Integrand size = 120, antiderivative size = 32 \[ \int \frac {-21-3 x+7 x^2-x^3+e^x (5+x)+\left (5+2 x-x^2\right ) \log (x)+\left (5-e^x+4 x-x^2+(-1-x) \log (x)\right ) \log \left (x^2\right )}{4 x-5 x^2+x^3+e^x \left (-x+x^2\right )+\left (-x+x^2\right ) \log (x)+\left (-4 x+e^x x+x^2+x \log (x)\right ) \log \left (x^2\right )} \, dx=-x+\log \left (\frac {\left (4-e^x-x-\log (x)\right ) \left (-1+x+\log \left (x^2\right )\right )^2}{x}\right ) \]
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {-21-3 x+7 x^2-x^3+e^x (5+x)+\left (5+2 x-x^2\right ) \log (x)+\left (5-e^x+4 x-x^2+(-1-x) \log (x)\right ) \log \left (x^2\right )}{4 x-5 x^2+x^3+e^x \left (-x+x^2\right )+\left (-x+x^2\right ) \log (x)+\left (-4 x+e^x x+x^2+x \log (x)\right ) \log \left (x^2\right )} \, dx=-x-\log (x)+\log \left (4-e^x-x-\log (x)\right )+2 \log \left (1-x-\log \left (x^2\right )\right ) \]
Integrate[(-21 - 3*x + 7*x^2 - x^3 + E^x*(5 + x) + (5 + 2*x - x^2)*Log[x] + (5 - E^x + 4*x - x^2 + (-1 - x)*Log[x])*Log[x^2])/(4*x - 5*x^2 + x^3 + E ^x*(-x + x^2) + (-x + x^2)*Log[x] + (-4*x + E^x*x + x^2 + x*Log[x])*Log[x^ 2]),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.
\(\displaystyle \int \frac {-x^3+7 x^2+\left (-x^2+2 x+5\right ) \log (x)+\left (-x^2+4 x-e^x+(-x-1) \log (x)+5\right ) \log \left (x^2\right )-3 x+e^x (x+5)-21}{x^3-5 x^2+e^x \left (x^2-x\right )+\left (x^2-x\right ) \log (x)+\left (x^2+e^x x-4 x+x \log (x)\right ) \log \left (x^2\right )+4 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^3+7 x^2+\left (-x^2+2 x+5\right ) \log (x)+\left (-x^2+4 x-e^x+(-x-1) \log (x)+5\right ) \log \left (x^2\right )-3 x+e^x (x+5)-21}{x \left (-x-e^x-\log (x)+4\right ) \left (-\log \left (x^2\right )-x+1\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-\log \left (x^2\right )+x+5}{x \left (\log \left (x^2\right )+x-1\right )}-\frac {x^2-5 x+x \log (x)-1}{x \left (x+e^x+\log (x)-4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \int \frac {1}{x+e^x+\log (x)-4}dx+\int \frac {1}{x \left (x+e^x+\log (x)-4\right )}dx-\int \frac {x}{x+e^x+\log (x)-4}dx-\int \frac {\log (x)}{x+e^x+\log (x)-4}dx+2 \log \left (-\log \left (x^2\right )-x+1\right )-\log (x)\) |
Int[(-21 - 3*x + 7*x^2 - x^3 + E^x*(5 + x) + (5 + 2*x - x^2)*Log[x] + (5 - E^x + 4*x - x^2 + (-1 - x)*Log[x])*Log[x^2])/(4*x - 5*x^2 + x^3 + E^x*(-x + x^2) + (-x + x^2)*Log[x] + (-4*x + E^x*x + x^2 + x*Log[x])*Log[x^2]),x]
3.11.64.3.1 Defintions of rubi rules used
Time = 2.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\ln \left (x \right )-x +2 \ln \left (-1+x +\ln \left (x^{2}\right )\right )+\ln \left (x +{\mathrm e}^{x}+\ln \left (x \right )-4\right )\) | \(27\) |
parallelrisch | \(-\ln \left (x \right )-x +2 \ln \left (-1+x +\ln \left (x^{2}\right )\right )+\ln \left (x +{\mathrm e}^{x}+\ln \left (x \right )-4\right )\) | \(27\) |
risch | \(-x -\ln \left (x \right )+\ln \left (x +{\mathrm e}^{x}+\ln \left (x \right )-4\right )+2 \ln \left (\ln \left (x \right )-\frac {i \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i x -2 i\right )}{4}\right )\) | \(77\) |
int((((-1-x)*ln(x)-exp(x)-x^2+4*x+5)*ln(x^2)+(-x^2+2*x+5)*ln(x)+(5+x)*exp( x)-x^3+7*x^2-3*x-21)/((x*ln(x)+exp(x)*x+x^2-4*x)*ln(x^2)+ln(x)*(x^2-x)+(x^ 2-x)*exp(x)+x^3-5*x^2+4*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-21-3 x+7 x^2-x^3+e^x (5+x)+\left (5+2 x-x^2\right ) \log (x)+\left (5-e^x+4 x-x^2+(-1-x) \log (x)\right ) \log \left (x^2\right )}{4 x-5 x^2+x^3+e^x \left (-x+x^2\right )+\left (-x+x^2\right ) \log (x)+\left (-4 x+e^x x+x^2+x \log (x)\right ) \log \left (x^2\right )} \, dx=-x + \log \left (x + e^{x} + \log \left (x\right ) - 4\right ) + 2 \, \log \left (x + 2 \, \log \left (x\right ) - 1\right ) - \log \left (x\right ) \]
