\(\int \frac {60 x-11 x^2+5 x^4-x^5+(5+e^x-x) x^{2 x}+e^x (12 x-x^2+x^4)+x^x (31-6 x+10 x^2-2 x^3+e^x (5+2 x^2)+(30+6 e^x-6 x) \log (x))+\log (-\frac {3}{5+e^x-x}) (-10 x-2 e^x x+2 x^2+x^x (-5-e^x+x+(-5-e^x+x) \log (x)))}{5 x^4+e^x x^4-x^5+(5+e^x-x) x^{2 x}+x^x (10 x^2+2 e^x x^2-2 x^3)} \, dx\) [7389]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F(-1)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 202, antiderivative size = 30 \[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=x-\frac {6-\log \left (\frac {3}{-5-e^x+x}\right )}{x^2+x^x} \]

[Out]

x-(6-ln(3/(-exp(x)+x-5)))/(x^2+exp(x*ln(x)))

Rubi [F]

\[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=\int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx \]

[In]

Int[(60*x - 11*x^2 + 5*x^4 - x^5 + (5 + E^x - x)*x^(2*x) + E^x*(12*x - x^2 + x^4) + x^x*(31 - 6*x + 10*x^2 - 2
*x^3 + E^x*(5 + 2*x^2) + (30 + 6*E^x - 6*x)*Log[x]) + Log[-3/(5 + E^x - x)]*(-10*x - 2*E^x*x + 2*x^2 + x^x*(-5
 - E^x + x + (-5 - E^x + x)*Log[x])))/(5*x^4 + E^x*x^4 - x^5 + (5 + E^x - x)*x^(2*x) + x^x*(10*x^2 + 2*E^x*x^2
 - 2*x^3)),x]

[Out]

