\(\int x^2 \cos ^2(x) \, dx\) [100]

Optimal result

Integrand size = 8, antiderivative size = 41 \[ \int x^2 \cos ^2(x) \, dx=-\frac {x}{4}+\frac {x^3}{6}+\frac {1}{2} x \cos ^2(x)-\frac {1}{4} \cos (x) \sin (x)+\frac {1}{2} x^2 \cos (x) \sin (x) \]

[Out]

-1/4*x+1/6*x^3+1/2*x*cos(x)^2-1/4*cos(x)*sin(x)+1/2*x^2*cos(x)*sin(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3392, 30, 2715, 8} \[ \int x^2 \cos ^2(x) \, dx=\frac {x^3}{6}+\frac {1}{2} x^2 \sin (x) \cos (x)-\frac {x}{4}+\frac {1}{2} x \cos ^2(x)-\frac {1}{4} \sin (x) \cos (x) \]

[In]

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

[Out]

-1/4*x + x^3/6 + (x*Cos[x]^2)/2 - (Cos[x]*Sin[x])/4 + (x^2*Cos[x]*Sin[x])/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x \cos ^2(x)+\frac {1}{2} x^2 \cos (x) \sin (x)+\frac {\int x^2 \, dx}{2}-\frac {1}{2} \int \cos ^2(x) \, dx \\ & = \frac {x^3}{6}+\frac {1}{2} x \cos ^2(x)-\frac {1}{4} \cos (x) \sin (x)+\frac {1}{2} x^2 \cos (x) \sin (x)-\frac {\int 1 \, dx}{4} \\ & = -\frac {x}{4}+\frac {x^3}{6}+\frac {1}{2} x \cos ^2(x)-\frac {1}{4} \cos (x) \sin (x)+\frac {1}{2} x^2 \cos (x) \sin (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int x^2 \cos ^2(x) \, dx=\frac {1}{24} \left (4 x^3+6 x \cos (2 x)+\left (-3+6 x^2\right ) \sin (2 x)\right ) \]

[In]

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

[Out]

(4*x^3 + 6*x*Cos[2*x] + (-3 + 6*x^2)*Sin[2*x])/24

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {x^{3}}{6}+\frac {x \cos \left (2 x \right )}{4}+\frac {\left (2 x^{2}-1\right ) \sin \left (2 x \right )}{8}\)	\(27\)
parallelrisch	\(\frac {x^{3}}{6}+\frac {x \cos \left (2 x \right )}{4}+\frac {x^{2} \sin \left (2 x \right )}{4}-\frac {\sin \left (2 x \right )}{8}\)	\(29\)
default	\(x^{2} \left (\frac {x}{2}+\frac {\cos \left (x \right ) \sin \left (x \right )}{2}\right )+\frac {\left (\cos ^{2}\left (x \right )\right ) x}{2}-\frac {\cos \left (x \right ) \sin \left (x \right )}{4}-\frac {x}{4}-\frac {x^{3}}{3}\)	\(37\)
norman	\(\frac {x^{2} \tan \left (\frac {x}{2}\right )+\frac {x}{4}+\frac {x^{3}}{6}+\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}-\frac {3 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}+\frac {x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}-x^{2} \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\frac {x^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{3}+\frac {x^{3} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{6}-\frac {\tan \left (\frac {x}{2}\right )}{2}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}\)	\(94\)

[In]

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

[Out]

1/6*x^3+1/4*x*cos(2*x)+1/8*(2*x^2-1)*sin(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int x^2 \cos ^2(x) \, dx=\frac {1}{6} \, x^{3} + \frac {1}{2} \, x \cos \left (x\right )^{2} + \frac {1}{4} \, {\left (2 \, x^{2} - 1\right )} \cos \left (x\right ) \sin \left (x\right ) - \frac {1}{4} \, x \]

[In]

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

[Out]

1/6*x^3 + 1/2*x*cos(x)^2 + 1/4*(2*x^2 - 1)*cos(x)*sin(x) - 1/4*x

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.37 \[ \int x^2 \cos ^2(x) \, dx=\frac {x^{3} \sin ^{2}{\left (x \right )}}{6} + \frac {x^{3} \cos ^{2}{\left (x \right )}}{6} + \frac {x^{2} \sin {\left (x \right )} \cos {\left (x \right )}}{2} - \frac {x \sin ^{2}{\left (x \right )}}{4} + \frac {x \cos ^{2}{\left (x \right )}}{4} - \frac {\sin {\left (x \right )} \cos {\left (x \right )}}{4} \]

[In]

integrate(x**2*cos(x)**2,x)

[Out]

x**3*sin(x)**2/6 + x**3*cos(x)**2/6 + x**2*sin(x)*cos(x)/2 - x*sin(x)**2/4 + x*cos(x)**2/4 - sin(x)*cos(x)/4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int x^2 \cos ^2(x) \, dx=\frac {1}{6} \, x^{3} + \frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{8} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]

[In]

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

[Out]

1/6*x^3 + 1/4*x*cos(2*x) + 1/8*(2*x^2 - 1)*sin(2*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int x^2 \cos ^2(x) \, dx=\frac {1}{6} \, x^{3} + \frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{8} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]

[In]

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

[Out]

1/6*x^3 + 1/4*x*cos(2*x) + 1/8*(2*x^2 - 1)*sin(2*x)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.68 \[ \int x^2 \cos ^2(x) \, dx=\frac {x\,\cos \left (2\,x\right )}{4}-\frac {\sin \left (2\,x\right )}{8}+\frac {x^2\,\sin \left (2\,x\right )}{4}+\frac {x^3}{6} \]

[In]

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

[Out]

(x*cos(2*x))/4 - sin(2*x)/8 + (x^2*sin(2*x))/4 + x^3/6