3.19.26 \(\int \frac {e^x (-5+e^3)+5 x+5 x^2+e^3 (-x-x^2)+e^{(-e^x+x+x^2)^x} (e^x-x-x^2+(-e^x+x+x^2)^x (x^2-e^x x^2+2 x^3+(-e^x x+x^2+x^3) \log (-e^x+x+x^2)))}{e^x x^2-x^3-x^4} \, dx\) [1826]

3.19.26.1 Optimal result

Integrand size = 134, antiderivative size = 27 \[ \int \frac {e^x \left (-5+e^3\right )+5 x+5 x^2+e^3 \left (-x-x^2\right )+e^{\left (-e^x+x+x^2\right )^x} \left (e^x-x-x^2+\left (-e^x+x+x^2\right )^x \left (x^2-e^x x^2+2 x^3+\left (-e^x x+x^2+x^3\right ) \log \left (-e^x+x+x^2\right )\right )\right )}{e^x x^2-x^3-x^4} \, dx=\frac {5-e^3-e^{\left (-e^x+x+x^2\right )^x}}{x} \]

output

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

3.19.26.2 Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^x \left (-5+e^3\right )+5 x+5 x^2+e^3 \left (-x-x^2\right )+e^{\left (-e^x+x+x^2\right )^x} \left (e^x-x-x^2+\left (-e^x+x+x^2\right )^x \left (x^2-e^x x^2+2 x^3+\left (-e^x x+x^2+x^3\right ) \log \left (-e^x+x+x^2\right )\right )\right )}{e^x x^2-x^3-x^4} \, dx=-\frac {-5+e^3+e^{\left (-e^x+x+x^2\right )^x}}{x} \]

input

Integrate[(E^x*(-5 + E^3) + 5*x + 5*x^2 + E^3*(-x - x^2) + E^(-E^x + x + x 
^2)^x*(E^x - x - x^2 + (-E^x + x + x^2)^x*(x^2 - E^x*x^2 + 2*x^3 + (-(E^x* 
x) + x^2 + x^3)*Log[-E^x + x + x^2])))/(E^x*x^2 - x^3 - x^4),x]
 

output

-((-5 + E^3 + E^(-E^x + x + x^2)^x)/x)
 

3.19.26.3 Rubi [F]

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.

\(\displaystyle \int \frac {5 x^2+e^3 \left (-x^2-x\right )+e^{\left (x^2+x-e^x\right )^x} \left (-x^2+\left (x^2+x-e^x\right )^x \left (2 x^3-e^x x^2+x^2+\left (x^3+x^2-e^x x\right ) \log \left (x^2+x-e^x\right )\right )+e^x-x\right )+5 x+\left (e^3-5\right ) e^x}{-x^4-x^3+e^x x^2} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {e^{\left (x^2+x-e^x\right )^x}}{-x^2-x+e^x}+\frac {e^{\left (x^2+x-e^x\right )^x+x}}{x^2 \left (-x^2-x+e^x\right )}+\frac {\left (e^3-5\right ) e^x}{x^2 \left (-x^2-x+e^x\right )}+\frac {5}{-x^2-x+e^x}+\frac {e^3 (x+1)}{x \left (x^2+x-e^x\right )}+\frac {e^{\left (x^2+x-e^x\right )^x}}{x \left (x^2+x-e^x\right )}-\frac {5}{x \left (x^2+x-e^x\right )}-\frac {e^{\left (x^2+x-e^x\right )^x} \left (x^2+x-e^x\right )^{x-1} \left (2 x^2+x^2 \log \left (x^2+x-e^x\right )+x \log \left (x^2+x-e^x\right )-e^x \log \left (x^2+x-e^x\right )-e^x x+x\right )}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -e^3 \int \frac {1}{-x^2-x+e^x}dx+5 \int \frac {1}{-x^2-x+e^x}dx-\int \frac {e^{\left (x^2+x-e^x\right )^x}}{-x^2-x+e^x}dx-\left (5-e^3\right ) \int \frac {e^x}{x^2 \left (-x^2-x+e^x\right )}dx+\int \frac {e^{\left (x^2+x-e^x\right )^x+x}}{x^2 \left (-x^2-x+e^x\right )}dx+e^3 \int \frac {1}{x \left (x^2+x-e^x\right )}dx-5 \int \frac {1}{x \left (x^2+x-e^x\right )}dx+\int \frac {e^{\left (x^2+x-e^x\right )^x}}{x \left (x^2+x-e^x\right )}dx-\log \left (x^2+x-e^x\right ) \int e^{\left (x^2+x-e^x\right )^x} \left (x^2+x-e^x\right )^{x-1}dx-\int e^{\left (x^2+x-e^x\right )^x} \left (x^2+x-e^x\right )^{x-1}dx+\int e^{\left (x^2+x-e^x\right )^x+x} \left (x^2+x-e^x\right )^{x-1}dx+\log \left (x^2+x-e^x\right ) \int \frac {e^{\left (x^2+x-e^x\right )^x+x} \left (x^2+x-e^x\right )^{x-1}}{x}dx-\log \left (x^2+x-e^x\right ) \int e^{\left (x^2+x-e^x\right )^x} x \left (x^2+x-e^x\right )^{x-1}dx-2 \int e^{\left (x^2+x-e^x\right )^x} x \left (x^2+x-e^x\right )^{x-1}dx+\int \int e^{\left (x^2+x-e^x\right )^x} \left (x^2+x-e^x\right )^{x-1}dxdx-\int \frac {\int e^{\left (x^2+x-e^x\right )^x} \left (x^2+x-e^x\right )^{x-1}dx}{-x^2-x+e^x}dx+\int \frac {x \int e^{\left (x^2+x-e^x\right )^x} \left (x^2+x-e^x\right )^{x-1}dx}{x^2+x-e^x}dx-\int \frac {x^2 \int e^{\left (x^2+x-e^x\right )^x} \left (x^2+x-e^x\right )^{x-1}dx}{x^2+x-e^x}dx-\int \int \frac {e^{\left (x^2+x-e^x\right )^x+x} \left (x^2+x-e^x\right )^{x-1}}{x}dxdx+\int \frac {\int \frac {e^{\left (x^2+x-e^x\right )^x+x} \left (x^2+x-e^x\right )^{x-1}}{x}dx}{-x^2-x+e^x}dx-\int \frac {x \int \frac {e^{\left (x^2+x-e^x\right )^x+x} \left (x^2+x-e^x\right )^{x-1}}{x}dx}{x^2+x-e^x}dx+\int \frac {x^2 \int \frac {e^{\left (x^2+x-e^x\right )^x+x} \left (x^2+x-e^x\right )^{x-1}}{x}dx}{x^2+x-e^x}dx+\int \int e^{\left (x^2+x-e^x\right )^x} x \left (x^2+x-e^x\right )^{x-1}dxdx-\int \frac {\int e^{\left (x^2+x-e^x\right )^x} x \left (x^2+x-e^x\right )^{x-1}dx}{-x^2-x+e^x}dx+\int \frac {x \int e^{\left (x^2+x-e^x\right )^x} x \left (x^2+x-e^x\right )^{x-1}dx}{x^2+x-e^x}dx-\int \frac {x^2 \int e^{\left (x^2+x-e^x\right )^x} x \left (x^2+x-e^x\right )^{x-1}dx}{x^2+x-e^x}dx\)

