\(\int x \sin ^2(x) \, dx\) [33]

Optimal result

Integrand size = 6, antiderivative size = 25 \[ \int x \sin ^2(x) \, dx=\frac {x^2}{4}-\frac {1}{2} x \cos (x) \sin (x)+\frac {\sin ^2(x)}{4} \]

[Out]

1/4*x^2-1/2*x*cos(x)*sin(x)+1/4*sin(x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3391, 30} \[ \int x \sin ^2(x) \, dx=\frac {x^2}{4}+\frac {\sin ^2(x)}{4}-\frac {1}{2} x \sin (x) \cos (x) \]

[In]

Int[x*Sin[x]^2,x]

[Out]

x^2/4 - (x*Cos[x]*Sin[x])/2 + Sin[x]^2/4

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3391

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] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2} x \cos (x) \sin (x)+\frac {\sin ^2(x)}{4}+\frac {\int x \, dx}{2} \\ & = \frac {x^2}{4}-\frac {1}{2} x \cos (x) \sin (x)+\frac {\sin ^2(x)}{4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int x \sin ^2(x) \, dx=\frac {x^2}{4}-\frac {1}{8} \cos (2 x)-\frac {1}{4} x \sin (2 x) \]

[In]

Integrate[x*Sin[x]^2,x]

[Out]

x^2/4 - Cos[2*x]/8 - (x*Sin[2*x])/4

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80

method	result	size
risch	\(\frac {x^{2}}{4}-\frac {\cos \left (2 x \right )}{8}-\frac {x \sin \left (2 x \right )}{4}\)	\(20\)
default	\(x \left (\frac {x}{2}-\frac {\cos \left (x \right ) \sin \left (x \right )}{2}\right )-\frac {x^{2}}{4}+\frac {\left (\sin ^{2}\left (x \right )\right )}{4}\)	\(25\)
meijerg	\(\frac {\sqrt {\pi }\, \left (\frac {2 x^{2}+1}{2 \sqrt {\pi }}-\frac {\cos \left (2 x \right )}{2 \sqrt {\pi }}-\frac {x \sin \left (2 x \right )}{\sqrt {\pi }}\right )}{4}\)	\(38\)
norman	\(\frac {\tan ^{2}\left (\frac {x}{2}\right )+\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) x +\frac {x^{2}}{4}-x \tan \left (\frac {x}{2}\right )+\frac {x^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}+\frac {x^{2} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}\)	\(61\)

[In]

int(x*sin(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x^2-1/8*cos(2*x)-1/4*x*sin(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \sin ^2(x) \, dx=-\frac {1}{2} \, x \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{4} \, x^{2} - \frac {1}{4} \, \cos \left (x\right )^{2} \]

[In]

integrate(x*sin(x)^2,x, algorithm="fricas")

[Out]

-1/2*x*cos(x)*sin(x) + 1/4*x^2 - 1/4*cos(x)^2

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int x \sin ^2(x) \, dx=\frac {x^{2} \sin ^{2}{\left (x \right )}}{4} + \frac {x^{2} \cos ^{2}{\left (x \right )}}{4} - \frac {x \sin {\left (x \right )} \cos {\left (x \right )}}{2} - \frac {\cos ^{2}{\left (x \right )}}{4} \]

[In]

integrate(x*sin(x)**2,x)

[Out]

x**2*sin(x)**2/4 + x**2*cos(x)**2/4 - x*sin(x)*cos(x)/2 - cos(x)**2/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \sin ^2(x) \, dx=\frac {1}{4} \, x^{2} - \frac {1}{4} \, x \sin \left (2 \, x\right ) - \frac {1}{8} \, \cos \left (2 \, x\right ) \]

[In]

integrate(x*sin(x)^2,x, algorithm="maxima")

[Out]

1/4*x^2 - 1/4*x*sin(2*x) - 1/8*cos(2*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \sin ^2(x) \, dx=\frac {1}{4} \, x^{2} - \frac {1}{4} \, x \sin \left (2 \, x\right ) - \frac {1}{8} \, \cos \left (2 \, x\right ) \]

[In]

integrate(x*sin(x)^2,x, algorithm="giac")

[Out]

1/4*x^2 - 1/4*x*sin(2*x) - 1/8*cos(2*x)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x \sin ^2(x) \, dx=\frac {{\sin \left (x\right )}^2}{4}-\frac {x\,\sin \left (2\,x\right )}{4}+\frac {x^2}{4} \]

[In]

int(x*sin(x)^2,x)

[Out]

sin(x)^2/4 - (x*sin(2*x))/4 + x^2/4