\(\int \frac {e^2 x-x^2+(x^2+2 x^3+e^4 (1+2 x)+e^2 (-2 x-4 x^2)) \log (x)+(2 e^2 x-x^2) \log (x) \log (\log (x))}{(e^4-2 e^2 x+x^2) \log (x)} \, dx\) [7965]

Optimal result

Integrand size = 83, antiderivative size = 21 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=x+x^2+\frac {x^2 \log (\log (x))}{e^2-x} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x \left (-\left (\left (e^2-x\right ) (1+x)\right )-x \log (\log (x))\right )}{-e^2+x} \]

[In]

Integrate[(E^2*x - x^2 + (x^2 + 2*x^3 + E^4*(1 + 2*x) + E^2*(-2*x - 4*x^2))*Log[x] + (2*E^2*x - x^2)*Log[x]*Lo
g[Log[x]])/((E^4 - 2*E^2*x + x^2)*Log[x]),x]

[Out]

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

Maple [A] (verified)

Time = 5.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81

method	result	size
risch	\(-\frac {-x^{2} {\mathrm e}^{2}-x^{2} \ln \left (\ln \left (x \right )\right )+x^{3}-{\mathrm e}^{2} x +x^{2}}{{\mathrm e}^{2}-x}\)	\(38\)
parallelrisch	\(\frac {x^{2} {\mathrm e}^{2}-x^{3}+x^{2} \ln \left (\ln \left (x \right )\right )+{\mathrm e}^{4}-x^{2}}{{\mathrm e}^{2}-x}\)	\(38\)

[In]

int(((2*exp(2)*x-x^2)*ln(x)*ln(ln(x))+((1+2*x)*exp(2)^2+(-4*x^2-2*x)*exp(2)+2*x^3+x^2)*ln(x)+exp(2)*x-x^2)/(ex
p(2)^2-2*exp(2)*x+x^2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x^{3} - x^{2} \log \left (\log \left (x\right )\right ) + x^{2} - {\left (x^{2} + x\right )} e^{2}}{x - e^{2}} \]

[In]

integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((1+2*x)*exp(2)^2+(-4*x^2-2*x)*exp(2)+2*x^3+x^2)*log(x)+exp(2)*
x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/log(x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=x^{2} + x - e^{2} \log {\left (\log {\left (x \right )} \right )} + \frac {\left (- x^{2} + x e^{2} - e^{4}\right ) \log {\left (\log {\left (x \right )} \right )}}{x - e^{2}} \]

[In]

integrate(((2*exp(2)*x-x**2)*ln(x)*ln(ln(x))+((1+2*x)*exp(2)**2+(-4*x**2-2*x)*exp(2)+2*x**3+x**2)*ln(x)+exp(2)
*x-x**2)/(exp(2)**2-2*exp(2)*x+x**2)/ln(x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x^{3} - x^{2} {\left (e^{2} - 1\right )} - x^{2} \log \left (\log \left (x\right )\right ) - x e^{2}}{x - e^{2}} \]

[In]

integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((1+2*x)*exp(2)^2+(-4*x^2-2*x)*exp(2)+2*x^3+x^2)*log(x)+exp(2)*
x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/log(x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=\frac {x^{3} - x^{2} e^{2} - x^{2} \log \left (\log \left (x\right )\right ) + x^{2} - x e^{2}}{x - e^{2}} \]

[In]

integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((1+2*x)*exp(2)^2+(-4*x^2-2*x)*exp(2)+2*x^3+x^2)*log(x)+exp(2)*
x-x^2)/(exp(2)^2-2*exp(2)*x+x^2)/log(x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.89 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^2 x-x^2+\left (x^2+2 x^3+e^4 (1+2 x)+e^2 \left (-2 x-4 x^2\right )\right ) \log (x)+\left (2 e^2 x-x^2\right ) \log (x) \log (\log (x))}{\left (e^4-2 e^2 x+x^2\right ) \log (x)} \, dx=x+x^2-\frac {x^2\,\ln \left (\ln \left (x\right )\right )}{x-{\mathrm {e}}^2} \]

[In]

int((x*exp(2) - x^2 + log(x)*(x^2 - exp(2)*(2*x + 4*x^2) + 2*x^3 + exp(4)*(2*x + 1)) + log(log(x))*log(x)*(2*x
*exp(2) - x^2))/(log(x)*(exp(4) - 2*x*exp(2) + x^2)),x)

[Out]

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