\(\int e^x x \sin (x) \, dx\) [47]

Optimal result

Integrand size = 7, antiderivative size = 30 \[ \int e^x x \sin (x) \, dx=\frac {1}{2} e^x \cos (x)-\frac {1}{2} e^x x \cos (x)+\frac {1}{2} e^x x \sin (x) \]

[Out]

1/2*exp(x)*cos(x)-1/2*exp(x)*x*cos(x)+1/2*exp(x)*x*sin(x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4517, 4553, 4518} \[ \int e^x x \sin (x) \, dx=\frac {1}{2} e^x x \sin (x)+\frac {1}{2} e^x \cos (x)-\frac {1}{2} e^x x \cos (x) \]

[In]

Int[E^x*x*Sin[x],x]

[Out]

(E^x*Cos[x])/2 - (E^x*x*Cos[x])/2 + (E^x*x*Sin[x])/2

Rule 4517

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(S
in[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] - Simp[e*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rule 4518

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(C
os[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] + Simp[e*F^(c*(a + b*x))*(Sin[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rule 4553

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]}, Dist[(f*x)^m, u, x] - Dist[f*m, Int[(f*x)^(m - 1)*u, x], x]] /
; FreeQ[{F, a, b, c, d, e, f}, x] && IGtQ[n, 0] && GtQ[m, 0]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int e^x x \sin (x) \, dx=\frac {1}{2} e^x (\cos (x)-x \cos (x)+x \sin (x)) \]

[In]

Integrate[E^x*x*Sin[x],x]

[Out]

(E^x*(Cos[x] - x*Cos[x] + x*Sin[x]))/2

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57

method	result	size
parallelrisch	\(-\frac {\left (\left (x -1\right ) \cos \left (x \right )-x \sin \left (x \right )\right ) {\mathrm e}^{x}}{2}\)	\(17\)
default	\(\left (-\frac {x}{2}+\frac {1}{2}\right ) {\mathrm e}^{x} \cos \left (x \right )+\frac {{\mathrm e}^{x} x \sin \left (x \right )}{2}\)	\(19\)
risch	\(\left (-\frac {1}{8}-\frac {i}{8}\right ) \left (-1+i+2 x \right ) {\mathrm e}^{\left (1+i\right ) x}+\left (-\frac {1}{8}+\frac {i}{8}\right ) \left (-1-i+2 x \right ) {\mathrm e}^{\left (1-i\right ) x}\)	\(36\)
norman	\(\frac {{\mathrm e}^{x} x \tan \left (\frac {x}{2}\right )-\frac {{\mathrm e}^{x} x}{2}-\frac {{\mathrm e}^{x} \tan \left (\frac {x}{2}\right )^{2}}{2}+\frac {{\mathrm e}^{x} x \tan \left (\frac {x}{2}\right )^{2}}{2}+\frac {{\mathrm e}^{x}}{2}}{1+\tan \left (\frac {x}{2}\right )^{2}}\)	\(51\)

[In]

int(exp(x)*x*sin(x),x,method=_RETURNVERBOSE)

[Out]

-1/2*((x-1)*cos(x)-x*sin(x))*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int e^x x \sin (x) \, dx=-\frac {1}{2} \, {\left (x - 1\right )} \cos \left (x\right ) e^{x} + \frac {1}{2} \, x e^{x} \sin \left (x\right ) \]

[In]

integrate(exp(x)*x*sin(x),x, algorithm="fricas")

[Out]

-1/2*(x - 1)*cos(x)*e^x + 1/2*x*e^x*sin(x)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int e^x x \sin (x) \, dx=\frac {x e^{x} \sin {\left (x \right )}}{2} - \frac {x e^{x} \cos {\left (x \right )}}{2} + \frac {e^{x} \cos {\left (x \right )}}{2} \]

[In]

integrate(exp(x)*x*sin(x),x)

[Out]

x*exp(x)*sin(x)/2 - x*exp(x)*cos(x)/2 + exp(x)*cos(x)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int e^x x \sin (x) \, dx=-\frac {1}{2} \, {\left (x - 1\right )} \cos \left (x\right ) e^{x} + \frac {1}{2} \, x e^{x} \sin \left (x\right ) \]

[In]

integrate(exp(x)*x*sin(x),x, algorithm="maxima")

[Out]

-1/2*(x - 1)*cos(x)*e^x + 1/2*x*e^x*sin(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.53 \[ \int e^x x \sin (x) \, dx=-\frac {1}{2} \, {\left ({\left (x - 1\right )} \cos \left (x\right ) - x \sin \left (x\right )\right )} e^{x} \]

[In]

integrate(exp(x)*x*sin(x),x, algorithm="giac")

[Out]

-1/2*((x - 1)*cos(x) - x*sin(x))*e^x

Mupad [B] (verification not implemented)

Time = 26.39 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.53 \[ \int e^x x \sin (x) \, dx=\frac {{\mathrm {e}}^x\,\left (\cos \left (x\right )-x\,\cos \left (x\right )+x\,\sin \left (x\right )\right )}{2} \]

[In]

int(x*exp(x)*sin(x),x)

[Out]

(exp(x)*(cos(x) - x*cos(x) + x*sin(x)))/2