3.28 \(\int t \cos (t) \sin (t) \, dt\)

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

[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 = {3443, 2635, 8} \[ -\frac {t}{4}+\frac {1}{2} t \sin ^2(t)+\frac {1}{4} \sin (t) \cos (t) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2635

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] && Integer
Q[2*n]

Rule 3443

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 \[ \frac {1}{8} \sin (2 t)-\frac {1}{4} t \cos (2 t) \]

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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fricas [A]  time = 0.43, size = 17, normalized size = 0.74 \[ -\frac {1}{2} \, t \cos \relax (t)^{2} + \frac {1}{4} \, \cos \relax (t) \sin \relax (t) + \frac {1}{4} \, t \]

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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giac [A]  time = 0.95, size = 14, normalized size = 0.61 \[ -\frac {1}{4} \, t \cos \left (2 \, t\right ) + \frac {1}{8} \, \sin \left (2 \, t\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*t*cos(2*t) + 1/8*sin(2*t)

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maple [A]  time = 0.00, size = 18, normalized size = 0.78 \[ -\frac {t \left (\cos ^{2}\relax (t )\right )}{2}+\frac {\cos \relax (t ) \sin \relax (t )}{4}+\frac {t}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 14, normalized size = 0.61 \[ -\frac {1}{4} \, t \cos \left (2 \, t\right ) + \frac {1}{8} \, \sin \left (2 \, t\right ) \]

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

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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sympy [A]  time = 0.33, size = 24, normalized size = 1.04 \[ \frac {t \sin ^{2}{\relax (t )}}{4} - \frac {t \cos ^{2}{\relax (t )}}{4} + \frac {\sin {\relax (t )} \cos {\relax (t )}}{4} \]

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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