3.89.43 $\int e^{-x}(4-e^x-x-4x^5+(2x-x^2)\log (\frac{e^{\frac{-4-2x+e^{3}x-x^5}{x}}}{x}))dx$

Optimal. Leaf size=35 $$-x+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)$$

________________________________________________________________________________________

Rubi [A]  time = 0.52, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 5, integrand size = 57, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.088, Rules used = {6742, 2194, 2176, 2196, 2554} $$\begin{array}{c}{e}^{-x}{x}^2\log \left(\frac{e^{-x^4-\frac{4}{x}+e^3-2}}{x}\right)-x\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(4 - E^x - x - 4*x^5 + (2*x - x^2)*Log[E^((-4 - 2*x + E^3*x - x^5)/x)/x])/E^x,x]

[Out]

-x + (x^2*Log[E^(-2 + E^3 - 4/x - x^4)/x])/E^x

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-1+4e^{-x}-e^{-x}x-4e^{-x}{x}^5+e^{-x}(2-x)x\log \left(\frac{e^{-2(1-\frac{e^3}{2})-\frac{4}{x}-x^4}}{x}\right))dx\\ & =-x+4\int e^{-x}dx-4\int e^{-x}{x}^{5}dx-\int e^{-x}xdx+\int e^{-x}(2-x)x\log \left(\frac{e^{-2(1-\frac{e^3}{2})-\frac{4}{x}-x^4}}{x}\right)dx\\ & =-4e^{-x}-x+e^{-x}x+4e^{-x}{x}^5+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)-20\int e^{-x}{x}^{4}dx-\int e^{-x}dx-\int e^{-x}(4-x-4x^5)dx\\ & =-3e^{-x}-x+e^{-x}x+20e^{-x}{x}^4+4e^{-x}{x}^5+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)-80\int e^{-x}{x}^{3}dx-\int (4e^{-x}-e^{-x}x-4e^{-x}{x}^5)dx\\ & =-3e^{-x}-x+e^{-x}x+80e^{-x}{x}^3+20e^{-x}{x}^4+4e^{-x}{x}^5+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)-4\int e^{-x}dx+4\int e^{-x}{x}^{5}dx-240\int e^{-x}{x}^{2}dx+\int e^{-x}xdx\\ & =e^{-x}-x+240e^{-x}{x}^2+80e^{-x}{x}^3+20e^{-x}{x}^4+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)+20\int e^{-x}{x}^{4}dx-480\int e^{-x}xdx+\int e^{-x}dx\\ & =-x+480e^{-x}x+240e^{-x}{x}^2+80e^{-x}{x}^3+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)+80\int e^{-x}{x}^{3}dx-480\int e^{-x}dx\\ & =480e^{-x}-x+480e^{-x}x+240e^{-x}{x}^2+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)+240\int e^{-x}{x}^{2}dx\\ & =480e^{-x}-x+480e^{-x}x+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)+480\int e^{-x}xdx\\ & =480e^{-x}-x+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)+480\int e^{-x}dx\\ & =-x+e^{-x}{x}^2\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right)\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 33, normalized size = 0.94 $$\begin{array}{c}x(-1+e^{-x}x\log \left(\frac{e^{-2+e^3-\frac{4}{x}-x^4}}{x}\right))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(4 - E^x - x - 4*x^5 + (2*x - x^2)*Log[E^((-4 - 2*x + E^3*x - x^5)/x)/x])/E^x,x]

[Out]

x*(-1 + (x*Log[E^(-2 + E^3 - 4/x - x^4)/x])/E^x)

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 39, normalized size = 1.11 $$\begin{array}{c}(x^2\log \left(\frac{e^{(-\frac{x^5-x{e}^3+2x+4}{x})}}{x}\right)-x{e}^x)e^{(-x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+2*x)*log(exp((x*exp(3)-x^5-2*x-4)/x)/x)-exp(x)-4*x^5-x+4)/exp(x),x, algorithm="fricas")

[Out]

(x^2*log(e^(-(x^5 - x*e^3 + 2*x + 4)/x)/x) - x*e^x)*e^(-x)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 50, normalized size = 1.43 $$\begin{array}{c}-x^6{e}^{(-x)}-x^2{e}^{(-x)}\log (x)-2x^2{e}^{(-x)}+x^2{e}^{(-x+3)}-4x{e}^{(-x)}-x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+2*x)*log(exp((x*exp(3)-x^5-2*x-4)/x)/x)-exp(x)-4*x^5-x+4)/exp(x),x, algorithm="giac")

[Out]

