Integrand size = 13, antiderivative size = 87 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=\frac {1}{250} e^{2 x} \cos (4 x)+\frac {2}{25} e^{2 x} x \cos (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)-\frac {11}{500} e^{2 x} \sin (4 x)+\frac {3}{50} e^{2 x} x \sin (4 x)+\frac {1}{10} e^{2 x} x^2 \sin (4 x) \] Output:
1/250*exp(2*x)*cos(4*x)+2/25*exp(2*x)*x*cos(4*x)-1/5*exp(2*x)*x^2*cos(4*x) -11/500*exp(2*x)*sin(4*x)+3/50*exp(2*x)*x*sin(4*x)+1/10*exp(2*x)*x^2*sin(4 *x)
Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.46 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=\frac {1}{500} e^{2 x} \left (\left (2+40 x-100 x^2\right ) \cos (4 x)+\left (-11+30 x+50 x^2\right ) \sin (4 x)\right ) \] Input:
Integrate[E^(2*x)*x^2*Sin[4*x],x]
Output:
(E^(2*x)*((2 + 40*x - 100*x^2)*Cos[4*x] + (-11 + 30*x + 50*x^2)*Sin[4*x])) /500
Time = 0.34 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4968, 27, 2010, 2009}
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.
\(\displaystyle \int e^{2 x} x^2 \sin (4 x) \, dx\) |
\(\Big \downarrow \) 4968 |
\(\displaystyle -2 \int -\frac {1}{10} x \left (2 e^{2 x} \cos (4 x)-e^{2 x} \sin (4 x)\right )dx+\frac {1}{10} e^{2 x} x^2 \sin (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int x \left (2 e^{2 x} \cos (4 x)-e^{2 x} \sin (4 x)\right )dx+\frac {1}{10} e^{2 x} x^2 \sin (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{5} \int \left (2 e^{2 x} x \cos (4 x)-e^{2 x} x \sin (4 x)\right )dx+\frac {1}{10} e^{2 x} x^2 \sin (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} e^{2 x} x^2 \sin (4 x)-\frac {1}{5} e^{2 x} x^2 \cos (4 x)+\frac {1}{5} \left (-\frac {11}{100} e^{2 x} \sin (4 x)+\frac {3}{10} e^{2 x} x \sin (4 x)+\frac {1}{50} e^{2 x} \cos (4 x)+\frac {2}{5} e^{2 x} x \cos (4 x)\right )\) |
Input:
Int[E^(2*x)*x^2*Sin[4*x],x]
Output:
-1/5*(E^(2*x)*x^2*Cos[4*x]) + (E^(2*x)*x^2*Sin[4*x])/10 + ((E^(2*x)*Cos[4* x])/50 + (2*E^(2*x)*x*Cos[4*x])/5 - (11*E^(2*x)*Sin[4*x])/100 + (3*E^(2*x) *x*Sin[4*x])/10)/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_.)*Sin[(d_.) + (e_.)* (x_)]^(n_.), x_Symbol] :> Module[{u = IntHide[F^(c*(a + b*x))*Sin[d + e*x]^ n, x]}, Simp[(f*x)^m u, x] - Simp[f*m Int[(f*x)^(m - 1)*u, x], x]] /; F reeQ[{F, a, b, c, d, e, f}, x] && IGtQ[n, 0] && GtQ[m, 0]
Time = 0.45 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.46
method | result | size |
default | \(\left (-\frac {1}{5} x^{2}+\frac {2}{25} x +\frac {1}{250}\right ) {\mathrm e}^{2 x} \cos \left (4 x \right )+\left (\frac {1}{10} x^{2}+\frac {3}{50} x -\frac {11}{500}\right ) {\mathrm e}^{2 x} \sin \left (4 x \right )\) | \(40\) |
risch | \(\left (-\frac {1}{500}-\frac {i}{1000}\right ) \left (50 x^{2}+20 i x -10 x -4 i-3\right ) {\mathrm e}^{\left (2+4 i\right ) x}+\left (-\frac {1}{500}+\frac {i}{1000}\right ) \left (50 x^{2}-20 i x -10 x +4 i-3\right ) {\mathrm e}^{\left (2-4 i\right ) x}\) | \(54\) |
parallelrisch | \(\frac {{\mathrm e}^{2 x} \left (\left (50 x^{2}-20 x -1\right ) \tan \left (2 x \right )^{2}+\left (50 x^{2}+30 x -11\right ) \tan \left (2 x \right )-50 x^{2}+20 x +1\right )}{250 \tan \left (2 x \right )^{2}+250}\) | \(59\) |
orering | \(\frac {\left (100 x^{3}+60 x^{2}-32 x -1\right ) {\mathrm e}^{2 x} \sin \left (4 x \right )}{500 x}-\frac {\left (50 x^{2}-20 x -1\right ) \left (2 \,{\mathrm e}^{2 x} x^{2} \sin \left (4 x \right )+2 \,{\mathrm e}^{2 x} x \sin \left (4 x \right )+4 \,{\mathrm e}^{2 x} x^{2} \cos \left (4 x \right )\right )}{1000 x^{2}}\) | \(83\) |
