\(\int x \cos (x^2) \, dx\) [766]

Optimal result

Integrand size = 6, antiderivative size = 8 \[ \int x \cos \left (x^2\right ) \, dx=\frac {\sin \left (x^2\right )}{2} \]

[Out]

1/2*sin(x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, 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 = {3461, 2717} \[ \int x \cos \left (x^2\right ) \, dx=\frac {\sin \left (x^2\right )}{2} \]

[In]

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

[Out]

Sin[x^2]/2

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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}& = \frac {1}{2} \text {Subst}\left (\int \cos (x) \, dx,x,x^2\right ) \\ & = \frac {\sin \left (x^2\right )}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int x \cos \left (x^2\right ) \, dx=\frac {\sin \left (x^2\right )}{2} \]

[In]

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

[Out]

Sin[x^2]/2

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\sin \left (x^{2}\right )}{2}\)	\(7\)
default	\(\frac {\sin \left (x^{2}\right )}{2}\)	\(7\)
meijerg	\(\frac {\sin \left (x^{2}\right )}{2}\)	\(7\)
risch	\(\frac {\sin \left (x^{2}\right )}{2}\)	\(7\)
parallelrisch	\(\frac {\sin \left (x^{2}\right )}{2}\)	\(7\)
norman	\(\frac {\tan \left (\frac {x^{2}}{2}\right )}{1+\tan \left (\frac {x^{2}}{2}\right )^{2}}\)	\(20\)
parts	\(\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}\right ) x}{2}-\frac {\pi \left (\frac {\operatorname {FresnelC}\left (\frac {\sqrt {2}\, x}{\sqrt {\pi }}\right ) \sqrt {2}\, x}{\sqrt {\pi }}-\frac {\sin \left (x^{2}\right )}{\pi }\right )}{2}\)	\(50\)

[In]

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

[Out]

1/2*sin(x^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int x \cos \left (x^2\right ) \, dx=\frac {1}{2} \, \sin \left (x^{2}\right ) \]

[In]

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

[Out]

1/2*sin(x^2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int x \cos \left (x^2\right ) \, dx=\frac {\sin {\left (x^{2} \right )}}{2} \]

[In]

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

[Out]

sin(x**2)/2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int x \cos \left (x^2\right ) \, dx=\frac {1}{2} \, \sin \left (x^{2}\right ) \]

[In]

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

[Out]

1/2*sin(x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int x \cos \left (x^2\right ) \, dx=\frac {1}{2} \, \sin \left (x^{2}\right ) \]

[In]

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

[Out]

1/2*sin(x^2)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int x \cos \left (x^2\right ) \, dx=\frac {\sin \left (x^2\right )}{2} \]

[In]

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

[Out]

sin(x^2)/2