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

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

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Rubi [F]  time = 0.62, 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^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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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))+\text {li}(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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 30, normalized size = 1.43 \begin {gather*} \frac {x \left (-\left (\left (e^2-x\right ) (1+x)\right )-x \log (\log (x))\right )}{-e^2+x} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.75, size = 33, normalized size = 1.57 \begin {gather*} \frac {x^{3} - x^{2} \log \left (\log \relax (x)\right ) + x^{2} - {\left (x^{2} + x\right )} e^{2}}{x - e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((2*x+1)*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)

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giac [A]  time = 0.14, size = 36, normalized size = 1.71 \begin {gather*} \frac {x^{3} - x^{2} e^{2} - x^{2} \log \left (\log \relax (x)\right ) + x^{2} - x e^{2}}{x - e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((2*x+1)*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)

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maple [A]  time = 0.35, size = 36, normalized size = 1.71

method	result	size
risch	\(\frac {\left ({\mathrm e}^{4}-{\mathrm e}^{2} x +x^{2}\right ) \ln \left (\ln \relax (x )\right )}{{\mathrm e}^{2}-x}+x^{2}+x -{\mathrm e}^{2} \ln \left (\ln \relax (x )\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(2)*x-x^2)*ln(x)*ln(ln(x))+((2*x+1)*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]

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

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maxima [A]  time = 0.40, size = 35, normalized size = 1.67 \begin {gather*} \frac {x^{3} - x^{2} {\left (e^{2} - 1\right )} - x^{2} \log \left (\log \relax (x)\right ) - x e^{2}}{x - e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-x^2)*log(x)*log(log(x))+((2*x+1)*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)

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mupad [B]  time = 5.33, size = 21, normalized size = 1.00 \begin {gather*} x+x^2-\frac {x^2\,\ln \left (\ln \relax (x)\right )}{x-{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.47, size = 34, normalized size = 1.62 \begin {gather*} x^{2} + x - e^{2} \log {\left (\log {\relax (x )} \right )} + \frac {\left (- x^{2} + x e^{2} - e^{4}\right ) \log {\left (\log {\relax (x )} \right )}}{x - e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(2)*x-x**2)*ln(x)*ln(ln(x))+((2*x+1)*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))

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