\(\int x \sin (2 x^2) \, dx\) [820]

Optimal result

Integrand size = 8, antiderivative size = 10 \[ \int x \sin \left (2 x^2\right ) \, dx=-\frac {1}{4} \cos \left (2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3460, 2718} \[ \int x \sin \left (2 x^2\right ) \, dx=-\frac {1}{4} \cos \left (2 x^2\right ) \]

[In]

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

[Out]

-1/4*Cos[2*x^2]

Rule 2718

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

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]))

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int x \sin \left (2 x^2\right ) \, dx=-\frac {1}{4} \cos \left (2 x^2\right ) \]

[In]

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

[Out]

-1/4*Cos[2*x^2]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(-\frac {\cos \left (2 x^{2}\right )}{4}\)	\(9\)
default	\(-\frac {\cos \left (2 x^{2}\right )}{4}\)	\(9\)
risch	\(-\frac {\cos \left (2 x^{2}\right )}{4}\)	\(9\)
parallelrisch	\(-\frac {\cos \left (2 x^{2}\right )}{4}-\frac {1}{4}\)	\(11\)
norman	\(-\frac {1}{2 \left (1+\tan \left (x^{2}\right )^{2}\right )}\)	\(13\)
meijerg	\(\frac {\sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (2 x^{2}\right )}{\sqrt {\pi }}\right )}{4}\)	\(21\)
parts	\(\frac {\sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 x}{\sqrt {\pi }}\right ) x}{2}-\frac {\pi \left (\frac {2 \,\operatorname {FresnelS}\left (\frac {2 x}{\sqrt {\pi }}\right ) x}{\sqrt {\pi }}+\frac {\cos \left (2 x^{2}\right )}{\pi }\right )}{4}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \sin \left (2 x^2\right ) \, dx=-\frac {1}{4} \, \cos \left (2 \, x^{2}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \sin \left (2 x^2\right ) \, dx=- \frac {\cos {\left (2 x^{2} \right )}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \sin \left (2 x^2\right ) \, dx=-\frac {1}{4} \, \cos \left (2 \, x^{2}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \sin \left (2 x^2\right ) \, dx=-\frac {1}{4} \, \cos \left (2 \, x^{2}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \sin \left (2 x^2\right ) \, dx=\frac {{\sin \left (x^2\right )}^2}{2} \]

[In]

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

[Out]

sin(x^2)^2/2