Integrand size = 87, antiderivative size = 32 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx=x (x-\log (x)) \left (-1-e^x-e^x \log \left (\frac {x^2}{5 \log (x)}\right )\right ) \] Output:
x*(-1-ln(1/5*x^2/ln(x))*exp(x)-exp(x))*(x-ln(x))
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx=-x (x-\log (x)) \left (1+e^x+e^x \log \left (\frac {x^2}{5 \log (x)}\right )\right ) \] Input:
Integrate[(E^x*x + (1 - 2*x + E^x*(-4*x - x^2))*Log[x] + (1 + E^x*(3 + x)) *Log[x]^2 + (E^x*(1 - 2*x - x^2)*Log[x] + E^x*(1 + x)*Log[x]^2)*Log[x^2/(5 *Log[x])])/Log[x],x]
Output:
-(x*(x - Log[x])*(1 + E^x + E^x*Log[x^2/(5*Log[x])]))
Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(32)=64\).
Time = 1.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {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.
\(\displaystyle \int \frac {\left (e^x \left (-x^2-2 x+1\right ) \log (x)+e^x (x+1) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )+\left (e^x \left (-x^2-4 x\right )-2 x+1\right ) \log (x)+e^x x+\left (e^x (x+3)+1\right ) \log ^2(x)}{\log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^x \left (x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+\log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x^2 (-\log (x))-x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-2 x \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )+\log (x) \log \left (\frac {x^2}{5 \log (x)}\right )+x+x \log ^2(x)+3 \log ^2(x)-4 x \log (x)\right )}{\log (x)}-2 x+\log (x)+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x^2-\frac {e^x \left (-x \log ^2(x) \log \left (\frac {x^2}{5 \log (x)}\right )+x^2 \log (x)+x^2 \log (x) \log \left (\frac {x^2}{5 \log (x)}\right )-x \log ^2(x)\right )}{\log (x)}+x \log (x)\) |
Input:
Int[(E^x*x + (1 - 2*x + E^x*(-4*x - x^2))*Log[x] + (1 + E^x*(3 + x))*Log[x ]^2 + (E^x*(1 - 2*x - x^2)*Log[x] + E^x*(1 + x)*Log[x]^2)*Log[x^2/(5*Log[x ])])/Log[x],x]
Output:
-x^2 + x*Log[x] - (E^x*(x^2*Log[x] - x*Log[x]^2 + x^2*Log[x]*Log[x^2/(5*Lo g[x])] - x*Log[x]^2*Log[x^2/(5*Log[x])]))/Log[x]
Time = 0.50 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.78
method | result | size |
parallelrisch | \(-\ln \left (\frac {x^{2}}{5 \ln \left (x \right )}\right ) x^{2} {\mathrm e}^{x}+{\mathrm e}^{x} \ln \left (x \right ) \ln \left (\frac {x^{2}}{5 \ln \left (x \right )}\right ) x -{\mathrm e}^{x} x^{2}+x \,{\mathrm e}^{x} \ln \left (x \right )-x^{2}+x \ln \left (x \right )\) | \(57\) |
risch | \(-\frac {i \pi x \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\mathrm e}^{x} \ln \left (x \right )}{2}-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x} \ln \left (x \right )}{2}+\left ({\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x} \ln \left (x \right )\right ) \ln \left (\ln \left (x \right )\right )+x^{2} \ln \left (5\right ) {\mathrm e}^{x}-{\mathrm e}^{x} x^{2}-{\mathrm e}^{x} \ln \left (x \right ) \ln \left (5\right ) x +x \,{\mathrm e}^{x} \ln \left (x \right )+x \ln \left (x \right )-x^{2}+2 x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-2 x^{2} {\mathrm e}^{x} \ln \left (x \right )+\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right )^{3} {\mathrm e}^{x}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x} \ln \left (x \right )-i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right )^{2} {\mathrm e}^{x}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right )^{2} {\mathrm e}^{x} \ln \left (x \right )}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\mathrm e}^{x}}{2}-\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x} \ln \left (x \right )}{2}+\frac {i \pi x \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\mathrm e}^{x} \ln \left (x \right )}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\mathrm e}^{x}}{2}-\frac {i \pi x \operatorname {csgn}\left (\frac {i x^{2}}{\ln \left (x \right )}\right )^{3} {\mathrm e}^{x} \ln \left (x \right )}{2}\) | \(437\) |
Input:
int((((1+x)*exp(x)*ln(x)^2+(-x^2-2*x+1)*exp(x)*ln(x))*ln(1/5*x^2/ln(x))+(( 3+x)*exp(x)+1)*ln(x)^2+((-x^2-4*x)*exp(x)+1-2*x)*ln(x)+exp(x)*x)/ln(x),x,m ethod=_RETURNVERBOSE)
Output:
-ln(1/5*x^2/ln(x))*x^2*exp(x)+exp(x)*ln(x)*ln(1/5*x^2/ln(x))*x-exp(x)*x^2+ x*exp(x)*ln(x)-x^2+x*ln(x)
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx=-x^{2} e^{x} - x^{2} + {\left (x e^{x} + x\right )} \log \left (x\right ) - {\left (x^{2} e^{x} - x e^{x} \log \left (x\right )\right )} \log \left (\frac {x^{2}}{5 \, \log \left (x\right )}\right ) \] Input:
integrate((((1+x)*exp(x)*log(x)^2+(-x^2-2*x+1)*exp(x)*log(x))*log(1/5*x^2/ log(x))+((3+x)*exp(x)+1)*log(x)^2+((-x^2-4*x)*exp(x)+1-2*x)*log(x)+exp(x)* x)/log(x),x, algorithm="fricas")
Output:
-x^2*e^x - x^2 + (x*e^x + x)*log(x) - (x^2*e^x - x*e^x*log(x))*log(1/5*x^2 /log(x))
Time = 2.49 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx=- x^{2} + x \log {\left (x \right )} + \left (- x^{2} \log {\left (\frac {x^{2}}{5 \log {\left (x \right )}} \right )} - x^{2} + x \log {\left (x \right )} \log {\left (\frac {x^{2}}{5 \log {\left (x \right )}} \right )} + x \log {\left (x \right )}\right ) e^{x} \] Input:
integrate((((1+x)*exp(x)*ln(x)**2+(-x**2-2*x+1)*exp(x)*ln(x))*ln(1/5*x**2/ ln(x))+((3+x)*exp(x)+1)*ln(x)**2+((-x**2-4*x)*exp(x)+1-2*x)*ln(x)+exp(x)*x )/ln(x),x)
Output:
-x**2 + x*log(x) + (-x**2*log(x**2/(5*log(x))) - x**2 + x*log(x)*log(x**2/ (5*log(x))) + x*log(x))*exp(x)
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (26) = 52\).
Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.53 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx={\left (x^{2} - x \log \left (x\right )\right )} e^{x} \log \left (\log \left (x\right )\right ) - x^{2} + {\left (x^{2} \log \left (5\right ) + 2 \, x \log \left (x\right )^{2} - {\left (2 \, x^{2} + x {\left (\log \left (5\right ) - 1\right )}\right )} \log \left (x\right ) + 2 \, x - 2\right )} e^{x} - {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 4 \, {\left (x - 1\right )} e^{x} + x \log \left (x\right ) \] Input:
integrate((((1+x)*exp(x)*log(x)^2+(-x^2-2*x+1)*exp(x)*log(x))*log(1/5*x^2/ log(x))+((3+x)*exp(x)+1)*log(x)^2+((-x^2-4*x)*exp(x)+1-2*x)*log(x)+exp(x)* x)/log(x),x, algorithm="maxima")
Output:
(x^2 - x*log(x))*e^x*log(log(x)) - x^2 + (x^2*log(5) + 2*x*log(x)^2 - (2*x ^2 + x*(log(5) - 1))*log(x) + 2*x - 2)*e^x - (x^2 - 2*x + 2)*e^x - 4*(x - 1)*e^x + x*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (26) = 52\).
Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx=-2 \, x^{2} e^{x} \log \left (x\right ) + 2 \, x e^{x} \log \left (x\right )^{2} + x^{2} e^{x} \log \left (5 \, \log \left (x\right )\right ) - x e^{x} \log \left (x\right ) \log \left (5 \, \log \left (x\right )\right ) - x^{2} e^{x} + x e^{x} \log \left (x\right ) - x^{2} + x \log \left (x\right ) \] Input:
integrate((((1+x)*exp(x)*log(x)^2+(-x^2-2*x+1)*exp(x)*log(x))*log(1/5*x^2/ log(x))+((3+x)*exp(x)+1)*log(x)^2+((-x^2-4*x)*exp(x)+1-2*x)*log(x)+exp(x)* x)/log(x),x, algorithm="giac")
Output:
-2*x^2*e^x*log(x) + 2*x*e^x*log(x)^2 + x^2*e^x*log(5*log(x)) - x*e^x*log(x )*log(5*log(x)) - x^2*e^x + x*e^x*log(x) - x^2 + x*log(x)
Time = 3.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.78 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx=\ln \left (\frac {x^2}{5\,\ln \left (x\right )}\right )\,\left (\frac {{\mathrm {e}}^x\,\left (x-x^3\right )}{x}-{\mathrm {e}}^x+x\,{\mathrm {e}}^x\,\ln \left (x\right )\right )-x^2\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (x+x\,{\mathrm {e}}^x\right )-x^2 \] Input:
int((log(x)^2*(exp(x)*(x + 3) + 1) - log(x^2/(5*log(x)))*(exp(x)*log(x)*(2 *x + x^2 - 1) - exp(x)*log(x)^2*(x + 1)) + x*exp(x) - log(x)*(2*x + exp(x) *(4*x + x^2) - 1))/log(x),x)
Output:
log(x^2/(5*log(x)))*((exp(x)*(x - x^3))/x - exp(x) + x*exp(x)*log(x)) - x^ 2*exp(x) + log(x)*(x + x*exp(x)) - x^2
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {e^x x+\left (1-2 x+e^x \left (-4 x-x^2\right )\right ) \log (x)+\left (1+e^x (3+x)\right ) \log ^2(x)+\left (e^x \left (1-2 x-x^2\right ) \log (x)+e^x (1+x) \log ^2(x)\right ) \log \left (\frac {x^2}{5 \log (x)}\right )}{\log (x)} \, dx=x \left (e^{x} \mathrm {log}\left (\frac {x^{2}}{5 \,\mathrm {log}\left (x \right )}\right ) \mathrm {log}\left (x \right )-e^{x} \mathrm {log}\left (\frac {x^{2}}{5 \,\mathrm {log}\left (x \right )}\right ) x +e^{x} \mathrm {log}\left (x \right )-e^{x} x +\mathrm {log}\left (x \right )-x \right ) \] Input:
int((((1+x)*exp(x)*log(x)^2+(-x^2-2*x+1)*exp(x)*log(x))*log(1/5*x^2/log(x) )+((3+x)*exp(x)+1)*log(x)^2+((-x^2-4*x)*exp(x)+1-2*x)*log(x)+exp(x)*x)/log (x),x)
Output:
x*(e**x*log(x**2/(5*log(x)))*log(x) - e**x*log(x**2/(5*log(x)))*x + e**x*l og(x) - e**x*x + log(x) - x)