3.74.60 \(\int \frac {e^{-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} (e^8 x \log ^2(x)+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}} (e^8 (4-x)+e^{8+\frac {x}{e^8}} x+(e^8 x+e^{\frac {x}{e^8}} (-e^8 x-x^2)) \log (x)))}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \]

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Rubi [F]  time = 41.96, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \left (e^8 x \log ^2(x)+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}} \left (e^8 (4-x)+e^{8+\frac {x}{e^8}} x+\left (e^8 x+e^{\frac {x}{e^8}} \left (-e^8 x-x^2\right )\right ) \log (x)\right )\right )}{x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-6 + E^((-4 + x - E^(x/E^8)*x)/Log[x]) + x)*(E^8*x*Log[x]^2 + E^((-4 + x - E^(x/E^8)*x)/Log[x])*(E^8*(
4 - x) + E^(8 + x/E^8)*x + (E^8*x + E^(x/E^8)*(-(E^8*x) - x^2))*Log[x])))/(x*Log[x]^2),x]

[Out]

Defer[Int][E^(2 + E^((-4 + x - E^(x/E^8)*x)/Log[x]) + x), x] - Defer[Int][E^(2 + E^((-4 + x - E^(x/E^8)*x)/Log
[x]) + x - (4 - x + E^(x/E^8)*x)/Log[x])/Log[x]^2, x] + Defer[Int][E^(2 + E^((-4 + x - E^(x/E^8)*x)/Log[x]) +
(1 + E^(-8))*x - (4 - x + E^(x/E^8)*x)/Log[x])/Log[x]^2, x] + 4*Defer[Int][E^(2 + E^((-4 + x - E^(x/E^8)*x)/Lo
g[x]) + x - (4 - x + E^(x/E^8)*x)/Log[x])/(x*Log[x]^2), x] + Defer[Int][E^(2 + E^((-4 + x - E^(x/E^8)*x)/Log[x
]) + x - (4 - x + E^(x/E^8)*x)/Log[x])/Log[x], x] - Defer[Int][E^(2 + E^((-4 + x - E^(x/E^8)*x)/Log[x]) + (1 +
E^(-8))*x - (4 - x + E^(x/E^8)*x)/Log[x])/Log[x], x] - Defer[Int][(E^(-6 + E^((-4 + x - E^(x/E^8)*x)/Log[x])
+ (1 + E^(-8))*x - (4 - x + E^(x/E^8)*x)/Log[x])*x)/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x}+\frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) \left (4 e^8-e^8 x+e^{8+\frac {x}{e^8}} x+e^8 x \log (x)-e^{8+\frac {x}{e^8}} x \log (x)-e^{\frac {x}{e^8}} x^2 \log (x)\right )}{x \log ^2(x)}\right ) \, dx\\ &=\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx+\int \frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) \left (4 e^8-e^8 x+e^{8+\frac {x}{e^8}} x+e^8 x \log (x)-e^{8+\frac {x}{e^8}} x \log (x)-e^{\frac {x}{e^8}} x^2 \log (x)\right )}{x \log ^2(x)} \, dx\\ &=\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx+\int \left (\frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) (4-x+x \log (x))}{x \log ^2(x)}-\frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x+\frac {x}{e^8}-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) \left (-e^8+e^8 \log (x)+x \log (x)\right )}{\log ^2(x)}\right ) \, dx\\ &=\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) (4-x+x \log (x))}{x \log ^2(x)} \, dx-\int \frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x+\frac {x}{e^8}-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) \left (-e^8+e^8 \log (x)+x \log (x)\right )}{\log ^2(x)} \, dx\\ &=\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx+\int \left (\frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) (4-x)}{x \log ^2(x)}+\frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log (x)}\right ) \, dx-\int \frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) \left (-e^8+e^8 \log (x)+x \log (x)\right )}{\log ^2(x)} \, dx\\ &=\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx-\int \left (-\frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log ^2(x)}+\frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) \left (e^8+x\right )}{\log (x)}\right ) \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) (4-x)}{x \log ^2(x)} \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log (x)} \, dx\\ &=\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx+\int \left (-\frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log ^2(x)}+\frac {4 \exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{x \log ^2(x)}\right ) \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log ^2(x)} \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log (x)} \, dx-\int \frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) \left (e^8+x\right )}{\log (x)} \, dx\\ &=4 \int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{x \log ^2(x)} \, dx+\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx-\int \left (\frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log (x)}+\frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) x}{\log (x)}\right ) \, dx-\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log ^2(x)} \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log ^2(x)} \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log (x)} \, dx\\ &=4 \int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{x \log ^2(x)} \, dx+\int e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \, dx-\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log ^2(x)} \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log ^2(x)} \, dx+\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log (x)} \, dx-\int \frac {\exp \left (2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right )}{\log (x)} \, dx-\int \frac {\exp \left (-6+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+\left (1+\frac {1}{e^8}\right ) x-\frac {4-x+e^{\frac {x}{e^8}} x}{\log (x)}\right ) x}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 25, normalized size = 1.00 \begin {gather*} e^{2+e^{\frac {-4+x-e^{\frac {x}{e^8}} x}{\log (x)}}+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-6 + E^((-4 + x - E^(x/E^8)*x)/Log[x]) + x)*(E^8*x*Log[x]^2 + E^((-4 + x - E^(x/E^8)*x)/Log[x])*
(E^8*(4 - x) + E^(8 + x/E^8)*x + (E^8*x + E^(x/E^8)*(-(E^8*x) - x^2))*Log[x])))/(x*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.64, size = 32, normalized size = 1.28 \begin {gather*} e^{\left (x + e^{\left (\frac {{\left ({\left (x - 4\right )} e^{8} - x e^{\left ({\left (x + 8 \, e^{8}\right )} e^{\left (-8\right )}\right )}\right )} e^{\left (-8\right )}}{\log \relax (x)}\right )} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x + e^(((x - 4)*e^8 - x*e^((x + 8*e^8)*e^(-8)))*e^(-8)/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*exp(4)^2-x^2)*exp(x/exp(4)^2)+x*exp(4)^2)*log(x)+x*exp(4)^2*exp(x/exp(4)^2)+(-x+4)*exp(4)^2)*
exp((-x*exp(x/exp(4)^2)+x-4)/log(x))+x*exp(4)^2*log(x)^2)*exp(exp((-x*exp(x/exp(4)^2)+x-4)/log(x))+2+x)/x/exp(
4)^2/log(x)^2,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.18, size = 24, normalized size = 0.96