x + 12*Defer[Int][x/(x^2 + x^x)^2, x] - 2*Log[-3/(5 + E^x - x)]*Defer[Int][x/(x^2 + x^x)^2, x] - 6*Defer[Int][
x^2/(x^2 + x^x)^2, x] + Log[-3/(5 + E^x - x)]*Defer[Int][x^2/(x^2 + x^x)^2, x] - 6*Log[x]*Defer[Int][x^2/(x^2
+ x^x)^2, x] + Log[-3/(5 + E^x - x)]*Log[x]*Defer[Int][x^2/(x^2 + x^x)^2, x] + 31*Defer[Int][1/((5 + E^x - x)*
(x^2 + x^x)), x] - 5*Log[-3/(5 + E^x - x)]*Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x] + 30*Log[x]*Defer[Int]
[1/((5 + E^x - x)*(x^2 + x^x)), x] - 5*Log[-3/(5 + E^x - x)]*Log[x]*Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)),
x] + 5*Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x] - Log[-3/(5 + E^x - x)]*Defer[Int][E^x/((5 + E^x - x)*(x
^2 + x^x)), x] + 6*Log[x]*Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x] - Log[-3/(5 + E^x - x)]*Log[x]*Defer[
Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x] - 6*Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x] + Log[-3/(5 + E^x -
x)]*Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x] - 6*Log[x]*Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x] + Log
[-3/(5 + E^x - x)]*Log[x]*Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x] - 2*Defer[Int][Defer[Int][x/(x^2 + x^x)
^2, x], x] + 12*Defer[Int][Defer[Int][x/(x^2 + x^x)^2, x]/(5 + E^x - x), x] - 2*Defer[Int][(x*Defer[Int][x/(x^
2 + x^x)^2, x])/(5 + E^x - x), x] + Defer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x], x] + Log[x]*Defer[Int][Defer[
Int][x^2/(x^2 + x^x)^2, x], x] - 6*Defer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x]/(5 + E^x - x), x] - 6*Log[x]*De
fer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x]/(5 + E^x - x), x] + 6*Defer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x]/x,
 x] - Log[-3/(5 + E^x - x)]*Defer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x]/x, x] + Defer[Int][(x*Defer[Int][x^2/(
x^2 + x^x)^2, x])/(5 + E^x - x), x] + Log[x]*Defer[Int][(x*Defer[Int][x^2/(x^2 + x^x)^2, x])/(5 + E^x - x), x]
 - 5*Defer[Int][Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x], x] - 5*Log[x]*Defer[Int][Defer[Int][1/((5 + E^x
- x)*(x^2 + x^x)), x], x] + 30*Defer[Int][Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x - x), x] + 30*
Log[x]*Defer[Int][Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x - x), x] - 30*Defer[Int][Defer[Int][1/
((5 + E^x - x)*(x^2 + x^x)), x]/x, x] + 5*Log[-3/(5 + E^x - x)]*Defer[Int][Defer[Int][1/((5 + E^x - x)*(x^2 +
x^x)), x]/x, x] - 5*Defer[Int][(x*Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x])/(5 + E^x - x), x] - 5*Log[x]*D
efer[Int][(x*Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x])/(5 + E^x - x), x] - Defer[Int][Defer[Int][E^x/((5 +
 E^x - x)*(x^2 + x^x)), x], x] - Log[x]*Defer[Int][Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x], x] + 6*Defe
r[Int][Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x - x), x] + 6*Log[x]*Defer[Int][Defer[Int][E^x/(
(5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x - x), x] - 6*Defer[Int][Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x]
/x, x] + Log[-3/(5 + E^x - x)]*Defer[Int][Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x]/x, x] - Defer[Int][(x
*Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x])/(5 + E^x - x), x] - Log[x]*Defer[Int][(x*Defer[Int][E^x/((5 +
 E^x - x)*(x^2 + x^x)), x])/(5 + E^x - x), x] + Defer[Int][Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x], x] +
Log[x]*Defer[Int][Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x], x] - 6*Defer[Int][Defer[Int][x/((5 + E^x - x)*
(x^2 + x^x)), x]/(5 + E^x - x), x] - 6*Log[x]*Defer[Int][Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x
 - x), x] + 6*Defer[Int][Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x]/x, x] - Log[-3/(5 + E^x - x)]*Defer[Int]
[Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x]/x, x] + Defer[Int][(x*Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)),
