Integrand size = 59, antiderivative size = 20 \[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=x^2 \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right ) \] Output:
ln(ln(1/4*x)+exp(exp(5)^2)-exp(1))*x^2
Time = 0.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=x^2 \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right ) \] Input:
Integrate[(x + (-2*E*x + 2*E^E^10*x + 2*x*Log[x/4])*Log[-E + E^E^10 + Log[ x/4]])/(-E + E^E^10 + Log[x/4]),x]
Output:
x^2*Log[-E + E^E^10 + Log[x/4]]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x+\left (2 e^{e^{10}} x-2 e x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (\log \left (\frac {x}{4}\right )+e^{e^{10}}-e\right )}{\log \left (\frac {x}{4}\right )+e^{e^{10}}-e} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x-\left (2 e^{e^{10}} x-2 e x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (\log \left (\frac {x}{4}\right )+e^{e^{10}}-e\right )}{e \left (1-e^{e^{10}-1}\right )-\log \left (\frac {x}{4}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int x \left (2 \log \left (\log \left (\frac {x}{4}\right )+e^{e^{10}}-e\right )+\frac {1}{\log \left (\frac {x}{4}\right )+e^{e^{10}}-e}\right )dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (2 x \log \left (\log \left (\frac {x}{4}\right )-e \left (1-e^{e^{10}-1}\right )\right )+\frac {x}{\log \left (\frac {x}{4}\right )-e \left (1-e^{e^{10}-1}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int x \log \left (\log \left (\frac {x}{4}\right )-e \left (1-e^{-1+e^{10}}\right )\right )dx+16 e^{2 e-2 e^{e^{10}}} \operatorname {ExpIntegralEi}\left (-2 \left (-\log \left (\frac {x}{4}\right )-e^{e^{10}}+e\right )\right )\) |
Input:
Int[(x + (-2*E*x + 2*E^E^10*x + 2*x*Log[x/4])*Log[-E + E^E^10 + Log[x/4]]) /(-E + E^E^10 + Log[x/4]),x]
Output:
$Aborted
Time = 0.66 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
method | result | size |
risch | \(x^{2} \ln \left (\ln \left (\frac {x}{4}\right )+{\mathrm e}^{{\mathrm e}^{10}}-{\mathrm e}\right )\) | \(18\) |
norman | \(x^{2} \ln \left (\ln \left (\frac {x}{4}\right )+{\mathrm e}^{{\mathrm e}^{10}}-{\mathrm e}\right )\) | \(20\) |
parallelrisch | \(x^{2} \ln \left (\ln \left (\frac {x}{4}\right )+{\mathrm e}^{{\mathrm e}^{10}}-{\mathrm e}\right )\) | \(20\) |
default | \(-{\mathrm e}^{4 \ln \left (2\right )-2 \,{\mathrm e}^{{\mathrm e}^{10}}+2 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (-2 \ln \left (x \right )+4 \ln \left (2\right )-2 \,{\mathrm e}^{{\mathrm e}^{10}}+2 \,{\mathrm e}\right )+x^{2} \ln \left (\ln \left (\frac {x}{4}\right )+{\mathrm e}^{{\mathrm e}^{10}}-{\mathrm e}\right )+16 \,{\mathrm e}^{-2 \,{\mathrm e}^{{\mathrm e}^{10}}+2 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (-2 \ln \left (\frac {x}{4}\right )-2 \,{\mathrm e}^{{\mathrm e}^{10}}+2 \,{\mathrm e}\right )\) | \(87\) |
Input:
int(((2*x*ln(1/4*x)+2*x*exp(exp(5)^2)-2*x*exp(1))*ln(ln(1/4*x)+exp(exp(5)^ 2)-exp(1))+x)/(ln(1/4*x)+exp(exp(5)^2)-exp(1)),x,method=_RETURNVERBOSE)