input

Int[(E^x*(-5 + E^3) + 5*x + 5*x^2 + E^3*(-x - x^2) + E^(-E^x + x + x^2)^x* 
(E^x - x - x^2 + (-E^x + x + x^2)^x*(x^2 - E^x*x^2 + 2*x^3 + (-(E^x*x) + x 
^2 + x^3)*Log[-E^x + x + x^2])))/(E^x*x^2 - x^3 - x^4),x]
 

output

$Aborted
 

3.19.26.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.19.26.4 Maple [A] (verified)

Time = 138.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(-\frac {-5+{\mathrm e}^{3}+{\mathrm e}^{{\mathrm e}^{x \ln \left (-{\mathrm e}^{x}+x^{2}+x \right )}}}{x}\)	\(24\)
risch	\(-\frac {{\mathrm e}^{3}}{x}+\frac {5}{x}-\frac {{\mathrm e}^{\left (-{\mathrm e}^{x}+x^{2}+x \right )^{x}}}{x}\)	\(31\)

input

int(((((-exp(x)*x+x^3+x^2)*ln(-exp(x)+x^2+x)-exp(x)*x^2+2*x^3+x^2)*exp(x*l 
n(-exp(x)+x^2+x))+exp(x)-x^2-x)*exp(exp(x*ln(-exp(x)+x^2+x)))+(exp(3)-5)*e 
xp(x)+(-x^2-x)*exp(3)+5*x^2+5*x)/(exp(x)*x^2-x^4-x^3),x,method=_RETURNVERB 
OSE)
 

output

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

3.19.26.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^x \left (-5+e^3\right )+5 x+5 x^2+e^3 \left (-x-x^2\right )+e^{\left (-e^x+x+x^2\right )^x} \left (e^x-x-x^2+\left (-e^x+x+x^2\right )^x \left (x^2-e^x x^2+2 x^3+\left (-e^x x+x^2+x^3\right ) \log \left (-e^x+x+x^2\right )\right )\right )}{e^x x^2-x^3-x^4} \, dx=-\frac {e^{3} + e^{\left ({\left (x^{2} + x - e^{x}\right )}^{x}\right )} - 5}{x} \]

input

integrate(((((-exp(x)*x+x^3+x^2)*log(-exp(x)+x^2+x)-exp(x)*x^2+2*x^3+x^2)* 
exp(x*log(-exp(x)+x^2+x))+exp(x)-x^2-x)*exp(exp(x*log(-exp(x)+x^2+x)))+(ex 
p(3)-5)*exp(x)+(-x^2-x)*exp(3)+5*x^2+5*x)/(exp(x)*x^2-x^4-x^3),x, algorith 
m=\
 