x])/(5 + E^x - x), x] + Log[x]*Defer[Int][(x*Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x])/(5 + E^x - x), x] -
 Defer[Int][Defer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x], x]/x, x] + 6*Defer[Int][Defer[Int][Defer[Int][x^2/(x^
2 + x^x)^2, x]/(5 + E^x - x), x]/x, x] - Defer[Int][Defer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x]/x, x], x] + 6*
Defer[Int][Defer[Int][Defer[Int][x^2/(x^2 + x^x)^2, x]/x, x]/(5 + E^x - x), x] - Defer[Int][(x*Defer[Int][Defe
r[Int][x^2/(x^2 + x^x)^2, x]/x, x])/(5 + E^x - x), x] - Defer[Int][Defer[Int][(x*Defer[Int][x^2/(x^2 + x^x)^2,
 x])/(5 + E^x - x), x]/x, x] + 5*Defer[Int][Defer[Int][Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x], x]/x, x]
- 30*Defer[Int][Defer[Int][Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x - x), x]/x, x] + 5*Defer[Int]
[Defer[Int][Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)), x]/x, x], x] - 30*Defer[Int][Defer[Int][Defer[Int][1/((5
 + E^x - x)*(x^2 + x^x)), x]/x, x]/(5 + E^x - x), x] + 5*Defer[Int][(x*Defer[Int][Defer[Int][1/((5 + E^x - x)*
(x^2 + x^x)), x]/x, x])/(5 + E^x - x), x] + 5*Defer[Int][Defer[Int][(x*Defer[Int][1/((5 + E^x - x)*(x^2 + x^x)
), x])/(5 + E^x - x), x]/x, x] + Defer[Int][Defer[Int][Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x], x]/x, x
] - 6*Defer[Int][Defer[Int][Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x - x), x]/x, x] + Defer[Int
][Defer[Int][Defer[Int][E^x/((5 + E^x - x)*(x^2 + x^x)), x]/x, x], x] - 6*Defer[Int][Defer[Int][Defer[Int][E^x
/((5 + E^x - x)*(x^2 + x^x)), x]/x, x]/(5 + E^x - x), x] + Defer[Int][(x*Defer[Int][Defer[Int][E^x/((5 + E^x -
 x)*(x^2 + x^x)), x]/x, x])/(5 + E^x - x), x] + Defer[Int][Defer[Int][(x*Defer[Int][E^x/((5 + E^x - x)*(x^2 +
x^x)), x])/(5 + E^x - x), x]/x, x] - Defer[Int][Defer[Int][Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x], x]/x,
 x] + 6*Defer[Int][Defer[Int][Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x]/(5 + E^x - x), x]/x, x] - Defer[Int
][Defer[Int][Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x]/x, x], x] + 6*Defer[Int][Defer[Int][Defer[Int][x/((5
 + E^x - x)*(x^2 + x^x)), x]/x, x]/(5 + E^x - x), x] - Defer[Int][(x*Defer[Int][Defer[Int][x/((5 + E^x - x)*(x
^2 + x^x)), x]/x, x])/(5 + E^x - x), x] - Defer[Int][Defer[Int][(x*Defer[Int][x/((5 + E^x - x)*(x^2 + x^x)), x
])/(5 + E^x - x), x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{\left (5+e^x-x\right ) \left (x^2+x^x\right )^2} \, dx \\ & = \int \left (1+\frac {x \left (-6+\log \left (-\frac {3}{5+e^x-x}\right )\right ) (-2+x+x \log (x))}{\left (x^2+x^x\right )^2}-\frac {-31-5 e^x+6 x+5 \log \left (-\frac {3}{5+e^x-x}\right )+e^x \log \left (-\frac {3}{5+e^x-x}\right )-x \log \left (-\frac {3}{5+e^x-x}\right )-30 \log (x)-6 e^x \log (x)+6 x \log (x)+5 \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)+e^x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)-x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}\right ) \, dx \\ & = x+\int \frac {x \left (-6+\log \left (-\frac {3}{5+e^x-x}\right )\right ) (-2+x+x \log (x))}{\left (x^2+x^x\right )^2} \, dx-\int \frac {-31-5 e^x+6 x+5 \log \left (-\frac {3}{5+e^x-x}\right )+e^x \log \left (-\frac {3}{5+e^x-x}\right )-x \log \left (-\frac {3}{5+e^x-x}\right )-30 \log (x)-6 e^x \log (x)+6 x \log (x)+5 \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)+e^x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)-x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx \\ & = x+\int \left (\frac {12 x}{\left (x^2+x^x\right )^2}-\frac {6 x^2}{\left (x^2+x^x\right )^2}-\frac {2 x \log \left (-\frac {3}{5+e^x-x}\right )}{\left (x^2+x^x\right )^2}+\frac {x^2 \log \left (-\frac {3}{5+e^x-x}\right )}{\left (x^2+x^x\right )^2}-\frac {6 x^2 \log (x)}{\left (x^2+x^x\right )^2}+\frac {x^2 \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (x^2+x^x\right )^2}\right ) \, dx-\int \frac {-31-5 