\(\int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx\) [9535]

Optimal result

Integrand size = 28, antiderivative size = 20 \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=2 \left (11-\frac {16 e^{-1+x}}{e^x-x}\right ) \]

[Out]

22-32/(exp(x)-x)*exp(-1+x)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6820, 12, 6843, 32} \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {32}{e \left (1-\frac {e^x}{x}\right )} \]

[In]

Int[(E^(-1 + x)*(-32 + 32*x))/(E^(2*x) - 2*E^x*x + x^2),x]

[Out]

32/(E*(1 - E^x/x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6843

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {32 e^{-1+x} (-1+x)}{\left (e^x-x\right )^2} \, dx \\ & = 32 \int \frac {e^{-1+x} (-1+x)}{\left (e^x-x\right )^2} \, dx \\ & = \frac {32 \text {Subst}\left (\int \frac {1}{(-1+x)^2} \, dx,x,\frac {e^x}{x}\right )}{e} \\ & = \frac {32}{e \left (1-\frac {e^x}{x}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=-\frac {32 x}{e \left (e^x-x\right )} \]

[In]

Integrate[(E^(-1 + x)*(-32 + 32*x))/(E^(2*x) - 2*E^x*x + x^2),x]

[Out]

(-32*x)/(E*(E^x - x))

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {32 \,{\mathrm e}^{-1} x}{x -{\mathrm e}^{x}}\)	\(14\)
parallelrisch	\(\frac {32 \,{\mathrm e}^{-1+x}}{x -{\mathrm e}^{x}}\)	\(15\)
norman	\(\frac {32 \,{\mathrm e}^{-1} {\mathrm e}^{x}}{x -{\mathrm e}^{x}}\)	\(17\)

[In]

int((32*x-32)*exp(-1+x)/(exp(x)^2-2*exp(x)*x+x^2),x,method=_RETURNVERBOSE)

[Out]

32*exp(-1)*x/(x-exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {32 \, x}{x e - e^{\left (x + 1\right )}} \]

[In]

integrate((32*x-32)*exp(-1+x)/(exp(x)^2-2*exp(x)*x+x^2),x, algorithm="fricas")

[Out]

32*x/(x*e - e^(x + 1))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=- \frac {32 x}{- e x + e e^{x}} \]

[In]

integrate((32*x-32)*exp(-1+x)/(exp(x)**2-2*exp(x)*x+x**2),x)

[Out]

-32*x/(-E*x + E*exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {32 \, x}{x e - e^{\left (x + 1\right )}} \]

[In]

integrate((32*x-32)*exp(-1+x)/(exp(x)^2-2*exp(x)*x+x^2),x, algorithm="maxima")

[Out]

32*x/(x*e - e^(x + 1))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {32 \, x}{x e - e^{\left (x + 1\right )}} \]

[In]

integrate((32*x-32)*exp(-1+x)/(exp(x)^2-2*exp(x)*x+x^2),x, algorithm="giac")

[Out]

32*x/(x*e - e^(x + 1))

Mupad [B] (verification not implemented)

Time = 17.32 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-1+x} (-32+32 x)}{e^{2 x}-2 e^x x+x^2} \, dx=-\frac {32\,x}{{\mathrm {e}}^{x+1}-x\,\mathrm {e}} \]

[In]

int((exp(x - 1)*(32*x - 32))/(exp(2*x) - 2*x*exp(x) + x^2),x)

[Out]

-(32*x)/(exp(x + 1) - x*exp(1))