Integrand size = 96, antiderivative size = 31 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=\frac {\left (-e^x+x\right ) \left (-x+\frac {3 x}{\log (x)}-\frac {16 x}{5+\log (x)}\right )}{x} \] Output:
(x-exp(x))*(3*x/ln(x)-x-16*x/(5+ln(x)))/x
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=\frac {\left (e^x-x\right ) \left (-15+18 \log (x)+\log ^2(x)\right )}{\log (x) (5+\log (x))} \] Input:
Integrate[(75*E^x - 75*x + (E^x*(30 - 75*x) + 45*x)*Log[x] + (-62*x + E^x* (-13 + 75*x))*Log[x]^2 + (-23*x + 23*E^x*x)*Log[x]^3 + (-x + E^x*x)*Log[x] ^4)/(25*x*Log[x]^2 + 10*x*Log[x]^3 + x*Log[x]^4),x]
Output:
((E^x - x)*(-15 + 18*Log[x] + Log[x]^2))/(Log[x]*(5 + Log[x]))
Leaf count is larger than twice the leaf count of optimal. \(65\) vs. \(2(31)=62\).
Time = 1.74 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, 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.
\(\displaystyle \int \frac {75 e^x-75 x+\left (e^x x-x\right ) \log ^4(x)+\left (23 e^x x-23 x\right ) \log ^3(x)+\left (e^x (75 x-13)-62 x\right ) \log ^2(x)+\left (e^x (30-75 x)+45 x\right ) \log (x)}{x \log ^4(x)+10 x \log ^3(x)+25 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {75 e^x-75 x+\left (e^x x-x\right ) \log ^4(x)+\left (23 e^x x-23 x\right ) \log ^3(x)+\left (e^x (75 x-13)-62 x\right ) \log ^2(x)+\left (e^x (30-75 x)+45 x\right ) \log (x)}{x \log ^2(x) (\log (x)+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {\log ^2(x)}{(\log (x)+5)^2}-\frac {75}{(\log (x)+5)^2 \log ^2(x)}+\frac {e^x \left (x \log ^4(x)+23 x \log ^3(x)+75 x \log ^2(x)-13 \log ^2(x)-75 x \log (x)+30 \log (x)+75\right )}{x (\log (x)+5)^2 \log ^2(x)}-\frac {23 \log (x)}{(\log (x)+5)^2}-\frac {62}{(\log (x)+5)^2}+\frac {45}{(\log (x)+5)^2 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x-\frac {e^x \left (-x \log ^4(x)-23 x \log ^3(x)-75 x \log ^2(x)+75 x \log (x)\right )}{x \log ^2(x) (\log (x)+5)^2}+\frac {3 x}{\log (x)}-\frac {16 x}{\log (x)+5}\) |
Input:
Int[(75*E^x - 75*x + (E^x*(30 - 75*x) + 45*x)*Log[x] + (-62*x + E^x*(-13 + 75*x))*Log[x]^2 + (-23*x + 23*E^x*x)*Log[x]^3 + (-x + E^x*x)*Log[x]^4)/(2 5*x*Log[x]^2 + 10*x*Log[x]^3 + x*Log[x]^4),x]
Output:
-x + (3*x)/Log[x] - (16*x)/(5 + Log[x]) - (E^x*(75*x*Log[x] - 75*x*Log[x]^ 2 - 23*x*Log[x]^3 - x*Log[x]^4))/(x*Log[x]^2*(5 + Log[x])^2)
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74
\[-\frac {x \left (\ln \left (x \right )^{2}+18 \ln \left (x \right )-15\right )}{\left (5+\ln \left (x \right )\right ) \ln \left (x \right )}+\frac {{\mathrm e}^{x} \ln \left (x \right )^{2}+18 \,{\mathrm e}^{x} \ln \left (x \right )-15 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (5+\ln \left (x \right )\right )}\]
Input:
int(((exp(x)*x-x)*ln(x)^4+(23*exp(x)*x-23*x)*ln(x)^3+((75*x-13)*exp(x)-62* x)*ln(x)^2+((-75*x+30)*exp(x)+45*x)*ln(x)+75*exp(x)-75*x)/(x*ln(x)^4+10*x* ln(x)^3+25*x*ln(x)^2),x)
Output:
-x*(ln(x)^2+18*ln(x)-15)/(5+ln(x))/ln(x)+(exp(x)*ln(x)^2+18*exp(x)*ln(x)-1 5*exp(x))/ln(x)/(5+ln(x))
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {{\left (x - e^{x}\right )} \log \left (x\right )^{2} + 18 \, {\left (x - e^{x}\right )} \log \left (x\right ) - 15 \, x + 15 \, e^{x}}{\log \left (x\right )^{2} + 5 \, \log \left (x\right )} \] Input:
integrate(((exp(x)*x-x)*log(x)^4+(23*exp(x)*x-23*x)*log(x)^3+((75*x-13)*ex p(x)-62*x)*log(x)^2+((-75*x+30)*exp(x)+45*x)*log(x)+75*exp(x)-75*x)/(x*log (x)^4+10*x*log(x)^3+25*x*log(x)^2),x, algorithm="fricas")
