3.12.72 \(\int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x)))}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx\) [1172]

3.12.72.1 Optimal result

Integrand size = 105, antiderivative size = 25 \[ \int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} \left (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x))\right )}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx=\frac {x^2}{\left (e^{e^{x^2}+x^{\frac {1}{x}}}-\frac {14 x}{5}\right )^2} \]

output

x^2/(exp(exp(ln(x)/x)+exp(x^2))-14/5*x)^2
 

3.12.72.2 Mathematica [A] (verified)

Time = 3.69 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} \left (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x))\right )}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx=\frac {25 x^2}{\left (5 e^{e^{x^2}+x^{\frac {1}{x}}}-14 x\right )^2} \]

input

Integrate[(E^(E^x^2 + x^x^(-1))*(250*x - 500*E^x^2*x^3 + x^x^(-1)*(-250 + 
250*Log[x])))/(125*E^(3*E^x^2 + 3*x^x^(-1)) - 1050*E^(2*E^x^2 + 2*x^x^(-1) 
)*x + 2940*E^(E^x^2 + x^x^(-1))*x^2 - 2744*x^3),x]
 

output

(25*x^2)/(5*E^(E^x^2 + x^x^(-1)) - 14*x)^2
 

3.12.72.3 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.

\(\displaystyle \int \frac {e^{x^{\frac {1}{x}}+e^{x^2}} \left (x^{\frac {1}{x}} (250 \log (x)-250)-500 e^{x^2} x^3+250 x\right )}{-2744 x^3+2940 e^{x^{\frac {1}{x}}+e^{x^2}} x^2-1050 e^{2 x^{\frac {1}{x}}+2 e^{x^2}} x+125 e^{3 x^{\frac {1}{x}}+3 e^{x^2}}} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {250 e^{x^{\frac {1}{x}}+e^{x^2}} \left (-x^{\frac {1}{x}}+x^{\frac {1}{x}} \log (x)-2 e^{x^2} x^3+x\right )}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 250 \int \frac {e^{x^{\frac {1}{x}}+e^{x^2}} \left (\log (x) x^{\frac {1}{x}}-x^{\frac {1}{x}}-2 e^{x^2} x^3+x\right )}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 250 \int \left (\frac {e^{x^{\frac {1}{x}}+e^{x^2}} x^{\frac {1}{x}} (\log (x)-1)}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}-\frac {e^{x^{\frac {1}{x}}+e^{x^2}} x \left (2 e^{x^2} x^2-1\right )}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 250 \left (\int \frac {e^{x^{\frac {1}{x}}+e^{x^2}} x}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}dx-\int \frac {e^{x^{\frac {1}{x}}+e^{x^2}} x^{\frac {1}{x}}}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}dx-\int \frac {\int \frac {e^{x^{\frac {1}{x}}+e^{x^2}} x^{\frac {1}{x}}}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}dx}{x}dx+\log (x) \int \frac {e^{x^{\frac {1}{x}}+e^{x^2}} x^{\frac {1}{x}}}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}dx-2 \int \frac {e^{x^{\frac {1}{x}}+x^2+e^{x^2}} x^3}{\left (5 e^{x^{\frac {1}{x}}+e^{x^2}}-14 x\right )^3}dx\right )\)

input

Int[(E^(E^x^2 + x^x^(-1))*(250*x - 500*E^x^2*x^3 + x^x^(-1)*(-250 + 250*Lo 
g[x])))/(125*E^(3*E^x^2 + 3*x^x^(-1)) - 1050*E^(2*E^x^2 + 2*x^x^(-1))*x + 
2940*E^(E^x^2 + x^x^(-1))*x^2 - 2744*x^3),x]
 

output

$Aborted
 

3.12.72.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293

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

3.12.72.4 Maple [A] (verified)

Time = 17.75 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {25 x^{2}}{\left (14 x -5 \,{\mathrm e}^{x^{\frac {1}{x}}+{\mathrm e}^{x^{2}}}\right )^{2}}\)	\(25\)
parallelrisch	\(\frac {25 x^{2}}{196 x^{2}-140 x \,{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (x \right )}{x}}+{\mathrm e}^{x^{2}}}+25 \,{\mathrm e}^{2 \,{\mathrm e}^{\frac {\ln \left (x \right )}{x}}+2 \,{\mathrm e}^{x^{2}}}}\)	\(47\)

input

int(((250*ln(x)-250)*exp(ln(x)/x)-500*x^3*exp(x^2)+250*x)*exp(exp(ln(x)/x) 
+exp(x^2))/(125*exp(exp(ln(x)/x)+exp(x^2))^3-1050*x*exp(exp(ln(x)/x)+exp(x 
^2))^2+2940*x^2*exp(exp(ln(x)/x)+exp(x^2))-2744*x^3),x,method=_RETURNVERBO 
SE)
 

