3.96.14 \(\int \frac {x^2+e^{2-x} (x^2+x^3)-x^2 \log (x)+e^{e^{5-x}} (1+e^{7-2 x} x+e^{2-x} (3+x)+(-3-e^{5-x} x) \log (x))}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} (-e^{2-x} x+x \log (x))} \, dx\)

Optimal. Leaf size=34 \[ 1+\log \left (\frac {\left (\frac {e^{e^{5-x}}}{x}+x\right ) \left (e^{2-x}-\log (x)\right )}{x^2}\right ) \]

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Rubi [F]  time = 6.92, 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 {x^2+e^{2-x} \left (x^2+x^3\right )-x^2 \log (x)+e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )}{-e^{2-x} x^3+x^3 \log (x)+e^{e^{5-x}} \left (-e^{2-x} x+x \log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^2 + E^(2 - x)*(x^2 + x^3) - x^2*Log[x] + E^E^(5 - x)*(1 + E^(7 - 2*x)*x + E^(2 - x)*(3 + x) + (-3 - E^(
5 - x)*x)*Log[x]))/(-(E^(2 - x)*x^3) + x^3*Log[x] + E^E^(5 - x)*(-(E^(2 - x)*x) + x*Log[x])),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (-x^2-e^{2-x} \left (x^2+x^3\right )+x^2 \log (x)-e^{e^{5-x}} \left (1+e^{7-2 x} x+e^{2-x} (3+x)+\left (-3-e^{5-x} x\right ) \log (x)\right )\right )}{x \left (e^{e^{5-x}}+x^2\right ) \left (e^2-e^x \log (x)\right )} \, dx\\ &=\int \left (-\frac {e^{5+e^{5-x}-x}}{e^{e^{5-x}}+x^2}-\frac {3 e^{e^{5-x}}+e^{e^{5-x}} x+x^2+x^3}{x \left (e^{e^{5-x}}+x^2\right )}-\frac {e^x (1+x \log (x))}{x \left (e^2-e^x \log (x)\right )}\right ) \, dx\\ &=-\int \frac {e^{5+e^{5-x}-x}}{e^{e^{5-x}}+x^2} \, dx-\int \frac {3 e^{e^{5-x}}+e^{e^{5-x}} x+x^2+x^3}{x \left (e^{e^{5-x}}+x^2\right )} \, dx-\int \frac {e^x (1+x \log (x))}{x \left (e^2-e^x \log (x)\right )} \, dx\\ &=-\int \frac {e^{5+e^{5-x}-x}}{e^{e^{5-x}}+x^2} \, dx-\int \frac {x^2 (1+x)+e^{e^{5-x}} (3+x)}{x \left (e^{e^{5-x}}+x^2\right )} \, dx-\operatorname {Subst}\left (\int \frac {1}{e^2-x} \, dx,x,e^x \log (x)\right )\\ &=\log \left (e^2-e^x \log (x)\right )-\int \frac {e^{5+e^{5-x}-x}}{e^{e^{5-x}}+x^2} \, dx-\int \left (\frac {3+x}{x}-\frac {2 x}{e^{e^{5-x}}+x^2}\right ) \, dx\\ &=\log \left (e^2-e^x \log (x)\right )+2 \int \frac {x}{e^{e^{5-x}}+x^2} \, dx-\int \frac {3+x}{x} \, dx-\int \frac {e^{5+e^{5-x}-x}}{e^{e^{5-x}}+x^2} \, dx\\ &=\log \left (e^2-e^x \log (x)\right )+2 \int \frac {x}{e^{e^{5-x}}+x^2} \, dx-\int \left (1+\frac {3}{x}\right ) \, dx-\int \frac {e^{5+e^{5-x}-x}}{e^{e^{5-x}}+x^2} \, dx\\ &=-x-3 \log (x)+\log \left (e^2-e^x \log (x)\right )+2 \int \frac {x}{e^{e^{5-x}}+x^2} \, dx-\int \frac {e^{5+e^{5-x}-x}}{e^{e^{5-x}}+x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 34, normalized size = 1.00 \begin {gather*} -x-3 \log (x)+\log \left (e^{e^{5-x}}+x^2\right )+\log \left (e^2-e^x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^(2 - x)*(x^2 + x^3) - x^2*Log[x] + E^E^(5 - x)*(1 + E^(7 - 2*x)*x + E^(2 - x)*(3 + x) + (-3
- E^(5 - x)*x)*Log[x]))/(-(E^(2 - x)*x^3) + x^3*Log[x] + E^E^(5 - x)*(-(E^(2 - x)*x) + x*Log[x])),x]