Output:
x^2*ln(ln(1/4*x)+exp(exp(10))-exp(1))
Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=x^{2} \log \left (-e + e^{\left (e^{10}\right )} + \log \left (\frac {1}{4} \, x\right )\right ) \] Input:
integrate(((2*x*log(1/4*x)+2*x*exp(exp(5)^2)-2*exp(1)*x)*log(log(1/4*x)+ex p(exp(5)^2)-exp(1))+x)/(log(1/4*x)+exp(exp(5)^2)-exp(1)),x, algorithm="fri cas")
Output:
x^2*log(-e + e^(e^10) + log(1/4*x))
Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=x^{2} \log {\left (\log {\left (\frac {x}{4} \right )} - e + e^{e^{10}} \right )} \] Input:
integrate(((2*x*ln(1/4*x)+2*x*exp(exp(5)**2)-2*exp(1)*x)*ln(ln(1/4*x)+exp( exp(5)**2)-exp(1))+x)/(ln(1/4*x)+exp(exp(5)**2)-exp(1)),x)
Output:
x**2*log(log(x/4) - E + exp(exp(10)))
\[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=\int { \frac {2 \, {\left (x e - x e^{\left (e^{10}\right )} - x \log \left (\frac {1}{4} \, x\right )\right )} \log \left (-e + e^{\left (e^{10}\right )} + \log \left (\frac {1}{4} \, x\right )\right ) - x}{e - e^{\left (e^{10}\right )} - \log \left (\frac {1}{4} \, x\right )} \,d x } \] Input:
integrate(((2*x*log(1/4*x)+2*x*exp(exp(5)^2)-2*exp(1)*x)*log(log(1/4*x)+ex p(exp(5)^2)-exp(1))+x)/(log(1/4*x)+exp(exp(5)^2)-exp(1)),x, algorithm="max ima")
Output:
x^2*log(-e + e^(e^10) - 2*log(2) + log(x)) - 16*e^(2*e - 2*e^(e^10))*exp_i ntegral_e(1, 2*e - 2*e^(e^10) - 2*log(1/4*x)) - integrate(-x/(e - e^(e^10) + 2*log(2) - log(x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).
Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75 \[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=\frac {1}{2} \, x^{2} \log \left (\frac {1}{4} \, \pi ^{2} {\left (\mathrm {sgn}\left (x\right ) - 1\right )}^{2} + {\left (e - e^{\left (e^{10}\right )} - \log \left (\frac {1}{4} \, {\left | x \right |}\right )\right )}^{2}\right ) \] Input:
integrate(((2*x*log(1/4*x)+2*x*exp(exp(5)^2)-2*exp(1)*x)*log(log(1/4*x)+ex p(exp(5)^2)-exp(1))+x)/(log(1/4*x)+exp(exp(5)^2)-exp(1)),x, algorithm="gia c")
Output:
1/2*x^2*log(1/4*pi^2*(sgn(x) - 1)^2 + (e - e^(e^10) - log(1/4*abs(x)))^2)
Time = 4.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=x^2\,\ln \left (\ln \left (\frac {x}{4}\right )-\mathrm {e}+{\mathrm {e}}^{{\mathrm {e}}^{10}}\right ) \] Input:
int((x + log(log(x/4) - exp(1) + exp(exp(10)))*(2*x*log(x/4) - 2*x*exp(1) + 2*x*exp(exp(10))))/(log(x/4) - exp(1) + exp(exp(10))),x)
Output:
x^2*log(log(x/4) - exp(1) + exp(exp(10)))
Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {x+\left (-2 e x+2 e^{e^{10}} x+2 x \log \left (\frac {x}{4}\right )\right ) \log \left (-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )\right )}{-e+e^{e^{10}}+\log \left (\frac {x}{4}\right )} \, dx=\mathrm {log}\left (e^{e^{10}}+\mathrm {log}\left (\frac {x}{4}\right )-e \right ) x^{2} \] Input:
int(((2*x*log(1/4*x)+2*x*exp(exp(5)^2)-2*exp(1)*x)*log(log(1/4*x)+exp(exp( 5)^2)-exp(1))+x)/(log(1/4*x)+exp(exp(5)^2)-exp(1)),x)
Output:
log(e**(e**10) + log(x/4) - e)*x**2