Integrand size = 102, antiderivative size = 28 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx=1+e^{\frac {4}{x+\log (x)}}-x-\log (x) \left (2-x^2+\log (x)\right ) \] Output:
exp(4/(x+ln(x)))+1-(2-x^2+ln(x))*ln(x)-x
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx=e^{\frac {4}{x+\log (x)}}-x-2 \log (x)+x^2 \log (x)-\log ^2(x) \] Input:
Integrate[(E^(4/(x + Log[x]))*(-4 - 4*x) - 2*x^2 - x^3 + x^4 + (-4*x - 4*x ^2 + 2*x^3 + 2*x^4)*Log[x] + (-2 - 5*x + x^2 + 4*x^3)*Log[x]^2 + (-2 + 2*x ^2)*Log[x]^3)/(x^3 + 2*x^2*Log[x] + x*Log[x]^2),x]
Output:
E^(4/(x + Log[x])) - x - 2*Log[x] + x^2*Log[x] - Log[x]^2
Time = 3.75 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {x^4-x^3-2 x^2+\left (2 x^2-2\right ) \log ^3(x)+\left (4 x^3+x^2-5 x-2\right ) \log ^2(x)+\left (2 x^4+2 x^3-4 x^2-4 x\right ) \log (x)+(-4 x-4) e^{\frac {4}{x+\log (x)}}}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(x+1) \left (x^3+2 x^3 \log (x)-2 x^2+4 x^2 \log ^2(x)+2 x \log ^3(x)-2 \log ^3(x)-3 x \log ^2(x)-2 \log ^2(x)-4 x \log (x)-4 e^{\frac {4}{x+\log (x)}}\right )}{x (x+\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2 (x+1) \log (x)}{(x+\log (x))^2}+\frac {x^2 (x+1)}{(x+\log (x))^2}-\frac {2 (x+1) \log ^3(x)}{x (x+\log (x))^2}+\frac {2 (x+1) \log ^3(x)}{(x+\log (x))^2}+\frac {4 x (x+1) \log ^2(x)}{(x+\log (x))^2}-\frac {2 (x+1) \log ^2(x)}{x (x+\log (x))^2}-\frac {3 (x+1) \log ^2(x)}{(x+\log (x))^2}-\frac {4 (x+1) \log (x)}{(x+\log (x))^2}-\frac {2 x (x+1)}{(x+\log (x))^2}-\frac {4 (x+1) e^{\frac {4}{x+\log (x)}}}{x (x+\log (x))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 x^2-5 x+2 (x+1)^2-\log ^2(x)-2 x \log (x)+e^{\frac {4}{x+\log (x)}}+(x+1)^2 \log (x)-3 \log (x)\) |
Input:
Int[(E^(4/(x + Log[x]))*(-4 - 4*x) - 2*x^2 - x^3 + x^4 + (-4*x - 4*x^2 + 2 *x^3 + 2*x^4)*Log[x] + (-2 - 5*x + x^2 + 4*x^3)*Log[x]^2 + (-2 + 2*x^2)*Lo g[x]^3)/(x^3 + 2*x^2*Log[x] + x*Log[x]^2),x]
Output:
E^(4/(x + Log[x])) - 5*x - 2*x^2 + 2*(1 + x)^2 - 3*Log[x] - 2*x*Log[x] + ( 1 + x)^2*Log[x] - Log[x]^2
Time = 3.63 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
risch | \(-\ln \left (x \right )^{2}+x^{2} \ln \left (x \right )-x -2 \ln \left (x \right )+{\mathrm e}^{\frac {4}{x +\ln \left (x \right )}}\) | \(30\) |
parallelrisch | \(-\ln \left (x \right )^{2}+x^{2} \ln \left (x \right )-x -2 \ln \left (x \right )+{\mathrm e}^{\frac {4}{x +\ln \left (x \right )}}\) | \(30\) |
Input:
int(((-4-4*x)*exp(4/(x+ln(x)))+(2*x^2-2)*ln(x)^3+(4*x^3+x^2-5*x-2)*ln(x)^2 +(2*x^4+2*x^3-4*x^2-4*x)*ln(x)+x^4-x^3-2*x^2)/(x*ln(x)^2+2*x^2*ln(x)+x^3), x,method=_RETURNVERBOSE)
Output:
-ln(x)^2+x^2*ln(x)-x-2*ln(x)+exp(4/(x+ln(x)))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx={\left (x^{2} - 2\right )} \log \left (x\right ) - \log \left (x\right )^{2} - x + e^{\left (\frac {4}{x + \log \left (x\right )}\right )} \] Input:
integrate(((-4-4*x)*exp(4/(x+log(x)))+(2*x^2-2)*log(x)^3+(4*x^3+x^2-5*x-2) *log(x)^2+(2*x^4+2*x^3-4*x^2-4*x)*log(x)+x^4-x^3-2*x^2)/(x*log(x)^2+2*x^2* log(x)+x^3),x, algorithm="fricas")
Output:
(x^2 - 2)*log(x) - log(x)^2 - x + e^(4/(x + log(x)))
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx=x^{2} \log {\left (x \right )} - x + e^{\frac {4}{x + \log {\left (x \right )}}} - \log {\left (x \right )}^{2} - 2 \log {\left (x \right )} \] Input:
integrate(((-4-4*x)*exp(4/(x+ln(x)))+(2*x**2-2)*ln(x)**3+(4*x**3+x**2-5*x- 2)*ln(x)**2+(2*x**4+2*x**3-4*x**2-4*x)*ln(x)+x**4-x**3-2*x**2)/(x*ln(x)**2 +2*x**2*ln(x)+x**3),x)
Output:
x**2*log(x) - x + exp(4/(x + log(x))) - log(x)**2 - 2*log(x)
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx={\left (x^{2} - 2\right )} \log \left (x\right ) - \log \left (x\right )^{2} - x + e^{\left (\frac {4}{x + \log \left (x\right )}\right )} \] Input:
integrate(((-4-4*x)*exp(4/(x+log(x)))+(2*x^2-2)*log(x)^3+(4*x^3+x^2-5*x-2) *log(x)^2+(2*x^4+2*x^3-4*x^2-4*x)*log(x)+x^4-x^3-2*x^2)/(x*log(x)^2+2*x^2* log(x)+x^3),x, algorithm="maxima")
Output:
(x^2 - 2)*log(x) - log(x)^2 - x + e^(4/(x + log(x)))
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx=x^{2} \log \left (x\right ) - \log \left (x\right )^{2} - x + e^{\left (\frac {4}{x + \log \left (x\right )}\right )} - 2 \, \log \left (x\right ) \] Input:
integrate(((-4-4*x)*exp(4/(x+log(x)))+(2*x^2-2)*log(x)^3+(4*x^3+x^2-5*x-2) *log(x)^2+(2*x^4+2*x^3-4*x^2-4*x)*log(x)+x^4-x^3-2*x^2)/(x*log(x)^2+2*x^2* log(x)+x^3),x, algorithm="giac")
Output:
x^2*log(x) - log(x)^2 - x + e^(4/(x + log(x))) - 2*log(x)
Time = 3.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx={\mathrm {e}}^{\frac {4}{x+\ln \left (x\right )}}-x-2\,\ln \left (x\right )+x^2\,\ln \left (x\right )-{\ln \left (x\right )}^2 \] Input:
int(-(log(x)*(4*x + 4*x^2 - 2*x^3 - 2*x^4) - log(x)^3*(2*x^2 - 2) + log(x) ^2*(5*x - x^2 - 4*x^3 + 2) + exp(4/(x + log(x)))*(4*x + 4) + 2*x^2 + x^3 - x^4)/(x*log(x)^2 + 2*x^2*log(x) + x^3),x)
Output:
exp(4/(x + log(x))) - x - 2*log(x) + x^2*log(x) - log(x)^2
Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {4}{x+\log (x)}} (-4-4 x)-2 x^2-x^3+x^4+\left (-4 x-4 x^2+2 x^3+2 x^4\right ) \log (x)+\left (-2-5 x+x^2+4 x^3\right ) \log ^2(x)+\left (-2+2 x^2\right ) \log ^3(x)}{x^3+2 x^2 \log (x)+x \log ^2(x)} \, dx=e^{\frac {4}{\mathrm {log}\left (x \right )+x}}-\mathrm {log}\left (x \right )^{2}+\mathrm {log}\left (x \right ) x^{2}-2 \,\mathrm {log}\left (x \right )-x \] Input:
int(((-4-4*x)*exp(4/(x+log(x)))+(2*x^2-2)*log(x)^3+(4*x^3+x^2-5*x-2)*log(x )^2+(2*x^4+2*x^3-4*x^2-4*x)*log(x)+x^4-x^3-2*x^2)/(x*log(x)^2+2*x^2*log(x) +x^3),x)
Output:
e**(4/(log(x) + x)) - log(x)**2 + log(x)*x**2 - 2*log(x) - x