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

Optimal. Leaf size=21 \[ 3-3 \log (x)+\log \left (\log \left (2 \left (e^{e^{\frac {1}{x}+x}}+x\right )\right )\right ) \]

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Rubi [F]  time = 5.54, 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^{e^{\frac {1}{x}+x}+\frac {1}{x}+x} \left (-1+x^2\right )+\left (-3 e^{e^{\frac {1}{x}+x}} x-3 x^2\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )}{\left (e^{e^{\frac {1}{x}+x}} x^2+x^3\right ) \log \left (2 e^{e^{\frac {1}{x}+x}}+2 x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 20, normalized size = 0.95 \begin {gather*} -3 \log (x)+\log \left (\log \left (2 \left (e^{e^{\frac {1}{x}+x}}+x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.72, size = 55, normalized size = 2.62 \begin {gather*} -3 \, \log \relax (x) + \log \left (\log \left (2 \, {\left (x e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {x^{2} + x e^{\left (\frac {x^{2} + 1}{x}\right )} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x) + log(log(2*(x*e^((x^2 + 1)/x) + e^((x^2 + x*e^((x^2 + 1)/x) + 1)/x))*e^(-(x^2 + 1)/x)))

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giac [B]  time = 0.38, size = 55, normalized size = 2.62 \begin {gather*} -3 \, \log \relax (x) + \log \left (\log \left (2 \, {\left (x e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {x^{2} + x e^{\left (\frac {x^{2} + 1}{x}\right )} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x) + log(log(2*(x*e^((x^2 + 1)/x) + e^((x^2 + x*e^((x^2 + 1)/x) + 1)/x))*e^(-(x^2 + 1)/x)))

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maple [A]  time = 0.29, size = 25, normalized size = 1.19

method	result	size
risch	\(-3 \ln \relax (x )+\ln \left (\ln \left (2 \,{\mathrm e}^{{\mathrm e}^{\frac {x^{2}+1}{x}}}+2 x \right )\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*ln(x)+ln(ln(2*exp(exp((x^2+1)/x))+2*x))

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maxima [A]  time = 0.49, size = 19, normalized size = 0.90 \begin {gather*} -3 \, \log \relax (x) + \log \left (\log \relax (2) + \log \left (x + e^{\left (e^{\left (x + \frac {1}{x}\right )}\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x) + log(log(2) + log(x + e^(e^(x + 1/x))))

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mupad [B]  time = 5.71, size = 21, normalized size = 1.00 \begin {gather*} \ln \left (\ln \left (2\,x+2\,{\mathrm {e}}^{{\mathrm {e}}^{1/x}\,{\mathrm {e}}^x}\right )\right )-3\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(2*x + 2*exp(exp(1/x)*exp(x)))) - 3*log(x)

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sympy [A]  time = 9.15, size = 22, normalized size = 1.05 \begin {gather*} - 3 \log {\relax (x )} + \log {\left (\log {\left (2 x + 2 e^{e^{\frac {1}{x}} e^{x}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x) + log(log(2*x + 2*exp(exp(1/x)*exp(x))))

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