\(\int x (\cos ^3(x^2)-\sin ^3(x^2)) \, dx\) [764]

Optimal result

Integrand size = 17, antiderivative size = 37 \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=\frac {\cos \left (x^2\right )}{2}-\frac {1}{6} \cos ^3\left (x^2\right )+\frac {\sin \left (x^2\right )}{2}-\frac {1}{6} \sin ^3\left (x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {14, 3461, 2713, 3460} \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=-\frac {1}{6} \sin ^3\left (x^2\right )+\frac {\sin \left (x^2\right )}{2}-\frac {1}{6} \cos ^3\left (x^2\right )+\frac {\cos \left (x^2\right )}{2} \]

[In]

Int[x*(Cos[x^2]^3 - Sin[x^2]^3),x]

[Out]

Cos[x^2]/2 - Cos[x^2]^3/6 + Sin[x^2]/2 - Sin[x^2]^3/6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=\frac {3 \cos \left (x^2\right )}{8}-\frac {1}{24} \cos \left (3 x^2\right )+\frac {\sin \left (x^2\right )}{2}-\frac {1}{6} \sin ^3\left (x^2\right ) \]

[In]

Integrate[x*(Cos[x^2]^3 - Sin[x^2]^3),x]

[Out]

(3*Cos[x^2])/8 - Cos[3*x^2]/24 + Sin[x^2]/2 - Sin[x^2]^3/6

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {\left (2+\cos \left (x^{2}\right )^{2}\right ) \sin \left (x^{2}\right )}{6}+\frac {\left (2+\sin \left (x^{2}\right )^{2}\right ) \cos \left (x^{2}\right )}{6}\)	\(30\)
default	\(\frac {\left (2+\cos \left (x^{2}\right )^{2}\right ) \sin \left (x^{2}\right )}{6}+\frac {\left (2+\sin \left (x^{2}\right )^{2}\right ) \cos \left (x^{2}\right )}{6}\)	\(30\)
risch	\(\frac {3 \cos \left (x^{2}\right )}{8}+\frac {3 \sin \left (x^{2}\right )}{8}-\frac {\cos \left (3 x^{2}\right )}{24}+\frac {\sin \left (3 x^{2}\right )}{24}\)	\(30\)
parts	\(\frac {\left (2+\cos \left (x^{2}\right )^{2}\right ) \sin \left (x^{2}\right )}{6}+\frac {\left (2+\sin \left (x^{2}\right )^{2}\right ) \cos \left (x^{2}\right )}{6}\)	\(30\)
norman	\(\frac {\tan \left (\frac {x^{2}}{2}\right )^{5}+2 \tan \left (\frac {x^{2}}{2}\right )^{2}+\frac {2 \tan \left (\frac {x^{2}}{2}\right )^{3}}{3}+\frac {2}{3}+\tan \left (\frac {x^{2}}{2}\right )}{{\left (1+\tan \left (\frac {x^{2}}{2}\right )^{2}\right )}^{3}}\)	\(50\)
parallelrisch	\(\frac {2+3 \tan \left (\frac {x^{2}}{2}\right )^{5}+2 \tan \left (\frac {x^{2}}{2}\right )^{3}+6 \tan \left (\frac {x^{2}}{2}\right )^{2}+3 \tan \left (\frac {x^{2}}{2}\right )}{3 {\left (1+\tan \left (\frac {x^{2}}{2}\right )^{2}\right )}^{3}}\)	\(55\)

[In]

int(x*(cos(x^2)^3-sin(x^2)^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=-\frac {1}{6} \, \cos \left (x^{2}\right )^{3} + \frac {1}{6} \, {\left (\cos \left (x^{2}\right )^{2} + 2\right )} \sin \left (x^{2}\right ) + \frac {1}{2} \, \cos \left (x^{2}\right ) \]

[In]

integrate(x*(cos(x^2)^3-sin(x^2)^3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=\frac {\sin ^{3}{\left (x^{2} \right )}}{3} + \frac {\sin ^{2}{\left (x^{2} \right )} \cos {\left (x^{2} \right )}}{2} + \frac {\sin {\left (x^{2} \right )} \cos ^{2}{\left (x^{2} \right )}}{2} + \frac {\cos ^{3}{\left (x^{2} \right )}}{3} \]

[In]

integrate(x*(cos(x**2)**3-sin(x**2)**3),x)

[Out]

sin(x**2)**3/3 + sin(x**2)**2*cos(x**2)/2 + sin(x**2)*cos(x**2)**2/2 + cos(x**2)**3/3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=-\frac {1}{24} \, \cos \left (3 \, x^{2}\right ) + \frac {3}{8} \, \cos \left (x^{2}\right ) + \frac {1}{24} \, \sin \left (3 \, x^{2}\right ) + \frac {3}{8} \, \sin \left (x^{2}\right ) \]

[In]

integrate(x*(cos(x^2)^3-sin(x^2)^3),x, algorithm="maxima")

[Out]

-1/24*cos(3*x^2) + 3/8*cos(x^2) + 1/24*sin(3*x^2) + 3/8*sin(x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=-\frac {1}{6} \, \cos \left (x^{2}\right )^{3} - \frac {1}{6} \, \sin \left (x^{2}\right )^{3} + \frac {1}{2} \, \cos \left (x^{2}\right ) + \frac {1}{2} \, \sin \left (x^{2}\right ) \]

[In]

integrate(x*(cos(x^2)^3-sin(x^2)^3),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 26.43 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int x \left (\cos ^3\left (x^2\right )-\sin ^3\left (x^2\right )\right ) \, dx=-\frac {{\cos \left (x^2\right )}^3}{6}+\frac {\sin \left (x^2\right )\,{\cos \left (x^2\right )}^2}{6}+\frac {\cos \left (x^2\right )}{2}+\frac {\sin \left (x^2\right )}{3} \]

[In]

int(x*(cos(x^2)^3 - sin(x^2)^3),x)

[Out]

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