3.908 \(\int \cos ^3(1+x) \sin ^3(1+x) \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{4} \sin ^4(x+1)-\frac {1}{6} \sin ^6(x+1) \]

[Out]

1/4*sin(1+x)^4-1/6*sin(1+x)^6

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2564, 14} \[ \frac {1}{4} \sin ^4(x+1)-\frac {1}{6} \sin ^6(x+1) \]

Antiderivative was successfully verified.

[In]

Int[Cos[1 + x]^3*Sin[1 + x]^3,x]

[Out]

Sin[1 + x]^4/4 - Sin[1 + x]^6/6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \cos ^3(1+x) \sin ^3(1+x) \, dx &=\operatorname {Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,\sin (1+x)\right )\\ &=\operatorname {Subst}\left (\int \left (x^3-x^5\right ) \, dx,x,\sin (1+x)\right )\\ &=\frac {1}{4} \sin ^4(1+x)-\frac {1}{6} \sin ^6(1+x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.19 \[ \frac {1}{8} \left (\frac {1}{24} \cos (6 (x+1))-\frac {3}{8} \cos (2 (x+1))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[1 + x]^3*Sin[1 + x]^3,x]

[Out]

((-3*Cos[2*(1 + x)])/8 + Cos[6*(1 + x)]/24)/8

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fricas [A]  time = 0.82, size = 17, normalized size = 0.81 \[ \frac {1}{6} \, \cos \left (x + 1\right )^{6} - \frac {1}{4} \, \cos \left (x + 1\right )^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1+x)^3*sin(1+x)^3,x, algorithm="fricas")

[Out]

1/6*cos(x + 1)^6 - 1/4*cos(x + 1)^4

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giac [A]  time = 0.14, size = 17, normalized size = 0.81 \[ -\frac {1}{6} \, \sin \left (x + 1\right )^{6} + \frac {1}{4} \, \sin \left (x + 1\right )^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1+x)^3*sin(1+x)^3,x, algorithm="giac")

[Out]

-1/6*sin(x + 1)^6 + 1/4*sin(x + 1)^4

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maple [A]  time = 0.08, size = 24, normalized size = 1.14 \[ -\frac {\left (\cos ^{4}\left (1+x \right )\right ) \left (\sin ^{2}\left (1+x \right )\right )}{6}-\frac {\left (\cos ^{4}\left (1+x \right )\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(1+x)^3*sin(1+x)^3,x)

[Out]

-1/6*cos(1+x)^4*sin(1+x)^2-1/12*cos(1+x)^4

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maxima [A]  time = 0.56, size = 17, normalized size = 0.81 \[ -\frac {1}{6} \, \sin \left (x + 1\right )^{6} + \frac {1}{4} \, \sin \left (x + 1\right )^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1+x)^3*sin(1+x)^3,x, algorithm="maxima")

[Out]

-1/6*sin(x + 1)^6 + 1/4*sin(x + 1)^4

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mupad [B]  time = 0.07, size = 18, normalized size = 0.86 \[ -\frac {{\sin \left (x+1\right )}^4\,\left (2\,{\sin \left (x+1\right )}^2-3\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x + 1)^3*sin(x + 1)^3,x)

[Out]

-(sin(x + 1)^4*(2*sin(x + 1)^2 - 3))/12

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sympy [A]  time = 1.78, size = 24, normalized size = 1.14 \[ - \frac {\sin ^{2}{\left (x + 1 \right )} \cos ^{4}{\left (x + 1 \right )}}{4} - \frac {\cos ^{6}{\left (x + 1 \right )}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1+x)**3*sin(1+x)**3,x)

[Out]

-sin(x + 1)**2*cos(x + 1)**4/4 - cos(x + 1)**6/12

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