\(\int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx\) [220]

Optimal result

Integrand size = 21, antiderivative size = 12 \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {1}{5} \cos ^5(x) \sin ^5(x) \]

[Out]

1/5*cos(x)^5*sin(x)^5

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {460} \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {1}{5} \sin ^5(x) \cos ^5(x) \]

[In]

Int[Cos[x]^4*(Cos[x] - Sin[x])*Sin[x]^4*(Cos[x] + Sin[x]),x]

[Out]

(Cos[x]^5*Sin[x]^5)/5

Rule 460

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(25\) vs. \(2(12)=24\).

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 2.08 \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {1}{256} \sin (2 x)-\frac {1}{512} \sin (6 x)+\frac {\sin (10 x)}{2560} \]

[In]

Integrate[Cos[x]^4*(Cos[x] - Sin[x])*Sin[x]^4*(Cos[x] + Sin[x]),x]

[Out]

Sin[2*x]/256 - Sin[6*x]/512 + Sin[10*x]/2560

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.67

method	result	size
risch	\(\frac {\sin \left (10 x \right )}{2560}-\frac {\sin \left (6 x \right )}{512}+\frac {\sin \left (2 x \right )}{256}\)	\(20\)
parallelrisch	\(\frac {\sin \left (10 x \right )}{2560}-\frac {\sin \left (6 x \right )}{512}+\frac {\sin \left (2 x \right )}{256}\)	\(20\)
default	\(-\frac {\sin \left (x \right )^{3} \cos \left (x \right )^{7}}{10}-\frac {3 \sin \left (x \right ) \cos \left (x \right )^{7}}{80}+\frac {\left (\cos \left (x \right )^{5}+\frac {5 \cos \left (x \right )^{3}}{4}+\frac {15 \cos \left (x \right )}{8}\right ) \sin \left (x \right )}{160}+\frac {\cos \left (x \right )^{5} \sin \left (x \right )^{5}}{10}+\frac {\sin \left (x \right )^{3} \cos \left (x \right )^{5}}{16}+\frac {\cos \left (x \right )^{5} \sin \left (x \right )}{32}-\frac {\left (\cos \left (x \right )^{3}+\frac {3 \cos \left (x \right )}{2}\right ) \sin \left (x \right )}{128}\)	\(80\)
parts	\(-\frac {\sin \left (x \right )^{3} \cos \left (x \right )^{7}}{10}-\frac {3 \sin \left (x \right ) \cos \left (x \right )^{7}}{80}+\frac {\left (\cos \left (x \right )^{5}+\frac {5 \cos \left (x \right )^{3}}{4}+\frac {15 \cos \left (x \right )}{8}\right ) \sin \left (x \right )}{160}+\frac {\cos \left (x \right )^{5} \sin \left (x \right )^{5}}{10}+\frac {\sin \left (x \right )^{3} \cos \left (x \right )^{5}}{16}+\frac {\cos \left (x \right )^{5} \sin \left (x \right )}{32}-\frac {\left (\cos \left (x \right )^{3}+\frac {3 \cos \left (x \right )}{2}\right ) \sin \left (x \right )}{128}\)	\(80\)

[In]

int(sin(x)^4*cos(x)^4*(cos(x)+sin(x))*(cos(x)-sin(x)),x,method=_RETURNVERBOSE)

[Out]

1/2560*sin(10*x)-1/512*sin(6*x)+1/256*sin(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.58 \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {1}{5} \, {\left (\cos \left (x\right )^{9} - 2 \, \cos \left (x\right )^{7} + \cos \left (x\right )^{5}\right )} \sin \left (x\right ) \]

[In]

integrate(sin(x)^4*cos(x)^4*(cos(x)+sin(x))*(cos(x)-sin(x)),x, algorithm="fricas")

[Out]

1/5*(cos(x)^9 - 2*cos(x)^7 + cos(x)^5)*sin(x)

Sympy [A] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {\sin ^{5}{\left (x \right )} \cos ^{5}{\left (x \right )}}{5} \]

[In]

integrate(sin(x)**4*cos(x)**4*(cos(x)+sin(x))*(cos(x)-sin(x)),x)

[Out]

sin(x)**5*cos(x)**5/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {1}{160} \, \sin \left (2 \, x\right )^{5} \]

[In]

integrate(sin(x)^4*cos(x)^4*(cos(x)+sin(x))*(cos(x)-sin(x)),x, algorithm="maxima")

[Out]

1/160*sin(2*x)^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.58 \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {1}{2560} \, \sin \left (10 \, x\right ) - \frac {1}{512} \, \sin \left (6 \, x\right ) + \frac {1}{256} \, \sin \left (2 \, x\right ) \]

[In]

integrate(sin(x)^4*cos(x)^4*(cos(x)+sin(x))*(cos(x)-sin(x)),x, algorithm="giac")

[Out]

1/2560*sin(10*x) - 1/512*sin(6*x) + 1/256*sin(2*x)

Mupad [B] (verification not implemented)

Time = 15.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \cos ^4(x) (\cos (x)-\sin (x)) \sin ^4(x) (\cos (x)+\sin (x)) \, dx=\frac {{\cos \left (x\right )}^5\,\sin \left (x\right )\,{\left ({\cos \left (x\right )}^2-1\right )}^2}{5} \]

[In]

int(cos(x)^4*sin(x)^4*(cos(x) + sin(x))*(cos(x) - sin(x)),x)

[Out]

(cos(x)^5*sin(x)*(cos(x)^2 - 1)^2)/5