3.1.28 \(\int t \cos (t) \sin (t) \, dt\) [28]

Optimal. Leaf size=23 \[ -\frac {t}{4}+\frac {1}{4} \cos (t) \sin (t)+\frac {1}{2} t \sin ^2(t) \]

[Out]

-1/4*t+1/4*cos(t)*sin(t)+1/2*t*sin(t)^2

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Rubi [A]
time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3524, 2715, 8} \begin {gather*} -\frac {t}{4}+\frac {1}{2} t \sin ^2(t)+\frac {1}{4} \sin (t) \cos (t) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[t*Cos[t]*Sin[t],t]

[Out]

-1/4*t + (Cos[t]*Sin[t])/4 + (t*Sin[t]^2)/2

Rule 8

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

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 3524

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[x^(m - n +
1)*(Sin[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sin[a + b*x^n]^
(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int t \cos (t) \sin (t) \, dt &=\frac {1}{2} t \sin ^2(t)-\frac {1}{2} \int \sin ^2(t) \, dt\\ &=\frac {1}{4} \cos (t) \sin (t)+\frac {1}{2} t \sin ^2(t)-\frac {\int 1 \, dt}{4}\\ &=-\frac {t}{4}+\frac {1}{4} \cos (t) \sin (t)+\frac {1}{2} t \sin ^2(t)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 18, normalized size = 0.78 \begin {gather*} -\frac {1}{4} t \cos (2 t)+\frac {1}{8} \sin (2 t) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[t*Cos[t]*Sin[t],t]

[Out]

-1/4*(t*Cos[2*t]) + Sin[2*t]/8

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Mathics [A]
time = 1.90, size = 14, normalized size = 0.61 \begin {gather*} -\frac {t \text {Cos}\left [2 t\right ]}{4}+\frac {\text {Sin}\left [2 t\right ]}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[t*Cos[t]*Sin[t],t]')

[Out]

-t Cos[2 t] / 4 + Sin[2 t] / 8

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Maple [A]
time = 0.02, size = 18, normalized size = 0.78

method	result	size
risch	\(-\frac {t \cos \left (2 t \right )}{4}+\frac {\sin \left (2 t \right )}{8}\)	\(15\)
default	\(-\frac {t \left (\cos ^{2}\left (t \right )\right )}{2}+\frac {\cos \left (t \right ) \sin \left (t \right )}{4}+\frac {t}{4}\)	\(18\)
meijerg	\(\frac {\sqrt {\pi }\, \left (-\frac {t \cos \left (2 t \right )}{\sqrt {\pi }}+\frac {\sin \left (2 t \right )}{2 \sqrt {\pi }}\right )}{4}\)	\(26\)
norman	\(\frac {-\frac {t}{4}-\frac {\left (\tan ^{3}\left (\frac {t}{2}\right )\right )}{2}+\frac {3 t \left (\tan ^{2}\left (\frac {t}{2}\right )\right )}{2}-\frac {t \left (\tan ^{4}\left (\frac {t}{2}\right )\right )}{4}+\frac {\tan \left (\frac {t}{2}\right )}{2}}{\left (1+\tan ^{2}\left (\frac {t}{2}\right )\right )^{2}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(t*cos(t)*sin(t),t,method=_RETURNVERBOSE)

[Out]

-1/2*t*cos(t)^2+1/4*cos(t)*sin(t)+1/4*t

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Maxima [A]
time = 0.27, size = 14, normalized size = 0.61 \begin {gather*} -\frac {1}{4} \, t \cos \left (2 \, t\right ) + \frac {1}{8} \, \sin \left (2 \, t\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t*cos(t)*sin(t),t, algorithm="maxima")

[Out]

-1/4*t*cos(2*t) + 1/8*sin(2*t)

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Fricas [A]
time = 0.34, size = 17, normalized size = 0.74 \begin {gather*} -\frac {1}{2} \, t \cos \left (t\right )^{2} + \frac {1}{4} \, \cos \left (t\right ) \sin \left (t\right ) + \frac {1}{4} \, t \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t*cos(t)*sin(t),t, algorithm="fricas")

[Out]

-1/2*t*cos(t)^2 + 1/4*cos(t)*sin(t) + 1/4*t

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Sympy [A]
time = 0.10, size = 24, normalized size = 1.04 \begin {gather*} \frac {t \sin ^{2}{\left (t \right )}}{4} - \frac {t \cos ^{2}{\left (t \right )}}{4} + \frac {\sin {\left (t \right )} \cos {\left (t \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t*cos(t)*sin(t),t)

[Out]

t*sin(t)**2/4 - t*cos(t)**2/4 + sin(t)*cos(t)/4

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Giac [A]
time = 0.00, size = 18, normalized size = 0.78 \begin {gather*} -\frac {2}{8} t \cos \left (2 t\right )+\frac {\sin \left (2 t\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t*cos(t)*sin(t),t)

[Out]

-1/4*t*cos(2*t) + 1/8*sin(2*t)

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Mupad [B]
time = 0.05, size = 18, normalized size = 0.78 \begin {gather*} \frac {\sin \left (2\,t\right )}{8}+\frac {t\,\left (2\,{\sin \left (t\right )}^2-1\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(t*cos(t)*sin(t),t)

[Out]

sin(2*t)/8 + (t*(2*sin(t)^2 - 1))/4

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