-x^6*e^(-x) - x^2*e^(-x)*log(x) - 2*x^2*e^(-x) + x^2*e^(-x + 3) - 4*x*e^(-x) - x

________________________________________________________________________________________

maple [C]  time = 0.25, size = 237, normalized size = 6.77

method	result	size
risch	$x^2{\mathrm{e}}^{-x}\ln \left({\mathrm{e}}^{\frac{x{\mathrm{e}}^3-x^5-2x-4}{x}}\right)-\frac{x(i\pi x\mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}\left(i{\mathrm{e}}^{\frac{x{\mathrm{e}}^3-x^5-2x-4}{x}}\right)\mathrm{csgn}\left(\frac{i{\mathrm{e}}^{\frac{x{\mathrm{e}}^3-x^5-2x-4}{x}}}{x}\right)-i\pi x\mathrm{csgn}\left(\frac{i}{x}\right)\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^{\frac{x{\mathrm{e}}^3-x^5-2x-4}{x}}}{x}\right)}^2-i\pi x\mathrm{csgn}\left(i{\mathrm{e}}^{\frac{x{\mathrm{e}}^3-x^5-2x-4}{x}}\right)\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^{\frac{x{\mathrm{e}}^3-x^5-2x-4}{x}}}{x}\right)}^2+i\pi x\mathrm{csgn}{\left(\frac{i{\mathrm{e}}^{\frac{x{\mathrm{e}}^3-x^5-2x-4}{x}}}{x}\right)}^3+2x\ln (x)+2{\mathrm{e}}^x){\mathrm{e}}^{-x}}{2}$	$237$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+2*x)*ln(exp((x*exp(3)-x^5-2*x-4)/x)/x)-exp(x)-4*x^5-x+4)/exp(x),x,method=_RETURNVERBOSE)

[Out]

x^2*exp(-x)*ln(exp((x*exp(3)-x^5-2*x-4)/x))-1/2*x*(I*Pi*x*csgn(I/x)*csgn(I*exp((x*exp(3)-x^5-2*x-4)/x))*csgn(I
/x*exp((x*exp(3)-x^5-2*x-4)/x))-I*Pi*x*csgn(I/x)*csgn(I/x*exp((x*exp(3)-x^5-2*x-4)/x))^2-I*Pi*x*csgn(I*exp((x*
exp(3)-x^5-2*x-4)/x))*csgn(I/x*exp((x*exp(3)-x^5-2*x-4)/x))^2+I*Pi*x*csgn(I/x*exp((x*exp(3)-x^5-2*x-4)/x))^3+2
*x*ln(x)+2*exp(x))*exp(-x)

________________________________________________________________________________________

maxima [B]  time = 0.42, size = 91, normalized size = 2.60 $$\begin{array}{c}-(x^6+4x^5+20x^4+80x^3-x^2(e^3-242)+x^2\log (x)+485x+477)e^{(-x)}+4(x^5+5x^4+20x^3+60x^2+120x+120)e^{(-x)}+(x+1)e^{(-x)}-x-4e^{(-x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+2*x)*log(exp((x*exp(3)-x^5-2*x-4)/x)/x)-exp(x)-4*x^5-x+4)/exp(x),x, algorithm="maxima")

[Out]

-(x^6 + 4*x^5 + 20*x^4 + 80*x^3 - x^2*(e^3 - 242) + x^2*log(x) + 485*x + 477)*e^(-x) + 4*(x^5 + 5*x^4 + 20*x^3
+ 60*x^2 + 120*x + 120)*e^(-x) + (x + 1)*e^(-x) - x - 4*e^(-x)

________________________________________________________________________________________

mupad [B]  time = 5.58, size = 121, normalized size = 3.46 $$\begin{array}{c}4{\mathrm{e}}^{-x}(x^5+5x^4+20x^3+60x^2+120x+120)-481{\mathrm{e}}^{-x}-485x{\mathrm{e}}^{-x}-x+{\mathrm{e}}^{-x}(x+1)-242x^2{\mathrm{e}}^{-x}-80x^3{\mathrm{e}}^{-x}-20x^4{\mathrm{e}}^{-x}-4x^5{\mathrm{e}}^{-x}-x^6{\mathrm{e}}^{-x}+x^2{\mathrm{e}}^{3-x}+x^2\ln \left(\frac{1}{x}\right){\mathrm{e}}^{-x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-x)*(x + exp(x) - log(exp(-(2*x - x*exp(3) + x^5 + 4)/x)/x)*(2*x - x^2) + 4*x^5 - 4),x)

[Out]

4*exp(-x)*(120*x + 60*x^2 + 20*x^3 + 5*x^4 + x^5 + 120) - 481*exp(-x) - 485*x*exp(-x) - x + exp(-x)*(x + 1) -
242*x^2*exp(-x) - 80*x^3*exp(-x) - 20*x^4*exp(-x) - 4*x^5*exp(-x) - x^6*exp(-x) + x^2*exp(3 - x) + x^2*log(1/x
)*exp(-x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Timed out}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+2*x)*ln(exp((x*exp(3)-x**5-2*x-4)/x)/x)-exp(x)-4*x**5-x+4)/exp(x),x)

[Out]

Timed out

________________________________________________________________________________________