[Out]

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

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fricas [A]  time = 0.64, size = 32, normalized size = 0.94 \begin {gather*} \log \left (x^{2} + e^{\left (e^{\left (-x + 5\right )}\right )}\right ) + \log \left (e^{3} \log \relax (x) - e^{\left (-x + 5\right )}\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(5-x)-3)*log(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*exp(exp(5-x))-x^2*log(x)+(x^3+x^2)*ex
p(2-x)+x^2)/((x*log(x)-x*exp(2-x))*exp(exp(5-x))+x^3*log(x)-x^3*exp(2-x)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \log \relax (x) - x^{2} - {\left (x^{3} + x^{2}\right )} e^{\left (-x + 2\right )} - {\left ({\left (x + 3\right )} e^{\left (-x + 2\right )} + x e^{\left (-2 \, x + 7\right )} - {\left (x e^{\left (-x + 5\right )} + 3\right )} \log \relax (x) + 1\right )} e^{\left (e^{\left (-x + 5\right )}\right )}}{x^{3} e^{\left (-x + 2\right )} - x^{3} \log \relax (x) + {\left (x e^{\left (-x + 2\right )} - x \log \relax (x)\right )} e^{\left (e^{\left (-x + 5\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(5-x)-3)*log(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*exp(exp(5-x))-x^2*log(x)+(x^3+x^2)*ex
p(2-x)+x^2)/((x*log(x)-x*exp(2-x))*exp(exp(5-x))+x^3*log(x)-x^3*exp(2-x)),x, algorithm="giac")

[Out]

integrate((x^2*log(x) - x^2 - (x^3 + x^2)*e^(-x + 2) - ((x + 3)*e^(-x + 2) + x*e^(-2*x + 7) - (x*e^(-x + 5) +
3)*log(x) + 1)*e^(e^(-x + 5)))/(x^3*e^(-x + 2) - x^3*log(x) + (x*e^(-x + 2) - x*log(x))*e^(e^(-x + 5))), x)

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maple [A]  time = 0.06, size = 30, normalized size = 0.88

method	result	size
risch	\(-3 \ln \relax (x )+\ln \left (-{\mathrm e}^{2-x}+\ln \relax (x )\right )+\ln \left (x^{2}+{\mathrm e}^{{\mathrm e}^{5-x}}\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*ln(x)+ln(-exp(2-x)+ln(x))+ln(x^2+exp(exp(5-x)))

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maxima [A]  time = 0.42, size = 39, normalized size = 1.15 \begin {gather*} -x + \log \left (x^{2} + e^{\left (e^{\left (-x + 5\right )}\right )}\right ) - 3 \, \log \relax (x) + \log \left (\frac {e^{x} \log \relax (x) - e^{2}}{\log \relax (x)}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x*exp(5-x)-3)*log(x)+x*exp(2-x)*exp(5-x)+(3+x)*exp(2-x)+1)*exp(exp(5-x))-x^2*log(x)+(x^3+x^2)*ex
p(2-x)+x^2)/((x*log(x)-x*exp(2-x))*exp(exp(5-x))+x^3*log(x)-x^3*exp(2-x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.19, size = 30, normalized size = 0.88 \begin {gather*} \ln \left (\left (\ln \relax (x)-{\mathrm {e}}^{-x}\,{\mathrm {e}}^2\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}+x^2\right )\right )-3\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2 - x)*(x^2 + x^3) - x^2*log(x) + exp(exp(5 - x))*(exp(2 - x)*(x + 3) - log(x)*(x*exp(5 - x) + 3) +
x*exp(2 - x)*exp(5 - x) + 1) + x^2)/(x^3*exp(2 - x) - x^3*log(x) + exp(exp(5 - x))*(x*exp(2 - x) - x*log(x))),
x)

[Out]

log((log(x) - exp(-x)*exp(2))*(exp(exp(-x)*exp(5)) + x^2)) - 3*log(x)

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sympy [A]  time = 0.45, size = 29, normalized size = 0.85 \begin {gather*} - 3 \log {\relax (x )} + \log {\left (x^{2} + e^{e^{3} e^{2 - x}} \right )} + \log {\left (e^{2 - x} - \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x) + log(x**2 + exp(exp(3)*exp(2 - x))) + log(exp(2 - x) - log(x))

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