3.95.77 \(\int \frac {2-e^{-1+x}-e^x+(e^{-1+x} x+e^x x) \log (x)-x \log ^2(x)}{(2 x-e^{-1+x} x-e^x x) \log (x)+(4 x+x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \log \left (-\frac {2}{4+x+\frac {2-e^{-1+x}-e^x}{\log (x)}}\right ) \]

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Rubi [F]  time = 2.74, 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 {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e \left (2-e^{-1+x} (1+e)+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)\right )}{x \log (x) \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx\\ &=e \int \frac {2-e^{-1+x} (1+e)+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{x \log (x) \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx\\ &=e \int \left (\frac {1-x \log (x)}{e x \log (x)}+\frac {-4+x+3 x \log (x)+x^2 \log (x)}{x \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )}\right ) \, dx\\ &=e \int \frac {-4+x+3 x \log (x)+x^2 \log (x)}{x \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx+\int \frac {1-x \log (x)}{x \log (x)} \, dx\\ &=e \int \left (\frac {4}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )}+\frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}+\frac {3 \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}+\frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}\right ) \, dx+\int \left (-1+\frac {1}{x \log (x)}\right ) \, dx\\ &=-x+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=-x+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-x+\log (\log (x))+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.15, size = 30, normalized size = 1.11 \begin {gather*} \log (\log (x))-\log \left (-2 e+e^x+e^{1+x}-4 e \log (x)-e x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - E^(-1 + x) - E^x + (E^(-1 + x)*x + E^x*x)*Log[x] - x*Log[x]^2)/((2*x - E^(-1 + x)*x - E^x*x)*Lo
g[x] + (4*x + x^2)*Log[x]^2),x]

[Out]

Log[Log[x]] - Log[-2*E + E^x + E^(1 + x) - 4*E*Log[x] - E*x*Log[x]]

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fricas [A]  time = 0.61, size = 40, normalized size = 1.48 \begin {gather*} -\log \left (x + 4\right ) - \log \left (\frac {{\left (x + 4\right )} e \log \relax (x) - {\left (e + 1\right )} e^{x} + 2 \, e}{x + 4}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)^2+(exp(x)*x+x*exp(x-1))*log(x)-exp(x)-exp(x-1)+2)/((x^2+4*x)*log(x)^2+(-exp(x)*x-x*exp(x-
1)+2*x)*log(x)),x, algorithm="fricas")

[Out]

-log(x + 4) - log(((x + 4)*e*log(x) - (e + 1)*e^x + 2*e)/(x + 4)) + log(log(x))

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giac [A]  time = 0.25, size = 34, normalized size = 1.26 \begin {gather*} -\log \left (x e \log \relax (x) + 4 \, e \log \relax (x) + 2 \, e - e^{\left (x + 1\right )} - e^{x}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)^2+(exp(x)*x+x*exp(x-1))*log(x)-exp(x)-exp(x-1)+2)/((x^2+4*x)*log(x)^2+(-exp(x)*x-x*exp(x-
1)+2*x)*log(x)),x, algorithm="giac")

[Out]

-log(x*e*log(x) + 4*e*log(x) + 2*e - e^(x + 1) - e^x) + log(log(x))

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maple [A]  time = 0.16, size = 32, normalized size = 1.19

method	result	size
risch	\(-\ln \left (4+x \right )+\ln \left (\ln \relax (x )\right )-\ln \left (\ln \relax (x )-\frac {{\mathrm e}^{x -1}+{\mathrm e}^{x}-2}{4+x}\right )\)	\(32\)
norman	\(-\ln \left (x \,{\mathrm e} \ln \relax (x )-{\mathrm e} \,{\mathrm e}^{x}+4 \,{\mathrm e} \ln \relax (x )+2 \,{\mathrm e}-{\mathrm e}^{x}\right )+\ln \left (\ln \relax (x )\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x)^2+(exp(x)*x+x*exp(x-1))*ln(x)-exp(x)-exp(x-1)+2)/((x^2+4*x)*ln(x)^2+(-exp(x)*x-x*exp(x-1)+2*x)*l
n(x)),x,method=_RETURNVERBOSE)

[Out]

-ln(4+x)+ln(ln(x))-ln(ln(x)-(exp(x-1)+exp(x)-2)/(4+x))

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maxima [A]  time = 0.39, size = 39, normalized size = 1.44 \begin {gather*} -\log \left (\frac {{\left (e + 1\right )} e^{x} - {\left (x e + 4 \, e\right )} \log \relax (x) - 2 \, e}{e + 1}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)^2+(exp(x)*x+x*exp(x-1))*log(x)-exp(x)-exp(x-1)+2)/((x^2+4*x)*log(x)^2+(-exp(x)*x-x*exp(x-
1)+2*x)*log(x)),x, algorithm="maxima")

[Out]

-log(((e + 1)*e^x - (x*e + 4*e)*log(x) - 2*e)/(e + 1)) + log(log(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - 1) + exp(x) + x*log(x)^2 - log(x)*(x*exp(x - 1) + x*exp(x)) - 2)/(log(x)*(x*exp(x - 1) - 2*x + x*
exp(x)) - log(x)^2*(4*x + x^2)),x)

[Out]

-int(-(exp(x - 1) + exp(x) + x*log(x)^2 - log(x)*(x*exp(x - 1) + x*exp(x)) - 2)/(log(x)*(x*exp(x - 1) - 2*x +
x*exp(x)) - log(x)^2*(4*x + x^2)), x)

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sympy [A]  time = 0.51, size = 37, normalized size = 1.37 \begin {gather*} - \log {\left (\frac {- e x \log {\relax (x )} - 4 e \log {\relax (x )} - 2 e}{1 + e} + e^{x} \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x)**2+(exp(x)*x+x*exp(x-1))*ln(x)-exp(x)-exp(x-1)+2)/((x**2+4*x)*ln(x)**2+(-exp(x)*x-x*exp(x-
1)+2*x)*ln(x)),x)

[Out]

-log((-E*x*log(x) - 4*E*log(x) - 2*E)/(1 + E) + exp(x)) + log(log(x))

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