Integrand size = 101, antiderivative size = 30 \begin {dmath*} \int \frac {\left (1+x+e^{4-x} \left (-x-x^2\right )\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )+\log \left (e^x x\right ) \left (-1+x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )\right )}{-1+e^{4-x} x} \, dx=\frac {x-x^2 \log \left (1-\frac {e^{-4+x}}{x}\right ) \log \left (e^x x\right )}{x} \end {dmath*}
Time = 0.80 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \begin {dmath*} \int \frac {\left (1+x+e^{4-x} \left (-x-x^2\right )\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )+\log \left (e^x x\right ) \left (-1+x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )\right )}{-1+e^{4-x} x} \, dx=-x \log \left (1-\frac {e^{-4+x}}{x}\right ) \log \left (e^x x\right ) \end {dmath*}
Integrate[((1 + x + E^(4 - x)*(-x - x^2))*Log[(E^(-4 + x)*(-1 + E^(4 - x)* x))/x] + Log[E^x*x]*(-1 + x + (1 - E^(4 - x)*x)*Log[(E^(-4 + x)*(-1 + E^(4 - x)*x))/x]))/(-1 + E^(4 - x)*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.
\(\displaystyle \int \frac {\left (e^{4-x} \left (-x^2-x\right )+x+1\right ) \log \left (\frac {e^{x-4} \left (e^{4-x} x-1\right )}{x}\right )+\log \left (e^x x\right ) \left (x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{x-4} \left (e^{4-x} x-1\right )}{x}\right )-1\right )}{e^{4-x} x-1} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-x \log \left (1-\frac {e^{x-4}}{x}\right )-\log \left (e^x x\right ) \log \left (1-\frac {e^{x-4}}{x}\right )-\log \left (1-\frac {e^{x-4}}{x}\right )-x \log \left (e^x x\right )+\frac {e^4 (x-1) x \log \left (e^x x\right )}{e^4 x-e^x}+\log \left (e^x x\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -e^4 \int \int \frac {x^2}{e^4 x-e^x}dxdx-e^4 \int \frac {\int \frac {x^2}{e^4 x-e^x}dx}{x}dx+e^4 \log \left (e^x x\right ) \int \frac {x^2}{e^4 x-e^x}dx-\frac {1}{2} \int \frac {e^x}{e^x-e^4 x}dx-\frac {1}{2} \int \frac {e^x}{x \left (e^4 x-e^x\right )}dx+\int \int \frac {e^x}{e^x-e^4 x}dxdx+\int \frac {\int \frac {e^x}{e^x-e^4 x}dx}{x}dx+e^4 \int \int \frac {x}{e^4 x-e^x}dxdx+e^4 \int \frac {\int \frac {x}{e^4 x-e^x}dx}{x}dx+\int \int \frac {e^x x}{e^4 x-e^x}dxdx+\int \frac {\int \frac {e^x x}{e^4 x-e^x}dx}{x}dx-\log \left (e^x x\right ) \int \frac {e^x}{e^x-e^4 x}dx-e^4 \log \left (e^x x\right ) \int \frac {x}{e^4 x-e^x}dx-\log \left (e^x x\right ) \int \frac {e^x x}{e^4 x-e^x}dx+\frac {x^3}{6}+\frac {x^2}{4}-\frac {1}{2} x^2 \log \left (1-\frac {e^{x-4}}{x}\right )-\frac {1}{2} x^2 \log \left (e^x x\right )-\frac {1}{2} (x+1)^2-x \log \left (1-\frac {e^{x-4}}{x}\right )-x \log \left (1-\frac {e^{x-4}}{x}\right ) \log \left (e^x x\right )+x \log \left (e^x x\right )+\frac {1}{2} (x+1)^2 \log \left (1-\frac {e^{x-4}}{x}\right )\) |
Int[((1 + x + E^(4 - x)*(-x - x^2))*Log[(E^(-4 + x)*(-1 + E^(4 - x)*x))/x] + Log[E^x*x]*(-1 + x + (1 - E^(4 - x)*x)*Log[(E^(-4 + x)*(-1 + E^(4 - x)* x))/x]))/(-1 + E^(4 - x)*x),x]
3.6.6.3.1 Defintions of rubi rules used
Time = 2.65 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\ln \left (\frac {\left (x \,{\mathrm e}^{-x +4}-1\right ) {\mathrm e}^{x -4}}{x}\right ) x \ln \left ({\mathrm e}^{x} x \right )\) | \(32\) |