integrate((((-1-x)*log(x)-exp(x)-x^2+4*x+5)*log(x^2)+(-x^2+2*x+5)*log(x)+( 5+x)*exp(x)-x^3+7*x^2-3*x-21)/((x*log(x)+exp(x)*x+x^2-4*x)*log(x^2)+log(x) *(x^2-x)+(x^2-x)*exp(x)+x^3-5*x^2+4*x),x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-21-3 x+7 x^2-x^3+e^x (5+x)+\left (5+2 x-x^2\right ) \log (x)+\left (5-e^x+4 x-x^2+(-1-x) \log (x)\right ) \log \left (x^2\right )}{4 x-5 x^2+x^3+e^x \left (-x+x^2\right )+\left (-x+x^2\right ) \log (x)+\left (-4 x+e^x x+x^2+x \log (x)\right ) \log \left (x^2\right )} \, dx=- x - \log {\left (x \right )} + 2 \log {\left (\frac {x}{2} + \log {\left (x \right )} - \frac {1}{2} \right )} + \log {\left (x + e^{x} + \log {\left (x \right )} - 4 \right )} \]
integrate((((-1-x)*ln(x)-exp(x)-x**2+4*x+5)*ln(x**2)+(-x**2+2*x+5)*ln(x)+( 5+x)*exp(x)-x**3+7*x**2-3*x-21)/((x*ln(x)+exp(x)*x+x**2-4*x)*ln(x**2)+ln(x )*(x**2-x)+(x**2-x)*exp(x)+x**3-5*x**2+4*x),x)
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-21-3 x+7 x^2-x^3+e^x (5+x)+\left (5+2 x-x^2\right ) \log (x)+\left (5-e^x+4 x-x^2+(-1-x) \log (x)\right ) \log \left (x^2\right )}{4 x-5 x^2+x^3+e^x \left (-x+x^2\right )+\left (-x+x^2\right ) \log (x)+\left (-4 x+e^x x+x^2+x \log (x)\right ) \log \left (x^2\right )} \, dx=-x + \log \left (x + e^{x} + \log \left (x\right ) - 4\right ) - \log \left (x\right ) + 2 \, \log \left (\frac {1}{2} \, x + \log \left (x\right ) - \frac {1}{2}\right ) \]
integrate((((-1-x)*log(x)-exp(x)-x^2+4*x+5)*log(x^2)+(-x^2+2*x+5)*log(x)+( 5+x)*exp(x)-x^3+7*x^2-3*x-21)/((x*log(x)+exp(x)*x+x^2-4*x)*log(x^2)+log(x) *(x^2-x)+(x^2-x)*exp(x)+x^3-5*x^2+4*x),x, algorithm=\
Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-21-3 x+7 x^2-x^3+e^x (5+x)+\left (5+2 x-x^2\right ) \log (x)+\left (5-e^x+4 x-x^2+(-1-x) \log (x)\right ) \log \left (x^2\right )}{4 x-5 x^2+x^3+e^x \left (-x+x^2\right )+\left (-x+x^2\right ) \log (x)+\left (-4 x+e^x x+x^2+x \log (x)\right ) \log \left (x^2\right )} \, dx=-x + \log \left (x + e^{x} + \log \left (x\right ) - 4\right ) - \log \left (x\right ) + 2 \, \log \left (-x - 2 \, \log \left (x\right ) + 1\right ) \]
integrate((((-1-x)*log(x)-exp(x)-x^2+4*x+5)*log(x^2)+(-x^2+2*x+5)*log(x)+( 5+x)*exp(x)-x^3+7*x^2-3*x-21)/((x*log(x)+exp(x)*x+x^2-4*x)*log(x^2)+log(x) *(x^2-x)+(x^2-x)*exp(x)+x^3-5*x^2+4*x),x, algorithm=\
Time = 11.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.88 \[ \int \frac {-21-3 x+7 x^2-x^3+e^x (5+x)+\left (5+2 x-x^2\right ) \log (x)+\left (5-e^x+4 x-x^2+(-1-x) \log (x)\right ) \log \left (x^2\right )}{4 x-5 x^2+x^3+e^x \left (-x+x^2\right )+\left (-x+x^2\right ) \log (x)+\left (-4 x+e^x x+x^2+x \log (x)\right ) \log \left (x^2\right )} \, dx=\ln \left (\frac {\left (x+2\right )\,\left (x+{\mathrm {e}}^x+\ln \left (x\right )-4\right )}{x}\right )-x-2\,\ln \left (\frac {x+x\,{\mathrm {e}}^x+1}{x}\right )-\ln \left (x+2\right )+2\,\ln \left (\frac {\left (x+\ln \left (x^2\right )-1\right )\,\left (x+x\,{\mathrm {e}}^x+1\right )}{x}\right ) \]
int(-(3*x - exp(x)*(x + 5) + log(x^2)*(exp(x) - 4*x + log(x)*(x + 1) + x^2 - 5) - log(x)*(2*x - x^2 + 5) - 7*x^2 + x^3 + 21)/(4*x + log(x^2)*(x*exp( x) - 4*x + x*log(x) + x^2) - exp(x)*(x - x^2) - log(x)*(x - x^2) - 5*x^2 + x^3),x)