Output:
-((x - e^x)*log(x)^2 + 18*(x - e^x)*log(x) - 15*x + 15*e^x)/(log(x)^2 + 5* log(x))
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=- x + \frac {- 13 x \log {\left (x \right )} + 15 x}{\log {\left (x \right )}^{2} + 5 \log {\left (x \right )}} + \frac {\left (\log {\left (x \right )}^{2} + 18 \log {\left (x \right )} - 15\right ) e^{x}}{\log {\left (x \right )}^{2} + 5 \log {\left (x \right )}} \] Input:
integrate(((exp(x)*x-x)*ln(x)**4+(23*exp(x)*x-23*x)*ln(x)**3+((75*x-13)*ex p(x)-62*x)*ln(x)**2+((-75*x+30)*exp(x)+45*x)*ln(x)+75*exp(x)-75*x)/(x*ln(x )**4+10*x*ln(x)**3+25*x*ln(x)**2),x)
Output:
-x + (-13*x*log(x) + 15*x)/(log(x)**2 + 5*log(x)) + (log(x)**2 + 18*log(x) - 15)*exp(x)/(log(x)**2 + 5*log(x))
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {x \log \left (x\right )^{2} - {\left (\log \left (x\right )^{2} + 18 \, \log \left (x\right ) - 15\right )} e^{x} + 18 \, x \log \left (x\right ) - 15 \, x}{\log \left (x\right )^{2} + 5 \, \log \left (x\right )} \] Input:
integrate(((exp(x)*x-x)*log(x)^4+(23*exp(x)*x-23*x)*log(x)^3+((75*x-13)*ex p(x)-62*x)*log(x)^2+((-75*x+30)*exp(x)+45*x)*log(x)+75*exp(x)-75*x)/(x*log (x)^4+10*x*log(x)^3+25*x*log(x)^2),x, algorithm="maxima")
Output:
-(x*log(x)^2 - (log(x)^2 + 18*log(x) - 15)*e^x + 18*x*log(x) - 15*x)/(log( x)^2 + 5*log(x))
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {x \log \left (x\right )^{2} - e^{x} \log \left (x\right )^{2} + 18 \, x \log \left (x\right ) - 18 \, e^{x} \log \left (x\right ) - 15 \, x + 15 \, e^{x}}{\log \left (x\right )^{2} + 5 \, \log \left (x\right )} \] Input:
integrate(((exp(x)*x-x)*log(x)^4+(23*exp(x)*x-23*x)*log(x)^3+((75*x-13)*ex p(x)-62*x)*log(x)^2+((-75*x+30)*exp(x)+45*x)*log(x)+75*exp(x)-75*x)/(x*log (x)^4+10*x*log(x)^3+25*x*log(x)^2),x, algorithm="giac")
Output:
-(x*log(x)^2 - e^x*log(x)^2 + 18*x*log(x) - 18*e^x*log(x) - 15*x + 15*e^x) /(log(x)^2 + 5*log(x))
Time = 2.62 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=-\frac {\left (x-{\mathrm {e}}^x\right )\,\left ({\ln \left (x\right )}^2+18\,\ln \left (x\right )-15\right )}{\ln \left (x\right )\,\left (\ln \left (x\right )+5\right )} \] Input:
int(-(75*x - 75*exp(x) + log(x)^4*(x - x*exp(x)) + log(x)^2*(62*x - exp(x) *(75*x - 13)) + log(x)^3*(23*x - 23*x*exp(x)) - log(x)*(45*x - exp(x)*(75* x - 30)))/(25*x*log(x)^2 + 10*x*log(x)^3 + x*log(x)^4),x)
Output:
-((x - exp(x))*(18*log(x) + log(x)^2 - 15))/(log(x)*(log(x) + 5))
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {75 e^x-75 x+\left (e^x (30-75 x)+45 x\right ) \log (x)+\left (-62 x+e^x (-13+75 x)\right ) \log ^2(x)+\left (-23 x+23 e^x x\right ) \log ^3(x)+\left (-x+e^x x\right ) \log ^4(x)}{25 x \log ^2(x)+10 x \log ^3(x)+x \log ^4(x)} \, dx=\frac {e^{x} \mathrm {log}\left (x \right )^{2}+18 e^{x} \mathrm {log}\left (x \right )-15 e^{x}-\mathrm {log}\left (x \right )^{2} x -18 \,\mathrm {log}\left (x \right ) x +15 x}{\mathrm {log}\left (x \right ) \left (\mathrm {log}\left (x \right )+5\right )} \] Input:
int(((exp(x)*x-x)*log(x)^4+(23*exp(x)*x-23*x)*log(x)^3+((75*x-13)*exp(x)-6 2*x)*log(x)^2+((-75*x+30)*exp(x)+45*x)*log(x)+75*exp(x)-75*x)/(x*log(x)^4+ 10*x*log(x)^3+25*x*log(x)^2),x)
Output:
(e**x*log(x)**2 + 18*e**x*log(x) - 15*e**x - log(x)**2*x - 18*log(x)*x + 1 5*x)/(log(x)*(log(x) + 5))