3.101.11 \(\int \frac {e^x x-x^2+(e^x (-3+x)+3 x-x^2) \log (x)+(x^2-e^x x^2) \log (x) \log (-10 e^{3/x} x \log (x)) \log (\log (-10 e^{3/x} x \log (x)))}{(3 e^{2 x} x^2-6 e^x x^3+3 x^4) \log (x) \log (-10 e^{3/x} x \log (x))} \, dx\)

Optimal. Leaf size=27 \[ \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{3 \left (e^x-x\right )} \]

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Rubi [F]  time = 4.84, 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^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (3 e^{2 x} x^2-6 e^x x^3+3 x^4\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*x - x^2 + (E^x*(-3 + x) + 3*x - x^2)*Log[x] + (x^2 - E^x*x^2)*Log[x]*Log[-10*E^(3/x)*x*Log[x]]*Log[Lo
g[-10*E^(3/x)*x*Log[x]]])/((3*E^(2*x)*x^2 - 6*E^x*x^3 + 3*x^4)*Log[x]*Log[-10*E^(3/x)*x*Log[x]]),x]

[Out]

-Defer[Int][1/((E^x - x)*x^2*Log[-10*E^(3/x)*x*Log[x]]), x] + Defer[Int][1/((E^x - x)*x*Log[-10*E^(3/x)*x*Log[
x]]), x]/3 + Defer[Int][1/((E^x - x)*x*Log[x]*Log[-10*E^(3/x)*x*Log[x]]), x]/3 + Defer[Int][Log[Log[-10*E^(3/x
)*x*Log[x]]]/(E^x - x)^2, x]/3 - Defer[Int][Log[Log[-10*E^(3/x)*x*Log[x]]]/(E^x - x), x]/3 - Defer[Int][(x*Log
[Log[-10*E^(3/x)*x*Log[x]]])/(E^x - x)^2, x]/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{3 \left (e^x-x\right )^2 x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx\\ &=\frac {1}{3} \int \frac {e^x x-x^2+\left (e^x (-3+x)+3 x-x^2\right ) \log (x)+\left (x^2-e^x x^2\right ) \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2 x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx\\ &=\frac {1}{3} \int \left (-\frac {(-1+x) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}-\frac {-x+3 \log (x)-x \log (x)+x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right ) x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {(-1+x) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2} \, dx\right )-\frac {1}{3} \int \frac {-x+3 \log (x)-x \log (x)+x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right ) \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right ) x^2 \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx\\ &=-\left (\frac {1}{3} \int \left (\frac {3}{\left (e^x-x\right ) x^2 \log \left (-10 e^{3/x} x \log (x)\right )}-\frac {1}{\left (e^x-x\right ) x \log \left (-10 e^{3/x} x \log (x)\right )}-\frac {1}{\left (e^x-x\right ) x \log (x) \log \left (-10 e^{3/x} x \log (x)\right )}+\frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{e^x-x}\right ) \, dx\right )-\frac {1}{3} \int \left (-\frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}+\frac {x \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {1}{\left (e^x-x\right ) x \log \left (-10 e^{3/x} x \log (x)\right )} \, dx+\frac {1}{3} \int \frac {1}{\left (e^x-x\right ) x \log (x) \log \left (-10 e^{3/x} x \log (x)\right )} \, dx+\frac {1}{3} \int \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2} \, dx-\frac {1}{3} \int \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{e^x-x} \, dx-\frac {1}{3} \int \frac {x \log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{\left (e^x-x\right )^2} \, dx-\int \frac {1}{\left (e^x-x\right ) x^2 \log \left (-10 e^{3/x} x \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 27, normalized size = 1.00 \begin {gather*} \frac {\log \left (\log \left (-10 e^{3/x} x \log (x)\right )\right )}{3 \left (e^x-x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*x - x^2 + (E^x*(-3 + x) + 3*x - x^2)*Log[x] + (x^2 - E^x*x^2)*Log[x]*Log[-10*E^(3/x)*x*Log[x]]*
Log[Log[-10*E^(3/x)*x*Log[x]]])/((3*E^(2*x)*x^2 - 6*E^x*x^3 + 3*x^4)*Log[x]*Log[-10*E^(3/x)*x*Log[x]]),x]

[Out]

Log[Log[-10*E^(3/x)*x*Log[x]]]/(3*(E^x - x))

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fricas [A]  time = 0.63, size = 23, normalized size = 0.85 \begin {gather*} -\frac {\log \left (\log \left (-10 \, x e^{\frac {3}{x}} \log \relax (x)\right )\right )}{3 \, {\left (x - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+x^2)*log(x)*log(-10*x*exp(3/x)*log(x))*log(log(-10*x*exp(3/x)*log(x)))+((x-3)*exp(x)-x
^2+3*x)*log(x)+exp(x)*x-x^2)/(3*exp(x)^2*x^2-6*exp(x)*x^3+3*x^4)/log(x)/log(-10*x*exp(3/x)*log(x)),x, algorith
m="fricas")

[Out]

-1/3*log(log(-10*x*e^(3/x)*log(x)))/(x - e^x)

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giac [A]  time = 0.15, size = 29, normalized size = 1.07 \begin {gather*} -\frac {\log \left (x \log \relax (x) + x \log \left (-10 \, \log \relax (x)\right ) + 3\right ) - \log \relax (x)}{3 \, {\left (x - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+x^2)*log(x)*log(-10*x*exp(3/x)*log(x))*log(log(-10*x*exp(3/x)*log(x)))+((x-3)*exp(x)-x
^2+3*x)*log(x)+exp(x)*x-x^2)/(3*exp(x)^2*x^2-6*exp(x)*x^3+3*x^4)/log(x)/log(-10*x*exp(3/x)*log(x)),x, algorith
m="giac")

