3.59.41 \(\int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} (e^{3 x} (1-x)-x^3+x^4+e^{2 x} (-3 x+3 x^2)+e^x (3 x^2-3 x^3)+(e^x (16 x^3-16 x^4)+e^x (-16 x^2+16 x^3) \log (x)) \log (-x+\log (x)))}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+(e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4) \log (x)} \, dx\)

Optimal. Leaf size=24 \[ e^{-\frac {8 x^2}{\left (e^x-x\right )^2}} \log (-x+\log (x)) \]

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Rubi [B]  time = 0.59, antiderivative size = 194, normalized size of antiderivative = 8.08, number of steps used = 1, number of rules used = 1, integrand size = 184, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {2288} \begin {gather*} \frac {e^{-\frac {8 x^2}{x^2-2 e^x x+e^{2 x}}} \left (e^x \left (x^3-x^4\right )-e^x \left (x^2-x^3\right ) \log (x)\right ) \log (\log (x)-x)}{\left (\frac {\left (e^x x-x+e^x-e^{2 x}\right ) x^2}{\left (x^2-2 e^x x+e^{2 x}\right )^2}+\frac {x}{x^2-2 e^x x+e^{2 x}}\right ) \left (-x^5+3 e^x x^4-3 e^{2 x} x^3+e^{3 x} x^2-\left (-x^4+3 e^x x^3-3 e^{2 x} x^2+e^{3 x} x\right ) \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3*x)*(1 - x) - x^3 + x^4 + E^(2*x)*(-3*x + 3*x^2) + E^x*(3*x^2 - 3*x^3) + (E^x*(16*x^3 - 16*x^4) + E^x
*(-16*x^2 + 16*x^3)*Log[x])*Log[-x + Log[x]])/(E^((8*x^2)/(E^(2*x) - 2*E^x*x + x^2))*(-(E^(3*x)*x^2) + 3*E^(2*
x)*x^3 - 3*E^x*x^4 + x^5 + (E^(3*x)*x - 3*E^(2*x)*x^2 + 3*E^x*x^3 - x^4)*Log[x])),x]

[Out]

((E^x*(x^3 - x^4) - E^x*(x^2 - x^3)*Log[x])*Log[-x + Log[x]])/(E^((8*x^2)/(E^(2*x) - 2*E^x*x + x^2))*((x^2*(E^
x - E^(2*x) - x + E^x*x))/(E^(2*x) - 2*E^x*x + x^2)^2 + x/(E^(2*x) - 2*E^x*x + x^2))*(E^(3*x)*x^2 - 3*E^(2*x)*
x^3 + 3*E^x*x^4 - x^5 - (E^(3*x)*x - 3*E^(2*x)*x^2 + 3*E^x*x^3 - x^4)*Log[x]))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^x \left (x^3-x^4\right )-e^x \left (x^2-x^3\right ) \log (x)\right ) \log (-x+\log (x))}{\left (\frac {x^2 \left (e^x-e^{2 x}-x+e^x x\right )}{\left (e^{2 x}-2 e^x x+x^2\right )^2}+\frac {x}{e^{2 x}-2 e^x x+x^2}\right ) \left (e^{3 x} x^2-3 e^{2 x} x^3+3 e^x x^4-x^5-\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 24, normalized size = 1.00 \begin {gather*} e^{-\frac {8 x^2}{\left (e^x-x\right )^2}} \log (-x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3*x)*(1 - x) - x^3 + x^4 + E^(2*x)*(-3*x + 3*x^2) + E^x*(3*x^2 - 3*x^3) + (E^x*(16*x^3 - 16*x^4)
+ E^x*(-16*x^2 + 16*x^3)*Log[x])*Log[-x + Log[x]])/(E^((8*x^2)/(E^(2*x) - 2*E^x*x + x^2))*(-(E^(3*x)*x^2) + 3
*E^(2*x)*x^3 - 3*E^x*x^4 + x^5 + (E^(3*x)*x - 3*E^(2*x)*x^2 + 3*E^x*x^3 - x^4)*Log[x])),x]

[Out]

Log[-x + Log[x]]/E^((8*x^2)/(E^x - x)^2)

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fricas [A]  time = 0.64, size = 29, normalized size = 1.21 \begin {gather*} e^{\left (-\frac {8 \, x^{2}}{x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}\right )} \log \left (-x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-16*x^2)*exp(x)*log(x)+(-16*x^4+16*x^3)*exp(x))*log(log(x)-x)+(-x+1)*exp(x)^3+(3*x^2-3*x)*e
xp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/((x*exp(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*log(x)-x^2*exp(x)^3+3*exp
(x)^2*x^3-3*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x, algorithm="fricas")

[Out]

e^(-8*x^2/(x^2 - 2*x*e^x + e^(2*x)))*log(-x + log(x))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-16*x^2)*exp(x)*log(x)+(-16*x^4+16*x^3)*exp(x))*log(log(x)-x)+(-x+1)*exp(x)^3+(3*x^2-3*x)*e
xp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/((x*exp(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*log(x)-x^2*exp(x)^3+3*exp
(x)^2*x^3-3*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-16,[1,23,7]%%%}+%%%{64,[1,23,6]%%%}+%%%{-96,[1,23,5]%%%
}+%%%{64,[1

