 1)/(b*n*(p + 1) )), x] - Simp[(m - n + 1)/(b*n*(p + 1)) Int[x^(m - n)*Sin[a + b*x^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]
Time = 0.66 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {x^{2} \sin \left (x \right )^{3}}{3}+\frac {2 x \left (2+\sin \left (x \right )^{2}\right ) \cos \left (x \right )}{9}-\frac {2 \sin \left (x \right )^{3}}{27}-\frac {4 \sin \left (x \right )}{9}\) | \(32\) |
risch | \(\frac {x \cos \left (x \right )}{2}+\frac {\left (x^{2}-2\right ) \sin \left (x \right )}{4}-\frac {x \cos \left (3 x \right )}{18}-\frac {\left (9 x^{2}-2\right ) \sin \left (3 x \right )}{108}\) | \(36\) |
parallelrisch | \(\frac {x \cos \left (x \right )}{2}+\frac {x^{2} \sin \left (x \right )}{4}-\frac {\sin \left (x \right )}{2}-\frac {x \cos \left (3 x \right )}{18}-\frac {x^{2} \sin \left (3 x \right )}{12}+\frac {\sin \left (3 x \right )}{54}\) | \(40\) |
norman | \(\frac {\frac {4 x}{9}-\frac {64 \tan \left (\frac {x}{2}\right )^{3}}{27}-\frac {8 \tan \left (\frac {x}{2}\right )^{5}}{9}+\frac {4 x \tan \left (\frac {x}{2}\right )^{2}}{3}-\frac {4 x \tan \left (\frac {x}{2}\right )^{4}}{3}-\frac {4 x \tan \left (\frac {x}{2}\right )^{6}}{9}+\frac {8 x^{2} \tan \left (\frac {x}{2}\right )^{3}}{3}-\frac {8 \tan \left (\frac {x}{2}\right )}{9}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3}}\) | \(76\) |
orering | \(\frac {40 \left (9 x^{4}-x^{2}-12\right ) \cos \left (x \right ) \sin \left (x \right )^{2}}{81 x^{3}}-\frac {2 \left (45 x^{4}+26 x^{2}-180\right ) \left (2 x \cos \left (x \right ) \sin \left (x \right )^{2}-x^{2} \sin \left (x \right )^{3}+2 x^{2} \cos \left (x \right )^{2} \sin \left (x \right )\right )}{81 x^{4}}+\frac {8 \left (3 x^{2}-5\right ) \left (2 \sin \left (x \right )^{2} \cos \left (x \right )-4 x \sin \left (x \right )^{3}+8 x \cos \left (x \right )^{2} \sin \left (x \right )-7 x^{2} \cos \left (x \right ) \sin \left (x \right )^{2}+2 x^{2} \cos \left (x \right )^{3}\right )}{27 x^{3}}-\frac {\left (9 x^{2}-20\right ) \left (12 \sin \left (x \right ) \cos \left (x \right )^{2}-6 \sin \left (x \right )^{3}-42 x \cos \left (x \right ) \sin \left (x \right )^{2}+12 x \cos \left (x \right )^{3}+7 x^{2} \sin \left (x \right )^{3}-20 x^{2} \cos \left (x \right )^{2} \sin \left (x \right )\right )}{81 x^{2}}\) | \(192\) |
Input:
int(x^2*cos(x)*sin(x)^2,x,method=_RETURNVERBOSE)
Output:
1/3*x^2*sin(x)^3+2/9*x*(2+sin(x)^2)*cos(x)-2/27*sin(x)^3-4/9*sin(x)
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=-\frac {2}{9} \, x \cos \left (x\right )^{3} + \frac {2}{3} \, x \cos \left (x\right ) - \frac {1}{27} \, {\left ({\left (9 \, x^{2} - 2\right )} \cos \left (x\right )^{2} - 9 \, x^{2} + 14\right )} \sin \left (x\right ) \] Input:
integrate(x^2*cos(x)*sin(x)^2,x, algorithm="fricas")
Output:
-2/9*x*cos(x)^3 + 2/3*x*cos(x) - 1/27*((9*x^2 - 2)*cos(x)^2 - 9*x^2 + 14)* sin(x)
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=\frac {x^{2} \sin ^{3}{\left (x \right )}}{3} + \frac {2 x \sin ^{2}{\left (x \right )} \cos {\left (x \right )}}{3} + \frac {4 x \cos ^{3}{\left (x \right )}}{9} - \frac {14 \sin ^{3}{\left (x \right )}}{27} - \frac {4 \sin {\left (x \right )} \cos ^{2}{\left (x \right )}}{9} \] Input:
integrate(x**2*cos(x)*sin(x)**2,x)
Output:
x**2*sin(x)**3/3 + 2*x*sin(x)**2*cos(x)/3 + 4*x*cos(x)**3/9 - 14*sin(x)**3 /27 - 4*sin(x)*cos(x)**2/9
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=-\frac {1}{18} \, x \cos \left (3 \, x\right ) + \frac {1}{2} \, x \cos \left (x\right ) - \frac {1}{108} \, {\left (9 \, x^{2} - 2\right )} \sin \left (3 \, x\right ) + \frac {1}{4} \, {\left (x^{2} - 2\right )} \sin \left (x\right ) \] Input:
integrate(x^2*cos(x)*sin(x)^2,x, algorithm="maxima")
Output:
-1/18*x*cos(3*x) + 1/2*x*cos(x) - 1/108*(9*x^2 - 2)*sin(3*x) + 1/4*(x^2 - 2)*sin(x)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=-\frac {1}{18} \, x \cos \left (3 \, x\right ) + \frac {1}{2} \, x \cos \left (x\right ) - \frac {1}{108} \, {\left (9 \, x^{2} - 2\right )} \sin \left (3 \, x\right ) + \frac {1}{4} \, {\left (x^{2} - 2\right )} \sin \left (x\right ) \] Input:
integrate(x^2*cos(x)*sin(x)^2,x, algorithm="giac")
Output:
-1/18*x*cos(3*x) + 1/2*x*cos(x) - 1/108*(9*x^2 - 2)*sin(3*x) + 1/4*(x^2 - 2)*sin(x)
Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=\frac {x^2\,{\sin \left (x\right )}^3}{3}+\frac {4\,x\,{\cos \left (x\right )}^3}{9}+\frac {2\,x\,\cos \left (x\right )\,{\sin \left (x\right )}^2}{3}-\frac {4\,{\cos \left (x\right )}^2\,\sin \left (x\right )}{9}-\frac {14\,{\sin \left (x\right )}^3}{27} \] Input:
int(x^2*cos(x)*sin(x)^2,x)
Output:
(4*x*cos(x)^3)/9 - (14*sin(x)^3)/27 + (x^2*sin(x)^3)/3 - (4*cos(x)^2*sin(x ))/9 + (2*x*cos(x)*sin(x)^2)/3
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int x^2 \cos (x) \sin ^2(x) \, dx=\frac {2 \cos \left (x \right ) \sin \left (x \right )^{2} x}{9}+\frac {4 \cos \left (x \right ) x}{9}+\frac {\sin \left (x \right )^{3} x^{2}}{3}-\frac {2 \sin \left (x \right )^{3}}{27}-\frac {4 \sin \left (x \right )}{9} \] Input:
int(x^2*cos(x)*sin(x)^2,x)
Output:
(6*cos(x)*sin(x)**2*x + 12*cos(x)*x + 9*sin(x)**3*x**2 - 2*sin(x)**3 - 12* sin(x))/27