risch | \(\text {Expression too large to display}\) | \(1357\) |
int((((-x*exp(-x+4)+1)*ln((x*exp(-x+4)-1)/x/exp(-x+4))+x-1)*ln(exp(x)*x)+( (-x^2-x)*exp(-x+4)+x+1)*ln((x*exp(-x+4)-1)/x/exp(-x+4)))/(x*exp(-x+4)-1),x ,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \begin {dmath*} \int \frac {\left (1+x+e^{4-x} \left (-x-x^2\right )\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )+\log \left (e^x x\right ) \left (-1+x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )\right )}{-1+e^{4-x} x} \, dx=-x \log \left (x e^{x}\right ) \log \left (\frac {{\left (x e^{4} - e^{x}\right )} e^{\left (-4\right )}}{x}\right ) \end {dmath*}
integrate((((-x*exp(-x+4)+1)*log((x*exp(-x+4)-1)/x/exp(-x+4))+x-1)*log(exp (x)*x)+((-x^2-x)*exp(-x+4)+x+1)*log((x*exp(-x+4)-1)/x/exp(-x+4)))/(x*exp(- x+4)-1),x, algorithm=\
Timed out. \begin {dmath*} \int \frac {\left (1+x+e^{4-x} \left (-x-x^2\right )\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )+\log \left (e^x x\right ) \left (-1+x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )\right )}{-1+e^{4-x} x} \, dx=\text {Timed out} \end {dmath*}
integrate((((-x*exp(-x+4)+1)*ln((x*exp(-x+4)-1)/x/exp(-x+4))+x-1)*ln(exp(x )*x)+((-x**2-x)*exp(-x+4)+x+1)*ln((x*exp(-x+4)-1)/x/exp(-x+4)))/(x*exp(-x+ 4)-1),x)
Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \begin {dmath*} \int \frac {\left (1+x+e^{4-x} \left (-x-x^2\right )\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )+\log \left (e^x x\right ) \left (-1+x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )\right )}{-1+e^{4-x} x} \, dx=x \log \left (x\right )^{2} + 4 \, x^{2} - {\left (x^{2} + x \log \left (x\right )\right )} \log \left (x e^{4} - e^{x}\right ) + {\left (x^{2} + 4 \, x\right )} \log \left (x\right ) \end {dmath*}
integrate((((-x*exp(-x+4)+1)*log((x*exp(-x+4)-1)/x/exp(-x+4))+x-1)*log(exp (x)*x)+((-x^2-x)*exp(-x+4)+x+1)*log((x*exp(-x+4)-1)/x/exp(-x+4)))/(x*exp(- x+4)-1),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \begin {dmath*} \int \frac {\left (1+x+e^{4-x} \left (-x-x^2\right )\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )+\log \left (e^x x\right ) \left (-1+x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )\right )}{-1+e^{4-x} x} \, dx=-x^{2} \log \left (x e^{4} - e^{x}\right ) + x^{2} \log \left (x\right ) - x \log \left (x e^{4} - e^{x}\right ) \log \left (x\right ) + x \log \left (x\right )^{2} + 4 \, x^{2} + 4 \, x \log \left (x\right ) \end {dmath*}
integrate((((-x*exp(-x+4)+1)*log((x*exp(-x+4)-1)/x/exp(-x+4))+x-1)*log(exp (x)*x)+((-x^2-x)*exp(-x+4)+x+1)*log((x*exp(-x+4)-1)/x/exp(-x+4)))/(x*exp(- x+4)-1),x, algorithm=\
-x^2*log(x*e^4 - e^x) + x^2*log(x) - x*log(x*e^4 - e^x)*log(x) + x*log(x)^ 2 + 4*x^2 + 4*x*log(x)
Time = 15.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \begin {dmath*} \int \frac {\left (1+x+e^{4-x} \left (-x-x^2\right )\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )+\log \left (e^x x\right ) \left (-1+x+\left (1-e^{4-x} x\right ) \log \left (\frac {e^{-4+x} \left (-1+e^{4-x} x\right )}{x}\right )\right )}{-1+e^{4-x} x} \, dx=-x\,\ln \left (\frac {x-{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}{x}\right )\,\left (x+\ln \left (x\right )\right ) \end {dmath*}