e^x+6 x-6 \left (5+e^x-x\right ) \log (x)+\left (5+e^x-x\right ) \log \left (-\frac {3}{5+e^x-x}\right ) (1+\log (x))}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx \\ & = x-2 \int \frac {x \log \left (-\frac {3}{5+e^x-x}\right )}{\left (x^2+x^x\right )^2} \, dx-6 \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx-6 \int \frac {x^2 \log (x)}{\left (x^2+x^x\right )^2} \, dx+12 \int \frac {x}{\left (x^2+x^x\right )^2} \, dx+\int \frac {x^2 \log \left (-\frac {3}{5+e^x-x}\right )}{\left (x^2+x^x\right )^2} \, dx+\int \frac {x^2 \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (x^2+x^x\right )^2} \, dx-\int \left (-\frac {31}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}-\frac {5 e^x}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}+\frac {6 x}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}+\frac {5 \log \left (-\frac {3}{5+e^x-x}\right )}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}+\frac {e^x \log \left (-\frac {3}{5+e^x-x}\right )}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}-\frac {x \log \left (-\frac {3}{5+e^x-x}\right )}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}-\frac {30 \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}-\frac {6 e^x \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}+\frac {6 x \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}+\frac {5 \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}+\frac {e^x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}-\frac {x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )}\right ) \, dx \\ & = x+2 \int \frac {\left (-1+e^x\right ) \left (-5-e^x+x\right ) \int \frac {x}{\left (x^2+x^x\right )^2} \, dx}{\left (5+e^x-x\right )^2} \, dx+5 \int \frac {e^x}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx-5 \int \frac {\log \left (-\frac {3}{5+e^x-x}\right )}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx-5 \int \frac {\log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx-6 \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx-6 \int \frac {x}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx+6 \int \frac {e^x \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx-6 \int \frac {x \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx+6 \int \frac {\int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx}{x} \, dx+12 \int \frac {x}{\left (x^2+x^x\right )^2} \, dx+30 \int \frac {\log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx+31 \int \frac {1}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx+\log \left (-\frac {3}{5+e^x-x}\right ) \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx-\left (2 \log \left (-\frac {3}{5+e^x-x}\right )\right ) \int \frac {x}{\left (x^2+x^x\right )^2} \, dx-(6 \log (x)) \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx+\left (\log \left (-\frac {3}{5+e^x-x}\right ) \log (x)\right ) \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx-\int \frac {e^x \log \left (-\frac {3}{5+e^x-x}\right )}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx+\int \frac {x \log \left (-\frac {3}{5+e^x-x}\right )}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx-\int \frac {e^x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx+\int \frac {x \log \left (-\frac {3}{5+e^x-x}\right ) \log (x)}{\left (5+e^x-x\right ) \left (x^2+x^x\right )} \, dx-\int \frac {\left (-1+e^x\right ) \left (-5-e^x+x\right ) \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx}{\left (5+e^x-x\right )^2} \, dx-\int \frac {\log \left (-\frac {3}{5+e^x-x}\right ) \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx}{x} \, dx-\int \frac {\left (-1+e^x\right ) \left (-5-e^x+x\right ) \log (x) \int \frac {x^2}{\left (x^2+x^x\right )^2} \, dx}{\left (5+e^x-x\right )^2} \, dx \\ & = \text {Too large to display} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {-6+x^3+x^{1+x}+\log \left (-\frac {3}{5+e^x-x}\right )}{x^2+x^x} \]