[Out]

-1/3*(log(x*log(x) + x*log(-10*log(x)) + 3) - log(x))/(x - e^x)

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maple [C]  time = 0.26, size = 195, normalized size = 7.22

method	result	size
risch	\(-\frac {\ln \left (\ln \relax (2)+\ln \relax (5)+i \pi +\ln \relax (x )+\ln \left ({\mathrm e}^{\frac {3}{x}}\right )+\ln \left (\ln \relax (x )\right )-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right )+\mathrm {csgn}\left (i {\mathrm e}^{\frac {3}{x}}\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right )+\mathrm {csgn}\left (i \ln \relax (x )\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right )+\mathrm {csgn}\left (i {\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right )\right )}{2}+i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right )^{2} \left (\mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {3}{x}} \ln \relax (x )\right )-1\right )\right )}{3 \left (x -{\mathrm e}^{x}\right )}\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x^2+x^2)*ln(x)*ln(-10*x*exp(3/x)*ln(x))*ln(ln(-10*x*exp(3/x)*ln(x)))+((x-3)*exp(x)-x^2+3*x)*ln(x
)+exp(x)*x-x^2)/(3*exp(x)^2*x^2-6*exp(x)*x^3+3*x^4)/ln(x)/ln(-10*x*exp(3/x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-1/3/(x-exp(x))*ln(ln(2)+ln(5)+I*Pi+ln(x)+ln(exp(3/x))+ln(ln(x))-1/2*I*Pi*csgn(I*exp(3/x)*ln(x))*(-csgn(I*exp(
3/x)*ln(x))+csgn(I*exp(3/x)))*(-csgn(I*exp(3/x)*ln(x))+csgn(I*ln(x)))-1/2*I*Pi*csgn(I*x*exp(3/x)*ln(x))*(-csgn
(I*x*exp(3/x)*ln(x))+csgn(I*x))*(-csgn(I*x*exp(3/x)*ln(x))+csgn(I*exp(3/x)*ln(x)))+I*Pi*csgn(I*x*exp(3/x)*ln(x
))^2*(csgn(I*x*exp(3/x)*ln(x))-1))

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maxima [A]  time = 0.55, size = 36, normalized size = 1.33 \begin {gather*} -\frac {\log \left (x {\left (\log \relax (5) + \log \relax (2)\right )} + x \log \relax (x) + x \log \left (-\log \relax (x)\right ) + 3\right ) - \log \relax (x)}{3 \, {\left (x - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+x^2)*log(x)*log(-10*x*exp(3/x)*log(x))*log(log(-10*x*exp(3/x)*log(x)))+((x-3)*exp(x)-x
^2+3*x)*log(x)+exp(x)*x-x^2)/(3*exp(x)^2*x^2-6*exp(x)*x^3+3*x^4)/log(x)/log(-10*x*exp(3/x)*log(x)),x, algorith
m="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x\,{\mathrm {e}}^x+\ln \relax (x)\,\left (3\,x+{\mathrm {e}}^x\,\left (x-3\right )-x^2\right )-x^2-\ln \left (\ln \left (-10\,x\,{\mathrm {e}}^{3/x}\,\ln \relax (x)\right )\right )\,\ln \left (-10\,x\,{\mathrm {e}}^{3/x}\,\ln \relax (x)\right )\,\ln \relax (x)\,\left (x^2\,{\mathrm {e}}^x-x^2\right )}{\ln \left (-10\,x\,{\mathrm {e}}^{3/x}\,\ln \relax (x)\right )\,\ln \relax (x)\,\left (3\,x^2\,{\mathrm {e}}^{2\,x}-6\,x^3\,{\mathrm {e}}^x+3\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(x) + log(x)*(3*x + exp(x)*(x - 3) - x^2) - x^2 - log(log(-10*x*exp(3/x)*log(x)))*log(-10*x*exp(3/x)
*log(x))*log(x)*(x^2*exp(x) - x^2))/(log(-10*x*exp(3/x)*log(x))*log(x)*(3*x^2*exp(2*x) - 6*x^3*exp(x) + 3*x^4)
),x)

[Out]

int((x*exp(x) + log(x)*(3*x + exp(x)*(x - 3) - x^2) - x^2 - log(log(-10*x*exp(3/x)*log(x)))*log(-10*x*exp(3/x)
*log(x))*log(x)*(x^2*exp(x) - x^2))/(log(-10*x*exp(3/x)*log(x))*log(x)*(3*x^2*exp(2*x) - 6*x^3*exp(x) + 3*x^4)
), x)

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sympy [A]  time = 2.83, size = 26, normalized size = 0.96 \begin {gather*} - \frac {\log {\left (\log {\left (- 10 x e^{\frac {3}{x}} \log {\relax (x )} \right )} \right )}}{3 x - 3 e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x**2+x**2)*ln(x)*ln(-10*x*exp(3/x)*ln(x))*ln(ln(-10*x*exp(3/x)*ln(x)))+((x-3)*exp(x)-x**2+
3*x)*ln(x)+exp(x)*x-x**2)/(3*exp(x)**2*x**2-6*exp(x)*x**3+3*x**4)/ln(x)/ln(-10*x*exp(3/x)*ln(x)),x)

[Out]

-log(log(-10*x*exp(3/x)*log(x)))/(3*x - 3*exp(x))

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