\(\int \cos ^3(1+x) \sin ^3(1+x) \, dx\) [908]

Optimal result

Integrand size = 13, antiderivative size = 21 \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=\frac {1}{4} \sin ^4(1+x)-\frac {1}{6} \sin ^6(1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2644, 14} \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=\frac {1}{4} \sin ^4(x+1)-\frac {1}{6} \sin ^6(x+1) \]

[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 2644

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*} \text {integral}& = \text {Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,\sin (1+x)\right ) \\ & = \text {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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=\frac {1}{8} \left (-\frac {3}{8} \cos (2 (1+x))+\frac {1}{24} \cos (6 (1+x))\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\sin \left (x +1\right )^{4}}{4}-\frac {\sin \left (x +1\right )^{6}}{6}\)	\(18\)
default	\(\frac {\sin \left (x +1\right )^{4}}{4}-\frac {\sin \left (x +1\right )^{6}}{6}\)	\(18\)
risch	\(\frac {\cos \left (6 x +6\right )}{192}-\frac {3 \cos \left (2 x +2\right )}{64}\)	\(18\)
parallelrisch	\(\frac {7}{40}+\frac {\cos \left (6 x +6\right )}{192}-\frac {3 \cos \left (2 x +2\right )}{64}\)	\(19\)
norman	\(\frac {4 \tan \left (\frac {x}{2}+\frac {1}{2}\right )^{4}+4 \tan \left (\frac {x}{2}+\frac {1}{2}\right )^{8}-\frac {8 \tan \left (\frac {x}{2}+\frac {1}{2}\right )^{6}}{3}}{\left (1+\tan \left (\frac {x}{2}+\frac {1}{2}\right )^{2}\right )^{6}}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=\frac {1}{6} \, \cos \left (x + 1\right )^{6} - \frac {1}{4} \, \cos \left (x + 1\right )^{4} \]

[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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=- \frac {\sin ^{2}{\left (x + 1 \right )} \cos ^{4}{\left (x + 1 \right )}}{4} - \frac {\cos ^{6}{\left (x + 1 \right )}}{12} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=-\frac {1}{6} \, \sin \left (x + 1\right )^{6} + \frac {1}{4} \, \sin \left (x + 1\right )^{4} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=-\frac {1}{6} \, \sin \left (x + 1\right )^{6} + \frac {1}{4} \, \sin \left (x + 1\right )^{4} \]

[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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \cos ^3(1+x) \sin ^3(1+x) \, dx=-\frac {{\sin \left (x+1\right )}^4\,\left (2\,{\sin \left (x+1\right )}^2-3\right )}{12} \]

[In]

int(cos(x + 1)^3*sin(x + 1)^3,x)

[Out]

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