[In]

Integrate[(60*x - 11*x^2 + 5*x^4 - x^5 + (5 + E^x - x)*x^(2*x) + E^x*(12*x - x^2 + x^4) + x^x*(31 - 6*x + 10*x
^2 - 2*x^3 + E^x*(5 + 2*x^2) + (30 + 6*E^x - 6*x)*Log[x]) + Log[-3/(5 + E^x - x)]*(-10*x - 2*E^x*x + 2*x^2 + x
^x*(-5 - E^x + x + (-5 - E^x + x)*Log[x])))/(5*x^4 + E^x*x^4 - x^5 + (5 + E^x - x)*x^(2*x) + x^x*(10*x^2 + 2*E
^x*x^2 - 2*x^3)),x]

[Out]

(-6 + x^3 + x^(1 + x) + Log[-3/(5 + E^x - x)])/(x^2 + x^x)

Maple [A] (verified)

Time = 32.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63

method result size
risch \(-\frac {\ln \left (-{\mathrm e}^{x}+x -5\right )}{x^{2}+x^{x}}+\frac {-12+2 x^{3}+2 x^{x} x +2 \ln \left (3\right )}{2 x^{2}+2 x^{x}}\) \(49\)
parallelrisch \(\frac {-12+2 x^{3}+10 x^{2}+2 \,{\mathrm e}^{x \ln \left (x \right )} x +10 \,{\mathrm e}^{x \ln \left (x \right )}+2 \ln \left (-\frac {3}{{\mathrm e}^{x}+5-x}\right )}{2 x^{2}+2 \,{\mathrm e}^{x \ln \left (x \right )}}\) \(55\)

[In]

int(((((-exp(x)+x-5)*ln(x)-exp(x)+x-5)*exp(x*ln(x))-2*exp(x)*x+2*x^2-10*x)*ln(-3/(exp(x)+5-x))+(exp(x)+5-x)*ex
p(x*ln(x))^2+((6*exp(x)-6*x+30)*ln(x)+(2*x^2+5)*exp(x)-2*x^3+10*x^2-6*x+31)*exp(x*ln(x))+(x^4-x^2+12*x)*exp(x)
-x^5+5*x^4-11*x^2+60*x)/((exp(x)+5-x)*exp(x*ln(x))^2+(2*exp(x)*x^2-2*x^3+10*x^2)*exp(x*ln(x))+exp(x)*x^4-x^5+5
*x^4),x,method=_RETURNVERBOSE)

[Out]

-1/(x^2+x^x)*ln(-exp(x)+x-5)+1/2*(-12+2*x^3+2*x^x*x+2*ln(3))/(x^2+x^x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {x^{3} + x x^{x} + \log \left (\frac {3}{x - e^{x} - 5}\right ) - 6}{x^{2} + x^{x}} \]

[In]

integrate(((((-exp(x)+x-5)*log(x)-exp(x)+x-5)*exp(x*log(x))-2*exp(x)*x+2*x^2-10*x)*log(-3/(exp(x)+5-x))+(exp(x
)+5-x)*exp(x*log(x))^2+((6*exp(x)-6*x+30)*log(x)+(2*x^2+5)*exp(x)-2*x^3+10*x^2-6*x+31)*exp(x*log(x))+(x^4-x^2+
12*x)*exp(x)-x^5+5*x^4-11*x^2+60*x)/((exp(x)+5-x)*exp(x*log(x))^2+(2*exp(x)*x^2-2*x^3+10*x^2)*exp(x*log(x))+ex
p(x)*x^4-x^5+5*x^4),x, algorithm="fricas")

[Out]

(x^3 + x*x^x + log(3/(x - e^x - 5)) - 6)/(x^2 + x^x)

Sympy [A] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=x + \frac {\log {\left (- \frac {3}{- x + e^{x} + 5} \right )}}{x^{2} + e^{x \log {\left (x \right )}}} - \frac {6}{x^{2} + e^{x \log {\left (x \right )}}} \]

[In]

integrate(((((-exp(x)+x-5)*ln(x)-exp(x)+x-5)*exp(x*ln(x))-2*exp(x)*x+2*x**2-10*x)*ln(-3/(exp(x)+5-x))+(exp(x)+
5-x)*exp(x*ln(x))**2+((6*exp(x)-6*x+30)*ln(x)+(2*x**2+5)*exp(x)-2*x**3+10*x**2-6*x+31)*exp(x*ln(x))+(x**4-x**2
+12*x)*exp(x)-x**5+5*x**4-11*x**2+60*x)/((exp(x)+5-x)*exp(x*ln(x))**2+(2*exp(x)*x**2-2*x**3+10*x**2)*exp(x*ln(
x))+exp(x)*x**4-x**5+5*x**4),x)