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maple [A]  time = 0.06, size = 34, normalized size = 1.42

method	result	size
risch	\({\mathrm e}^{\frac {8 x^{2}}{-{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x -x^{2}}} \ln \left (\ln \relax (x )-x \right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^3-16*x^2)*exp(x)*ln(x)+(-16*x^4+16*x^3)*exp(x))*ln(ln(x)-x)+(1-x)*exp(x)^3+(3*x^2-3*x)*exp(x)^2+(-
3*x^3+3*x^2)*exp(x)+x^4-x^3)/((x*exp(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*ln(x)-x^2*exp(x)^3+3*exp(x)^2*x^3-3
*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x,method=_RETURNVERBOSE)

[Out]

exp(8*x^2/(-exp(2*x)+2*exp(x)*x-x^2))*ln(ln(x)-x)

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maxima [A]  time = 0.58, size = 44, normalized size = 1.83 \begin {gather*} e^{\left (-\frac {8 \, e^{\left (2 \, x\right )}}{x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}} - \frac {16 \, e^{x}}{x - e^{x}} - 8\right )} \log \left (-x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-16*x^2)*exp(x)*log(x)+(-16*x^4+16*x^3)*exp(x))*log(log(x)-x)+(-x+1)*exp(x)^3+(3*x^2-3*x)*e
xp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/((x*exp(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*log(x)-x^2*exp(x)^3+3*exp
(x)^2*x^3-3*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x, algorithm="maxima")

[Out]

e^(-8*e^(2*x)/(x^2 - 2*x*e^x + e^(2*x)) - 16*e^x/(x - e^x) - 8)*log(-x + log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{-\frac {8\,x^2}{{\mathrm {e}}^{2\,x}-2\,x\,{\mathrm {e}}^x+x^2}}\,\left ({\mathrm {e}}^{2\,x}\,\left (3\,x-3\,x^2\right )-{\mathrm {e}}^x\,\left (3\,x^2-3\,x^3\right )-\ln \left (\ln \relax (x)-x\right )\,\left ({\mathrm {e}}^x\,\left (16\,x^3-16\,x^4\right )-{\mathrm {e}}^x\,\ln \relax (x)\,\left (16\,x^2-16\,x^3\right )\right )+{\mathrm {e}}^{3\,x}\,\left (x-1\right )+x^3-x^4\right )}{3\,x^3\,{\mathrm {e}}^{2\,x}-x^2\,{\mathrm {e}}^{3\,x}-3\,x^4\,{\mathrm {e}}^x+\ln \relax (x)\,\left (x\,{\mathrm {e}}^{3\,x}+3\,x^3\,{\mathrm {e}}^x-3\,x^2\,{\mathrm {e}}^{2\,x}-x^4\right )+x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(8*x^2)/(exp(2*x) - 2*x*exp(x) + x^2))*(exp(2*x)*(3*x - 3*x^2) - exp(x)*(3*x^2 - 3*x^3) - log(log(x
) - x)*(exp(x)*(16*x^3 - 16*x^4) - exp(x)*log(x)*(16*x^2 - 16*x^3)) + exp(3*x)*(x - 1) + x^3 - x^4))/(3*x^3*ex
p(2*x) - x^2*exp(3*x) - 3*x^4*exp(x) + log(x)*(x*exp(3*x) + 3*x^3*exp(x) - 3*x^2*exp(2*x) - x^4) + x^5),x)

[Out]

int(-(exp(-(8*x^2)/(exp(2*x) - 2*x*exp(x) + x^2))*(exp(2*x)*(3*x - 3*x^2) - exp(x)*(3*x^2 - 3*x^3) - log(log(x
) - x)*(exp(x)*(16*x^3 - 16*x^4) - exp(x)*log(x)*(16*x^2 - 16*x^3)) + exp(3*x)*(x - 1) + x^3 - x^4))/(3*x^3*ex
p(2*x) - x^2*exp(3*x) - 3*x^4*exp(x) + log(x)*(x*exp(3*x) + 3*x^3*exp(x) - 3*x^2*exp(2*x) - x^4) + x^5), x)

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sympy [A]  time = 73.72, size = 27, normalized size = 1.12 \begin {gather*} e^{- \frac {8 x^{2}}{x^{2} - 2 x e^{x} + e^{2 x}}} \log {\left (- x + \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**3-16*x**2)*exp(x)*ln(x)+(-16*x**4+16*x**3)*exp(x))*ln(ln(x)-x)+(-x+1)*exp(x)**3+(3*x**2-3*x
)*exp(x)**2+(-3*x**3+3*x**2)*exp(x)+x**4-x**3)/((x*exp(x)**3-3*exp(x)**2*x**2+3*exp(x)*x**3-x**4)*ln(x)-x**2*e
xp(x)**3+3*exp(x)**2*x**3-3*exp(x)*x**4+x**5)/exp(8*x**2/(exp(x)**2-2*exp(x)*x+x**2)),x)

[Out]

exp(-8*x**2/(x**2 - 2*x*exp(x) + exp(2*x)))*log(-x + log(x))

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