3.64.12 \(\int \frac {(-8 x^2+2 x^3) \log ^3(x)+e^{\frac {e^2+(-e^{e+x} x+x^2) \log ^2(x)}{x \log ^2(x)}} (-2 e^2-e^2 \log (x)+(x^2-e^{e+x} x^2) \log ^3(x))}{x^2 \log ^3(x)} \, dx\)

Optimal. Leaf size=30 \[ e^{-e^{e+x}+x+\frac {e^2}{x \log ^2(x)}}+(4-x)^2 \]

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Rubi [A]  time = 1.87, antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 2, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6688, 6706} \begin {gather*} x^2-8 x+e^{x-e^{x+e}+\frac {e^2}{x \log ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-8+2 x+\frac {e^{-e^{e+x}+x+\frac {e^2}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)-\left (-1+e^{e+x}\right ) x^2 \log ^3(x)\right )}{x^2 \log ^3(x)}\right ) \, dx\\ &=-8 x+x^2+\int \frac {e^{-e^{e+x}+x+\frac {e^2}{x \log ^2(x)}} \left (-2 e^2-e^2 \log (x)-\left (-1+e^{e+x}\right ) x^2 \log ^3(x)\right )}{x^2 \log ^3(x)} \, dx\\ &=e^{-e^{e+x}+x+\frac {e^2}{x \log ^2(x)}}-8 x+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.12, size = 29, normalized size = 0.97 \begin {gather*} e^{-e^{e+x}+x+\frac {e^2}{x \log ^2(x)}}-8 x+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.82, size = 36, normalized size = 1.20 \begin {gather*} x^{2} - 8 \, x + e^{\left (\frac {{\left (x^{2} - x e^{\left (x + e\right )}\right )} \log \relax (x)^{2} + e^{2}}{x \log \relax (x)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.16, size = 40, normalized size = 1.33

method	result	size
risch	\(x^{2}-8 x +{\mathrm e}^{\frac {-\ln \relax (x )^{2} {\mathrm e}^{x +{\mathrm e}} x +x^{2} \ln \relax (x )^{2}+{\mathrm e}^{2}}{\ln \relax (x )^{2} x}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2-8*x+exp((-ln(x)^2*exp(x+exp(1))*x+x^2*ln(x)^2+exp(2))/ln(x)^2/x)

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maxima [A]  time = 0.47, size = 27, normalized size = 0.90 \begin {gather*} x^{2} - 8 \, x + e^{\left (x + \frac {e^{2}}{x \log \relax (x)^{2}} - e^{\left (x + e\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.25, size = 29, normalized size = 0.97 \begin {gather*} x^2-8\,x+{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x\,{\ln \relax (x)}^2}}\,{\mathrm {e}}^{-{\mathrm {e}}^{\mathrm {e}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(2) - log(x)^2*(x*exp(x + exp(1)) - x^2))/(x*log(x)^2))*(2*exp(2) + exp(2)*log(x) - log(x)^3*(x^
2 - x^2*exp(x + exp(1)))) + log(x)^3*(8*x^2 - 2*x^3))/(x^2*log(x)^3),x)

[Out]

x^2 - 8*x + exp(exp(2)/(x*log(x)^2))*exp(-exp(exp(1))*exp(x))*exp(x)

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sympy [A]  time = 0.55, size = 34, normalized size = 1.13 \begin {gather*} x^{2} - 8 x + e^{\frac {\left (x^{2} - x e^{x + e}\right ) \log {\relax (x )}^{2} + e^{2}}{x \log {\relax (x )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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