Integrand size = 10, antiderivative size = 18 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=2 x+2 \cos (x) \sin (x)-4 \cos ^3(x) \sin (x) \] Output:
2*x+2*cos(x)*sin(x)-4*cos(x)^3*sin(x)
Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=4 \left (\frac {x}{2}-\frac {1}{8} \sin (4 x)\right ) \] Input:
Integrate[16*Cos[x]^2*Sin[x]^2,x]
Output:
4*(x/2 - Sin[4*x]/8)
Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {27, 3042, 3048, 3042, 3115, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int 16 \sin ^2(x) \cos ^2(x) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 16 \int \cos ^2(x) \sin ^2(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 16 \int \cos (x)^2 \sin (x)^2dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle 16 \left (\frac {1}{4} \int \cos ^2(x)dx-\frac {1}{4} \sin (x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 16 \left (\frac {1}{4} \int \sin \left (x+\frac {\pi }{2}\right )^2dx-\frac {1}{4} \sin (x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle 16 \left (\frac {1}{4} \left (\frac {\int 1dx}{2}+\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin (x) \cos ^3(x)\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle 16 \left (\frac {1}{4} \left (\frac {x}{2}+\frac {1}{2} \sin (x) \cos (x)\right )-\frac {1}{4} \sin (x) \cos ^3(x)\right )\) |
Input:
Int[16*Cos[x]^2*Sin[x]^2,x]
Output:
16*(-1/4*(Cos[x]^3*Sin[x]) + (x/2 + (Cos[x]*Sin[x])/2)/4)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Time = 0.30 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61
method | result | size |
risch | \(2 x -\frac {\sin \left (4 x \right )}{2}\) | \(11\) |
parallelrisch | \(2 x -\frac {\sin \left (4 x \right )}{2}\) | \(11\) |
default | \(2 x +2 \cos \left (x \right ) \sin \left (x \right )-4 \cos \left (x \right )^{3} \sin \left (x \right )\) | \(19\) |
orering | \(16 \cos \left (x \right )^{2} \sin \left (x \right )^{2} x +2 \sin \left (x \right )^{3} \cos \left (x \right )-2 \cos \left (x \right )^{3} \sin \left (x \right )+\frac {x \left (-192 \cos \left (x \right )^{2} \sin \left (x \right )^{2}+32 \sin \left (x \right )^{4}+32 \cos \left (x \right )^{4}\right )}{16}\) | \(55\) |
norman | \(\frac {2 x +28 \tan \left (\frac {x}{2}\right )^{3}-28 \tan \left (\frac {x}{2}\right )^{5}+4 \tan \left (\frac {x}{2}\right )^{7}+12 x \tan \left (\frac {x}{2}\right )^{4}+8 x \tan \left (\frac {x}{2}\right )^{6}+2 x \tan \left (\frac {x}{2}\right )^{8}+8 \tan \left (\frac {x}{2}\right )^{2} x -4 \tan \left (\frac {x}{2}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{4}}\) | \(82\) |
Input:
int(16*cos(x)^2*sin(x)^2,x,method=_RETURNVERBOSE)
Output:
2*x-1/2*sin(4*x)
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=-2 \, {\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + 2 \, x \] Input:
integrate(16*cos(x)^2*sin(x)^2,x, algorithm="fricas")
Output:
-2*(2*cos(x)^3 - cos(x))*sin(x) + 2*x
Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=2 x - \sin {\left (2 x \right )} \cos {\left (2 x \right )} \] Input:
integrate(16*cos(x)**2*sin(x)**2,x)
Output:
2*x - sin(2*x)*cos(2*x)
Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=2 \, x - \frac {1}{2} \, \sin \left (4 \, x\right ) \] Input:
integrate(16*cos(x)^2*sin(x)^2,x, algorithm="maxima")
Output:
2*x - 1/2*sin(4*x)
Time = 0.14 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=2 \, x - \frac {1}{2} \, \sin \left (4 \, x\right ) \] Input:
integrate(16*cos(x)^2*sin(x)^2,x, algorithm="giac")
Output:
2*x - 1/2*sin(4*x)
Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=4\,\cos \left (x\right )\,{\sin \left (x\right )}^3-2\,\cos \left (x\right )\,\sin \left (x\right )+2\,x \] Input:
int(16*cos(x)^2*sin(x)^2,x)
Output:
2*x - 2*cos(x)*sin(x) + 4*cos(x)*sin(x)^3
Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int 16 \cos ^2(x) \sin ^2(x) \, dx=4 \cos \left (x \right ) \sin \left (x \right )^{3}-2 \cos \left (x \right ) \sin \left (x \right )+2 x \] Input:
int(16*cos(x)^2*sin(x)^2,x)
Output:
2*(2*cos(x)*sin(x)**3 - cos(x)*sin(x) + x)