[Out]

x + log(-3/(-x + exp(x) + 5))/(x**2 + exp(x*log(x))) - 6/(x**2 + exp(x*log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {x^{3} + x x^{x} + \log \left (3\right ) - \log \left (x - e^{x} - 5\right ) - 6}{x^{2} + x^{x}} \]

[In]

integrate(((((-exp(x)+x-5)*log(x)-exp(x)+x-5)*exp(x*log(x))-2*exp(x)*x+2*x^2-10*x)*log(-3/(exp(x)+5-x))+(exp(x
)+5-x)*exp(x*log(x))^2+((6*exp(x)-6*x+30)*log(x)+(2*x^2+5)*exp(x)-2*x^3+10*x^2-6*x+31)*exp(x*log(x))+(x^4-x^2+
12*x)*exp(x)-x^5+5*x^4-11*x^2+60*x)/((exp(x)+5-x)*exp(x*log(x))^2+(2*exp(x)*x^2-2*x^3+10*x^2)*exp(x*log(x))+ex
p(x)*x^4-x^5+5*x^4),x, algorithm="maxima")

[Out]

(x^3 + x*x^x + log(3) - log(x - e^x - 5) - 6)/(x^2 + x^x)

Giac [F(-1)]

Timed out. \[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=\text {Timed out} \]

[In]

integrate(((((-exp(x)+x-5)*log(x)-exp(x)+x-5)*exp(x*log(x))-2*exp(x)*x+2*x^2-10*x)*log(-3/(exp(x)+5-x))+(exp(x
)+5-x)*exp(x*log(x))^2+((6*exp(x)-6*x+30)*log(x)+(2*x^2+5)*exp(x)-2*x^3+10*x^2-6*x+31)*exp(x*log(x))+(x^4-x^2+
12*x)*exp(x)-x^5+5*x^4-11*x^2+60*x)/((exp(x)+5-x)*exp(x*log(x))^2+(2*exp(x)*x^2-2*x^3+10*x^2)*exp(x*log(x))+ex
p(x)*x^4-x^5+5*x^4),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 13.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {60 x-11 x^2+5 x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+e^x \left (12 x-x^2+x^4\right )+x^x \left (31-6 x+10 x^2-2 x^3+e^x \left (5+2 x^2\right )+\left (30+6 e^x-6 x\right ) \log (x)\right )+\log \left (-\frac {3}{5+e^x-x}\right ) \left (-10 x-2 e^x x+2 x^2+x^x \left (-5-e^x+x+\left (-5-e^x+x\right ) \log (x)\right )\right )}{5 x^4+e^x x^4-x^5+\left (5+e^x-x\right ) x^{2 x}+x^x \left (10 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {\ln \left (-\frac {3}{{\mathrm {e}}^x-x+5}\right )+x\,x^x+x^3-6}{x^x+x^2} \]

[In]

int((60*x + exp(x)*(12*x - x^2 + x^4) + exp(2*x*log(x))*(exp(x) - x + 5) - log(-3/(exp(x) - x + 5))*(10*x + 2*
x*exp(x) - 2*x^2 + exp(x*log(x))*(exp(x) - x + log(x)*(exp(x) - x + 5) + 5)) + exp(x*log(x))*(log(x)*(6*exp(x)
 - 6*x + 30) - 6*x + exp(x)*(2*x^2 + 5) + 10*x^2 - 2*x^3 + 31) - 11*x^2 + 5*x^4 - x^5)/(x^4*exp(x) + exp(x*log
(x))*(2*x^2*exp(x) + 10*x^2 - 2*x^3) + exp(2*x*log(x))*(exp(x) - x + 5) + 5*x^4 - x^5),x)

[Out]

(log(-3/(exp(x) - x + 5)) + x*x^x + x^3 - 6)/(x^x + x^2)