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{-\frac {x \,{\mathrm e}^{x \,{\mathrm e}^{-8}}-x +4}{\ln \relax (x )}}+2+x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(-(x*exp(x*exp(-8))-x+4)/ln(x))+2+x)

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maxima [A]  time = 0.64, size = 30, normalized size = 1.20 \begin {gather*} e^{\left (x + e^{\left (-\frac {x e^{\left (x e^{\left (-8\right )}\right )}}{\log \relax (x)} + \frac {x}{\log \relax (x)} - \frac {4}{\log \relax (x)}\right )} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x + e^(-x*e^(x*e^(-8))/log(x) + x/log(x) - 4/log(x)) + 2)

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mupad [B]  time = 5.22, size = 34, normalized size = 1.36 \begin {gather*} {\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {4}{\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-8}}}{\ln \relax (x)}}\,{\mathrm {e}}^{\frac {x}{\ln \relax (x)}}}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-8)*exp(x + exp(-(x*exp(x*exp(-8)) - x + 4)/log(x)) + 2)*(exp(-(x*exp(x*exp(-8)) - x + 4)/log(x))*(lo
g(x)*(x*exp(8) - exp(x*exp(-8))*(x*exp(8) + x^2)) - exp(8)*(x - 4) + x*exp(8)*exp(x*exp(-8))) + x*exp(8)*log(x
)^2))/(x*log(x)^2),x)

[Out]

exp(2)*exp(exp(-4/log(x))*exp(-(x*exp(x*exp(-8)))/log(x))*exp(x/log(x)))*exp(x)

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sympy [A]  time = 64.39, size = 20, normalized size = 0.80 \begin {gather*} e^{x + e^{\frac {- x e^{\frac {x}{e^{8}}} + x - 4}{\log {\relax (x )}}} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x + exp((-x*exp(x*exp(-8)) + x - 4)/log(x)) + 2)

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