output

-(e^3 + e^((x^2 + x - e^x)^x) - 5)/x
 

3.19.26.6 Sympy [F(-1)]

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

input

integrate(((((-exp(x)*x+x**3+x**2)*ln(-exp(x)+x**2+x)-exp(x)*x**2+2*x**3+x 
**2)*exp(x*ln(-exp(x)+x**2+x))+exp(x)-x**2-x)*exp(exp(x*ln(-exp(x)+x**2+x) 
))+(exp(3)-5)*exp(x)+(-x**2-x)*exp(3)+5*x**2+5*x)/(exp(x)*x**2-x**4-x**3), 
x)
 

output

Timed out
 

3.19.26.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^x \left (-5+e^3\right )+5 x+5 x^2+e^3 \left (-x-x^2\right )+e^{\left (-e^x+x+x^2\right )^x} \left (e^x-x-x^2+\left (-e^x+x+x^2\right )^x \left (x^2-e^x x^2+2 x^3+\left (-e^x x+x^2+x^3\right ) \log \left (-e^x+x+x^2\right )\right )\right )}{e^x x^2-x^3-x^4} \, dx=-\frac {e^{3} + e^{\left ({\left (x^{2} + x - e^{x}\right )}^{x}\right )} - 5}{x} \]

input

integrate(((((-exp(x)*x+x^3+x^2)*log(-exp(x)+x^2+x)-exp(x)*x^2+2*x^3+x^2)* 
exp(x*log(-exp(x)+x^2+x))+exp(x)-x^2-x)*exp(exp(x*log(-exp(x)+x^2+x)))+(ex 
p(3)-5)*exp(x)+(-x^2-x)*exp(3)+5*x^2+5*x)/(exp(x)*x^2-x^4-x^3),x, algorith 
m=\
 

output

-(e^3 + e^((x^2 + x - e^x)^x) - 5)/x
 

3.19.26.8 Giac [F]

\[ \int \frac {e^x \left (-5+e^3\right )+5 x+5 x^2+e^3 \left (-x-x^2\right )+e^{\left (-e^x+x+x^2\right )^x} \left (e^x-x-x^2+\left (-e^x+x+x^2\right )^x \left (x^2-e^x x^2+2 x^3+\left (-e^x x+x^2+x^3\right ) \log \left (-e^x+x+x^2\right )\right )\right )}{e^x x^2-x^3-x^4} \, dx=\int { -\frac {5 \, x^{2} - {\left (x^{2} + x\right )} e^{3} + {\left ({\left (2 \, x^{3} - x^{2} e^{x} + x^{2} + {\left (x^{3} + x^{2} - x e^{x}\right )} \log \left (x^{2} + x - e^{x}\right )\right )} {\left (x^{2} + x - e^{x}\right )}^{x} - x^{2} - x + e^{x}\right )} e^{\left ({\left (x^{2} + x - e^{x}\right )}^{x}\right )} + {\left (e^{3} - 5\right )} e^{x} + 5 \, x}{x^{4} + x^{3} - x^{2} e^{x}} \,d x } \]

input

integrate(((((-exp(x)*x+x^3+x^2)*log(-exp(x)+x^2+x)-exp(x)*x^2+2*x^3+x^2)* 
exp(x*log(-exp(x)+x^2+x))+exp(x)-x^2-x)*exp(exp(x*log(-exp(x)+x^2+x)))+(ex 
p(3)-5)*exp(x)+(-x^2-x)*exp(3)+5*x^2+5*x)/(exp(x)*x^2-x^4-x^3),x, algorith 
m=\
 

output

integrate(-(5*x^2 - (x^2 + x)*e^3 + ((2*x^3 - x^2*e^x + x^2 + (x^3 + x^2 - 
 x*e^x)*log(x^2 + x - e^x))*(x^2 + x - e^x)^x - x^2 - x + e^x)*e^((x^2 + x 
 - e^x)^x) + (e^3 - 5)*e^x + 5*x)/(x^4 + x^3 - x^2*e^x), x)
 

3.19.26.9 Mupad [B] (verification not implemented)

Time = 14.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^x \left (-5+e^3\right )+5 x+5 x^2+e^3 \left (-x-x^2\right )+e^{\left (-e^x+x+x^2\right )^x} \left (e^x-x-x^2+\left (-e^x+x+x^2\right )^x \left (x^2-e^x x^2+2 x^3+\left (-e^x x+x^2+x^3\right ) \log \left (-e^x+x+x^2\right )\right )\right )}{e^x x^2-x^3-x^4} \, dx=-\frac {{\mathrm {e}}^{{\left (x-{\mathrm {e}}^x+x^2\right )}^x}+{\mathrm {e}}^3-5}{x} \]

input

int(-(5*x + exp(x)*(exp(3) - 5) - exp(exp(x*log(x - exp(x) + x^2)))*(x - e 
xp(x) + x^2 - exp(x*log(x - exp(x) + x^2))*(log(x - exp(x) + x^2)*(x^2 - x 
*exp(x) + x^3) - x^2*exp(x) + x^2 + 2*x^3)) - exp(3)*(x + x^2) + 5*x^2)/(x 
^3 - x^2*exp(x) + x^4),x)
 

output

-(exp((x - exp(x) + x^2)^x) + exp(3) - 5)/x