\(\int x \cos (x^2) \sin (x^2) \, dx\) [54]

Optimal result

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

[Out]

1/4*sin(x^2)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3522} \[ \int x \cos \left (x^2\right ) \sin \left (x^2\right ) \, dx=\frac {1}{4} \sin ^2\left (x^2\right ) \]

[In]

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

[Out]

Sin[x^2]^2/4

Rule 3522

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[Sin[a + b*
x^n]^(p + 1)/(b*n*(p + 1)), x] /; FreeQ[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int x \cos \left (x^2\right ) \sin \left (x^2\right ) \, dx=-\frac {1}{4} \cos ^2\left (x^2\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(-\frac {\left (\cos ^{2}\left (x^{2}\right )\right )}{4}\)	\(9\)
default	\(-\frac {\left (\cos ^{2}\left (x^{2}\right )\right )}{4}\)	\(9\)
risch	\(-\frac {\cos \left (2 x^{2}\right )}{8}\)	\(9\)
parallelrisch	\(-\frac {\cos \left (2 x^{2}\right )}{8}+\frac {1}{8}\)	\(11\)
meijerg	\(\frac {\sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (2 x^{2}\right )}{\sqrt {\pi }}\right )}{8}\)	\(21\)
norman	\(\frac {\tan ^{2}\left (\frac {x^{2}}{2}\right )}{\left (1+\tan ^{2}\left (\frac {x^{2}}{2}\right )\right )^{2}}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \cos \left (x^2\right ) \sin \left (x^2\right ) \, dx=- \frac {\cos ^{2}{\left (x^{2} \right )}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \cos \left (x^2\right ) \sin \left (x^2\right ) \, dx=-\frac {1}{4} \, \cos \left (x^{2}\right )^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int x \cos \left (x^2\right ) \sin \left (x^2\right ) \, dx=-\frac {1}{4} \, \cos \left (x^{2}\right )^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

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

[In]

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

[Out]

sin(x^2)^2/4