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]

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

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Rubi [A]  time = 0.0127517, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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]

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 8

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

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*}

Mathematica [A]  time = 0.0028449, 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]

-(t*Cos[2*t])/4 + Sin[2*t]/8

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Maple [A]  time = 0.001, size = 18, normalized size = 0.8 \begin{align*} -{\frac{t \left ( \cos \left ( t \right ) \right ) ^{2}}{2}}+{\frac{\cos \left ( t \right ) \sin \left ( t \right ) }{4}}+{\frac{t}{4}} \end{align*}

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 = 0.938485, size = 19, normalized size = 0.83 \begin{align*} -\frac{1}{4} \, t \cos \left (2 \, t\right ) + \frac{1}{8} \, \sin \left (2 \, t\right ) \end{align*}

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 = 1.95604, size = 61, normalized size = 2.65 \begin{align*} -\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{align*}

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 = 0.313081, size = 24, normalized size = 1.04 \begin{align*} \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{align*}

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 = 1.05105, size = 19, normalized size = 0.83 \begin{align*} -\frac{1}{4} \, t \cos \left (2 \, t\right ) + \frac{1}{8} \, \sin \left (2 \, t\right ) \end{align*}

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)