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

Optimal. Leaf size=26 \[ x \log \left (\frac {\log \left (x-\frac {x^3}{\left (-e^x+x\right )^2}\right )}{\log (\log (x))}\right ) \]

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Rubi [A]  time = 6.59, antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 17, number of rules used = 3, integrand size = 245, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6688, 6742, 2549} \begin {gather*} x \log \left (\frac {\log \left (\frac {e^x \left (e^x-2 x\right ) x}{\left (e^x-x\right )^2}\right )}{\log (\log (x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.74, size = 31, normalized size = 1.19 \begin {gather*} x \log \left (\frac {\log \left (\frac {e^x \left (e^x-2 x\right ) x}{\left (e^x-x\right )^2}\right )}{\log (\log (x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.28, size = 42, normalized size = 1.62 \begin {gather*} x \log \left (\frac {\log \left (-\frac {2 \, x^{2} e^{x} - x e^{\left (2 \, x\right )}}{x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}\right )}{\log \left (\log \relax (x)\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Simplification assuming t_nos
tep near 0S

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left ({\mathrm e}^{2 x}-3 \,{\mathrm e}^{x} x +2 x^{2}\right ) \ln \relax (x ) \ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right ) \ln \left (\ln \relax (x )\right ) \ln \left (\frac {\ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right )}{\ln \left (\ln \relax (x )\right )}\right )+\left ({\mathrm e}^{2 x}-3 \,{\mathrm e}^{x} x +2 x^{3}\right ) \ln \relax (x ) \ln \left (\ln \relax (x )\right )+\left (-{\mathrm e}^{2 x}+3 \,{\mathrm e}^{x} x -2 x^{2}\right ) \ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right )}{\left ({\mathrm e}^{2 x}-3 \,{\mathrm e}^{x} x +2 x^{2}\right ) \ln \relax (x ) \ln \left (\frac {x \,{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x^{2}}{{\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}}\right ) \ln \left (\ln \relax (x )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.64, size = 31, normalized size = 1.19 \begin {gather*} x \log \left (x + \log \relax (x) - 2 \, \log \left (-x + e^{x}\right ) + \log \left (-2 \, x + e^{x}\right )\right ) - x \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.54, size = 40, normalized size = 1.54 \begin {gather*} x\,\ln \left (\frac {\ln \left (\frac {x\,{\mathrm {e}}^{2\,x}-2\,x^2\,{\mathrm {e}}^x}{{\mathrm {e}}^{2\,x}-2\,x\,{\mathrm {e}}^x+x^2}\right )}{\ln \left (\ln \relax (x)\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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