Integrand size = 167, antiderivative size = 26 \[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx=x+\log ^2\left (x+\frac {x}{x+e^{-x} \log (4)}\right )-\log (\log (x)) \] Output:
x-ln(ln(x))+ln(x+x/(exp(ln(2*ln(2))-x)+x))^2
\[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx=\int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx \] Input:
Integrate[(-x - x^2 + (x^2 + x^3)*Log[x] + (Log[4]^2*(-1 + x*Log[x]))/E^(2 *x) + (Log[4]*(-1 - 2*x + (x + 2*x^2)*Log[x]))/E^x + (2*x^2*Log[x] + ((2 + 6*x)*Log[4]*Log[x])/E^x + (2*Log[4]^2*Log[x])/E^(2*x))*Log[(x + x^2 + (x* Log[4])/E^x)/(x + Log[4]/E^x)])/((x^2 + x^3)*Log[x] + ((x + 2*x^2)*Log[4]* Log[x])/E^x + (x*Log[4]^2*Log[x])/E^(2*x)),x]
Output:
Integrate[(-x - x^2 + (x^2 + x^3)*Log[x] + (Log[4]^2*(-1 + x*Log[x]))/E^(2 *x) + (Log[4]*(-1 - 2*x + (x + 2*x^2)*Log[x]))/E^x + (2*x^2*Log[x] + ((2 + 6*x)*Log[4]*Log[x])/E^x + (2*Log[4]^2*Log[x])/E^(2*x))*Log[(x + x^2 + (x* Log[4])/E^x)/(x + Log[4]/E^x)])/((x^2 + x^3)*Log[x] + ((x + 2*x^2)*Log[4]* Log[x])/E^x + (x*Log[4]^2*Log[x])/E^(2*x)), x]
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.
| \(\displaystyle \int \frac {-x^2+\left (2 x^2 \log (x)+2 e^{-2 x} \log ^2(4) \log (x)+e^{-x} (6 x+2) \log (4) \log (x)\right ) \log \left (\frac {x^2+x+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )+e^{-x} \log (4) \left (\left (2 x^2+x\right ) \log (x)-2 x-1\right )+\left (x^3+x^2\right ) \log (x)-x+e^{-2 x} \log ^2(4) (x \log (x)-1)}{e^{-x} \left (2 x^2+x\right ) \log (4) \log (x)+\left (x^3+x^2\right ) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x^2 \log (x)-x+x \log (x)+2 x \log (x) \log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right )-1}{x (x+1) \log (x)}+\frac {2 (x+1) \log (4) \log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right )}{x \left (e^x x+\log (4)\right )}-\frac {2 (x+2) \log (4) \log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right )}{(x+1) \left (e^x x+e^x+\log (4)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle x-\log (\log (x))+2 \log (4) \log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right ) \int \frac {1}{e^x x+\log (4)}dx+2 \log (4) \log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right ) \int \frac {1}{x \left (e^x x+\log (4)\right )}dx-2 \log (4) \log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right ) \int \frac {1}{e^x x+e^x+\log (4)}dx-2 \log (4) \log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right ) \int \frac {1}{(x+1) \left (e^x x+e^x+\log (4)\right )}dx+2 \int \frac {\log \left (\frac {x \left (e^x (x+1)+\log (4)\right )}{e^x x+\log (4)}\right )}{x+1}dx-2 \log (4) \int \frac {\int \frac {1}{e^x x+\log (4)}dx}{x+1}dx-2 \log ^2(4) \int \frac {\int \frac {1}{e^x x+\log (4)}dx}{e^x x+\log (4)}dx-2 \log ^2(4) \int \frac {\int \frac {1}{e^x x+\log (4)}dx}{x \left (e^x x+\log (4)\right )}dx+2 \log ^2(4) \int \frac {\int \frac {1}{e^x x+\log (4)}dx}{e^x x+e^x+\log (4)}dx+2 \log ^2(4) \int \frac {\int \frac {1}{e^x x+\log (4)}dx}{(x+1) \left (e^x x+e^x+\log (4)\right )}dx+2 \log (4) \int \frac {\int \frac {1}{e^x (x+1)+\log (4)}dx}{x+1}dx+2 \log ^2(4) \int \frac {\int \frac {1}{e^x (x+1)+\log (4)}dx}{e^x x+\log (4)}dx+2 \log ^2(4) \int \frac {\int \frac {1}{e^x (x+1)+\log (4)}dx}{x \left (e^x x+\log (4)\right )}dx-2 \log ^2(4) \int \frac {\int \frac {1}{e^x (x+1)+\log (4)}dx}{e^x x+e^x+\log (4)}dx-2 \log ^2(4) \int \frac {\int \frac {1}{e^x (x+1)+\log (4)}dx}{(x+1) \left (e^x x+e^x+\log (4)\right )}dx+2 \log (4) \int \frac {\int \frac {1}{(x+1) \left (e^x (x+1)+\log (4)\right )}dx}{x+1}dx+2 \log ^2(4) \int \frac {\int \frac {1}{(x+1) \left (e^x (x+1)+\log (4)\right )}dx}{e^x x+\log (4)}dx+2 \log ^2(4) \int \frac {\int \frac {1}{(x+1) \left (e^x (x+1)+\log (4)\right )}dx}{x \left (e^x x+\log (4)\right )}dx-2 \log ^2(4) \int \frac {\int \frac {1}{(x+1) \left (e^x (x+1)+\log (4)\right )}dx}{e^x x+e^x+\log (4)}dx-2 \log ^2(4) \int \frac {\int \frac {1}{(x+1) \left (e^x (x+1)+\log (4)\right )}dx}{(x+1) \left (e^x x+e^x+\log (4)\right )}dx-2 \log (4) \int \frac {\int \frac {1}{e^x x^2+\log (4) x}dx}{x+1}dx-2 \log ^2(4) \int \frac {\int \frac {1}{e^x x^2+\log (4) x}dx}{e^x x+\log (4)}dx-2 \log ^2(4) \int \frac {\int \frac {1}{e^x x^2+\log (4) x}dx}{x \left (e^x x+\log (4)\right )}dx+2 \log ^2(4) \int \frac {\int \frac {1}{e^x x^2+\log (4) x}dx}{e^x x+e^x+\log (4)}dx+2 \log ^2(4) \int \frac {\int \frac {1}{e^x x^2+\log (4) x}dx}{(x+1) \left (e^x x+e^x+\log (4)\right )}dx\) |
Input:
Int[(-x - x^2 + (x^2 + x^3)*Log[x] + (Log[4]^2*(-1 + x*Log[x]))/E^(2*x) + (Log[4]*(-1 - 2*x + (x + 2*x^2)*Log[x]))/E^x + (2*x^2*Log[x] + ((2 + 6*x)* Log[4]*Log[x])/E^x + (2*Log[4]^2*Log[x])/E^(2*x))*Log[(x + x^2 + (x*Log[4] )/E^x)/(x + Log[4]/E^x)])/((x^2 + x^3)*Log[x] + ((x + 2*x^2)*Log[4]*Log[x] )/E^x + (x*Log[4]^2*Log[x])/E^(2*x)),x]
Output:
$Aborted
Time = 3.55 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54
| method | result | size |
| parallelrisch | \({\ln \left (\frac {x \left ({\mathrm e}^{\ln \left (2 \ln \left (2\right )\right )-x}+x +1\right )}{{\mathrm e}^{\ln \left (2 \ln \left (2\right )\right )-x}+x}\right )}^{2}-\ln \left (\ln \left (x \right )\right )+x\) | \(40\) |
| risch | \(\text {Expression too large to display}\) | \(1475\) |
Input:
int(((2*ln(x)*exp(ln(2*ln(2))-x)^2+(6*x+2)*ln(x)*exp(ln(2*ln(2))-x)+2*x^2* ln(x))*ln((x*exp(ln(2*ln(2))-x)+x^2+x)/(exp(ln(2*ln(2))-x)+x))+(x*ln(x)-1) *exp(ln(2*ln(2))-x)^2+((2*x^2+x)*ln(x)-2*x-1)*exp(ln(2*ln(2))-x)+(x^3+x^2) *ln(x)-x^2-x)/(x*ln(x)*exp(ln(2*ln(2))-x)^2+(2*x^2+x)*ln(x)*exp(ln(2*ln(2) )-x)+(x^3+x^2)*ln(x)),x,method=_RETURNVERBOSE)
Output:
ln(x*(exp(ln(2*ln(2))-x)+x+1)/(exp(ln(2*ln(2))-x)+x))^2-ln(ln(x))+x
Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx=\log \left (\frac {x^{2} + x e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} + x}{x + e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )}}\right )^{2} + x - \log \left (\log \left (x\right )\right ) \] Input:
integrate(((2*log(x)*exp(log(2*log(2))-x)^2+(6*x+2)*log(x)*exp(log(2*log(2 ))-x)+2*x^2*log(x))*log((x*exp(log(2*log(2))-x)+x^2+x)/(exp(log(2*log(2))- x)+x))+(x*log(x)-1)*exp(log(2*log(2))-x)^2+((2*x^2+x)*log(x)-2*x-1)*exp(lo g(2*log(2))-x)+(x^3+x^2)*log(x)-x^2-x)/(x*log(x)*exp(log(2*log(2))-x)^2+(2 *x^2+x)*log(x)*exp(log(2*log(2))-x)+(x^3+x^2)*log(x)),x, algorithm="fricas ")
Output:
log((x^2 + x*e^(-x + log(2*log(2))) + x)/(x + e^(-x + log(2*log(2)))))^2 + x - log(log(x))
Time = 0.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx=x + \log {\left (\frac {x^{2} + x + 2 x e^{- x} \log {\left (2 \right )}}{x + 2 e^{- x} \log {\left (2 \right )}} \right )}^{2} - \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate(((2*ln(x)*exp(ln(2*ln(2))-x)**2+(6*x+2)*ln(x)*exp(ln(2*ln(2))-x) +2*x**2*ln(x))*ln((x*exp(ln(2*ln(2))-x)+x**2+x)/(exp(ln(2*ln(2))-x)+x))+(x *ln(x)-1)*exp(ln(2*ln(2))-x)**2+((2*x**2+x)*ln(x)-2*x-1)*exp(ln(2*ln(2))-x )+(x**3+x**2)*ln(x)-x**2-x)/(x*ln(x)*exp(ln(2*ln(2))-x)**2+(2*x**2+x)*ln(x )*exp(ln(2*ln(2))-x)+(x**3+x**2)*ln(x)),x)
Output:
x + log((x**2 + x + 2*x*exp(-x)*log(2))/(x + 2*exp(-x)*log(2)))**2 - log(l og(x))
\[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx=\int { -\frac {x^{2} - {\left ({\left (2 \, x^{2} + x\right )} \log \left (x\right ) - 2 \, x - 1\right )} e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} - {\left (x \log \left (x\right ) - 1\right )} e^{\left (-2 \, x + 2 \, \log \left (2 \, \log \left (2\right )\right )\right )} - {\left (x^{3} + x^{2}\right )} \log \left (x\right ) - 2 \, {\left (x^{2} \log \left (x\right ) + {\left (3 \, x + 1\right )} e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right ) + e^{\left (-2 \, x + 2 \, \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right )\right )} \log \left (\frac {x^{2} + x e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} + x}{x + e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )}}\right ) + x}{{\left (2 \, x^{2} + x\right )} e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right ) + x e^{\left (-2 \, x + 2 \, \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right ) + {\left (x^{3} + x^{2}\right )} \log \left (x\right )} \,d x } \] Input:
integrate(((2*log(x)*exp(log(2*log(2))-x)^2+(6*x+2)*log(x)*exp(log(2*log(2 ))-x)+2*x^2*log(x))*log((x*exp(log(2*log(2))-x)+x^2+x)/(exp(log(2*log(2))- x)+x))+(x*log(x)-1)*exp(log(2*log(2))-x)^2+((2*x^2+x)*log(x)-2*x-1)*exp(lo g(2*log(2))-x)+(x^3+x^2)*log(x)-x^2-x)/(x*log(x)*exp(log(2*log(2))-x)^2+(2 *x^2+x)*log(x)*exp(log(2*log(2))-x)+(x^3+x^2)*log(x)),x, algorithm="maxima ")
Output:
-integrate(-(4*(x*log(x) - 1)*e^(-2*x)*log(2)^2 + 2*((2*x^2 + x)*log(x) - 2*x - 1)*e^(-x)*log(2) - x^2 + (x^3 + x^2)*log(x) + 2*(2*(3*x + 1)*e^(-x)* log(2)*log(x) + 4*e^(-2*x)*log(2)^2*log(x) + x^2*log(x))*log((2*x*e^(-x)*l og(2) + x^2 + x)/(2*e^(-x)*log(2) + x)) - x)/(4*x*e^(-2*x)*log(2)^2*log(x) + 2*(2*x^2 + x)*e^(-x)*log(2)*log(x) + (x^3 + x^2)*log(x)), x)
\[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx=\int { -\frac {x^{2} - {\left ({\left (2 \, x^{2} + x\right )} \log \left (x\right ) - 2 \, x - 1\right )} e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} - {\left (x \log \left (x\right ) - 1\right )} e^{\left (-2 \, x + 2 \, \log \left (2 \, \log \left (2\right )\right )\right )} - {\left (x^{3} + x^{2}\right )} \log \left (x\right ) - 2 \, {\left (x^{2} \log \left (x\right ) + {\left (3 \, x + 1\right )} e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right ) + e^{\left (-2 \, x + 2 \, \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right )\right )} \log \left (\frac {x^{2} + x e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} + x}{x + e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )}}\right ) + x}{{\left (2 \, x^{2} + x\right )} e^{\left (-x + \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right ) + x e^{\left (-2 \, x + 2 \, \log \left (2 \, \log \left (2\right )\right )\right )} \log \left (x\right ) + {\left (x^{3} + x^{2}\right )} \log \left (x\right )} \,d x } \] Input:
integrate(((2*log(x)*exp(log(2*log(2))-x)^2+(6*x+2)*log(x)*exp(log(2*log(2 ))-x)+2*x^2*log(x))*log((x*exp(log(2*log(2))-x)+x^2+x)/(exp(log(2*log(2))- x)+x))+(x*log(x)-1)*exp(log(2*log(2))-x)^2+((2*x^2+x)*log(x)-2*x-1)*exp(lo g(2*log(2))-x)+(x^3+x^2)*log(x)-x^2-x)/(x*log(x)*exp(log(2*log(2))-x)^2+(2 *x^2+x)*log(x)*exp(log(2*log(2))-x)+(x^3+x^2)*log(x)),x, algorithm="giac")
Output:
integrate(-(x^2 - ((2*x^2 + x)*log(x) - 2*x - 1)*e^(-x + log(2*log(2))) - (x*log(x) - 1)*e^(-2*x + 2*log(2*log(2))) - (x^3 + x^2)*log(x) - 2*(x^2*lo g(x) + (3*x + 1)*e^(-x + log(2*log(2)))*log(x) + e^(-2*x + 2*log(2*log(2)) )*log(x))*log((x^2 + x*e^(-x + log(2*log(2))) + x)/(x + e^(-x + log(2*log( 2))))) + x)/((2*x^2 + x)*e^(-x + log(2*log(2)))*log(x) + x*e^(-2*x + 2*log (2*log(2)))*log(x) + (x^3 + x^2)*log(x)), x)
Time = 3.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx={\ln \left (\frac {x^2\,{\mathrm {e}}^x+2\,x\,\ln \left (2\right )+x\,{\mathrm {e}}^x}{2\,\ln \left (2\right )+x\,{\mathrm {e}}^x}\right )}^2+x-\ln \left (\ln \left (x\right )\right ) \] Input:
int(-(x - log(x)*(x^2 + x^3) - exp(2*log(2*log(2)) - 2*x)*(x*log(x) - 1) + exp(log(2*log(2)) - x)*(2*x - log(x)*(x + 2*x^2) + 1) - log((x + x*exp(lo g(2*log(2)) - x) + x^2)/(x + exp(log(2*log(2)) - x)))*(2*x^2*log(x) + 2*ex p(2*log(2*log(2)) - 2*x)*log(x) + exp(log(2*log(2)) - x)*log(x)*(6*x + 2)) + x^2)/(log(x)*(x^2 + x^3) + exp(log(2*log(2)) - x)*log(x)*(x + 2*x^2) + x*exp(2*log(2*log(2)) - 2*x)*log(x)),x)
Output:
x - log(log(x)) + log((x^2*exp(x) + 2*x*log(2) + x*exp(x))/(2*log(2) + x*e xp(x)))^2
Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-x-x^2+\left (x^2+x^3\right ) \log (x)+e^{-2 x} \log ^2(4) (-1+x \log (x))+e^{-x} \log (4) \left (-1-2 x+\left (x+2 x^2\right ) \log (x)\right )+\left (2 x^2 \log (x)+e^{-x} (2+6 x) \log (4) \log (x)+2 e^{-2 x} \log ^2(4) \log (x)\right ) \log \left (\frac {x+x^2+e^{-x} x \log (4)}{x+e^{-x} \log (4)}\right )}{\left (x^2+x^3\right ) \log (x)+e^{-x} \left (x+2 x^2\right ) \log (4) \log (x)+e^{-2 x} x \log ^2(4) \log (x)} \, dx=-\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+\mathrm {log}\left (\frac {e^{x} x^{2}+e^{x} x +2 \,\mathrm {log}\left (2\right ) x}{e^{x} x +2 \,\mathrm {log}\left (2\right )}\right )^{2}+x \] Input:
int(((2*log(x)*exp(log(2*log(2))-x)^2+(6*x+2)*log(x)*exp(log(2*log(2))-x)+ 2*x^2*log(x))*log((x*exp(log(2*log(2))-x)+x^2+x)/(exp(log(2*log(2))-x)+x)) +(x*log(x)-1)*exp(log(2*log(2))-x)^2+((2*x^2+x)*log(x)-2*x-1)*exp(log(2*lo g(2))-x)+(x^3+x^2)*log(x)-x^2-x)/(x*log(x)*exp(log(2*log(2))-x)^2+(2*x^2+x )*log(x)*exp(log(2*log(2))-x)+(x^3+x^2)*log(x)),x)
Output:
- log(log(x)) + log((e**x*x**2 + e**x*x + 2*log(2)*x)/(e**x*x + 2*log(2)) )**2 + x