norman | \(\frac {\frac {2 \,{\mathrm e}^{2 x} x}{25}-\frac {{\mathrm e}^{2 x} x^{2}}{5}-\frac {11 \,{\mathrm e}^{2 x} \tan \left (2 x \right )}{250}-\frac {{\mathrm e}^{2 x} \tan \left (2 x \right )^{2}}{250}+\frac {3 \,{\mathrm e}^{2 x} x \tan \left (2 x \right )}{25}-\frac {2 \,{\mathrm e}^{2 x} x \tan \left (2 x \right )^{2}}{25}+\frac {{\mathrm e}^{2 x} x^{2} \tan \left (2 x \right )}{5}+\frac {{\mathrm e}^{2 x} x^{2} \tan \left (2 x \right )^{2}}{5}+\frac {{\mathrm e}^{2 x}}{250}}{1+\tan \left (2 x \right )^{2}}\) | \(109\) |
Input:
int(exp(2*x)*x^2*sin(4*x),x,method=_RETURNVERBOSE)
Output:
(-1/5*x^2+2/25*x+1/250)*exp(2*x)*cos(4*x)+(1/10*x^2+3/50*x-11/500)*exp(2*x )*sin(4*x)
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.47 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=-\frac {1}{250} \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac {1}{500} \, {\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \] Input:
integrate(exp(2*x)*x^2*sin(4*x),x, algorithm="fricas")
Output:
-1/250*(50*x^2 - 20*x - 1)*cos(4*x)*e^(2*x) + 1/500*(50*x^2 + 30*x - 11)*e ^(2*x)*sin(4*x)
Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=\frac {x^{2} e^{2 x} \sin {\left (4 x \right )}}{10} - \frac {x^{2} e^{2 x} \cos {\left (4 x \right )}}{5} + \frac {3 x e^{2 x} \sin {\left (4 x \right )}}{50} + \frac {2 x e^{2 x} \cos {\left (4 x \right )}}{25} - \frac {11 e^{2 x} \sin {\left (4 x \right )}}{500} + \frac {e^{2 x} \cos {\left (4 x \right )}}{250} \] Input:
integrate(exp(2*x)*x**2*sin(4*x),x)
Output:
x**2*exp(2*x)*sin(4*x)/10 - x**2*exp(2*x)*cos(4*x)/5 + 3*x*exp(2*x)*sin(4* x)/50 + 2*x*exp(2*x)*cos(4*x)/25 - 11*exp(2*x)*sin(4*x)/500 + exp(2*x)*cos (4*x)/250
Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.47 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=-\frac {1}{250} \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac {1}{500} \, {\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \] Input:
integrate(exp(2*x)*x^2*sin(4*x),x, algorithm="maxima")
Output:
-1/250*(50*x^2 - 20*x - 1)*cos(4*x)*e^(2*x) + 1/500*(50*x^2 + 30*x - 11)*e ^(2*x)*sin(4*x)
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.45 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=-\frac {1}{500} \, {\left (2 \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) - {\left (50 \, x^{2} + 30 \, x - 11\right )} \sin \left (4 \, x\right )\right )} e^{\left (2 \, x\right )} \] Input:
integrate(exp(2*x)*x^2*sin(4*x),x, algorithm="giac")
Output:
-1/500*(2*(50*x^2 - 20*x - 1)*cos(4*x) - (50*x^2 + 30*x - 11)*sin(4*x))*e^ (2*x)
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.59 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=\frac {{\mathrm {e}}^{2\,x}\,\left (2\,\cos \left (4\,x\right )-11\,\sin \left (4\,x\right )+40\,x\,\cos \left (4\,x\right )+30\,x\,\sin \left (4\,x\right )-100\,x^2\,\cos \left (4\,x\right )+50\,x^2\,\sin \left (4\,x\right )\right )}{500} \] Input:
int(x^2*sin(4*x)*exp(2*x),x)
Output:
(exp(2*x)*(2*cos(4*x) - 11*sin(4*x) + 40*x*cos(4*x) + 30*x*sin(4*x) - 100* x^2*cos(4*x) + 50*x^2*sin(4*x)))/500
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int e^{2 x} x^2 \sin (4 x) \, dx=\frac {e^{2 x} \left (-100 \cos \left (4 x \right ) x^{2}+40 \cos \left (4 x \right ) x +2 \cos \left (4 x \right )+50 \sin \left (4 x \right ) x^{2}+30 \sin \left (4 x \right ) x -11 \sin \left (4 x \right )\right )}{500} \] Input:
int(exp(2*x)*x^2*sin(4*x),x)
Output:
(e**(2*x)*( - 100*cos(4*x)*x**2 + 40*cos(4*x)*x + 2*cos(4*x) + 50*sin(4*x) *x**2 + 30*sin(4*x)*x - 11*sin(4*x)))/500