output

25*x^2/(14*x-5*exp(x^(1/x)+exp(x^2)))^2
 

3.12.72.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} \left (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x))\right )}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx=\frac {25 \, x^{2}}{196 \, x^{2} - 140 \, x e^{\left (x^{\left (\frac {1}{x}\right )} + e^{\left (x^{2}\right )}\right )} + 25 \, e^{\left (2 \, x^{\left (\frac {1}{x}\right )} + 2 \, e^{\left (x^{2}\right )}\right )}} \]

input

integrate(((250*log(x)-250)*exp(log(x)/x)-500*x^3*exp(x^2)+250*x)*exp(exp( 
log(x)/x)+exp(x^2))/(125*exp(exp(log(x)/x)+exp(x^2))^3-1050*x*exp(exp(log( 
x)/x)+exp(x^2))^2+2940*x^2*exp(exp(log(x)/x)+exp(x^2))-2744*x^3),x, algori 
thm=\
 

output

25*x^2/(196*x^2 - 140*x*e^(x^(1/x) + e^(x^2)) + 25*e^(2*x^(1/x) + 2*e^(x^2 
)))
 

3.12.72.6 Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} \left (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x))\right )}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx=\frac {x^{2}}{\frac {196 x^{2}}{25} - \frac {28 x e^{e^{x^{2}} + e^{\frac {\log {\left (x \right )}}{x}}}}{5} + e^{2 e^{x^{2}} + 2 e^{\frac {\log {\left (x \right )}}{x}}}} \]

input

integrate(((250*ln(x)-250)*exp(ln(x)/x)-500*x**3*exp(x**2)+250*x)*exp(exp( 
ln(x)/x)+exp(x**2))/(125*exp(exp(ln(x)/x)+exp(x**2))**3-1050*x*exp(exp(ln( 
x)/x)+exp(x**2))**2+2940*x**2*exp(exp(ln(x)/x)+exp(x**2))-2744*x**3),x)
 

output

x**2/(196*x**2/25 - 28*x*exp(exp(x**2) + exp(log(x)/x))/5 + exp(2*exp(x**2 
) + 2*exp(log(x)/x)))
 

3.12.72.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} \left (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x))\right )}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx=\frac {25 \, x^{2}}{196 \, x^{2} - 140 \, x e^{\left (x^{\left (\frac {1}{x}\right )} + e^{\left (x^{2}\right )}\right )} + 25 \, e^{\left (2 \, x^{\left (\frac {1}{x}\right )} + 2 \, e^{\left (x^{2}\right )}\right )}} \]

input

integrate(((250*log(x)-250)*exp(log(x)/x)-500*x^3*exp(x^2)+250*x)*exp(exp( 
log(x)/x)+exp(x^2))/(125*exp(exp(log(x)/x)+exp(x^2))^3-1050*x*exp(exp(log( 
x)/x)+exp(x^2))^2+2940*x^2*exp(exp(log(x)/x)+exp(x^2))-2744*x^3),x, algori 
thm=\
 

output

25*x^2/(196*x^2 - 140*x*e^(x^(1/x) + e^(x^2)) + 25*e^(2*x^(1/x) + 2*e^(x^2 
)))
 

3.12.72.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} \left (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x))\right )}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx=\text {Exception raised: NotImplementedError} \]

input

integrate(((250*log(x)-250)*exp(log(x)/x)-500*x^3*exp(x^2)+250*x)*exp(exp( 
log(x)/x)+exp(x^2))/(125*exp(exp(log(x)/x)+exp(x^2))^3-1050*x*exp(exp(log( 
x)/x)+exp(x^2))^2+2940*x^2*exp(exp(log(x)/x)+exp(x^2))-2744*x^3),x, algori 
thm=\
 

output

Exception raised: NotImplementedError >> unable to parse Giac output: Exch 
ange 58065148 31 Vector [13,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, 
1,1,1,1,1,1,4] 70 Vector [13,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 
,1,1,1,1,1,1,4] 70/
 

3.12.72.9 Mupad [B] (verification not implemented)

Time = 13.93 (sec) , antiderivative size = 108, normalized size of antiderivative = 4.32 \[ \int \frac {e^{e^{x^2}+x^{\frac {1}{x}}} \left (250 x-500 e^{x^2} x^3+x^{\frac {1}{x}} (-250+250 \log (x))\right )}{125 e^{3 e^{x^2}+3 x^{\frac {1}{x}}}-1050 e^{2 e^{x^2}+2 x^{\frac {1}{x}}} x+2940 e^{e^{x^2}+x^{\frac {1}{x}}} x^2-2744 x^3} \, dx=-\frac {25\,x^5\,\left (x-2\,x^3\,{\mathrm {e}}^{x^2}-x^{1/x}+x^{1/x}\,\ln \left (x\right )\right )}{\left (25\,{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}+2\,x^{1/x}}-140\,x\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}+x^{1/x}}+196\,x^2\right )\,\left (2\,x^6\,{\mathrm {e}}^{x^2}-x^4+x^{1/x}\,x^3-x^{1/x}\,x^3\,\ln \left (x\right )\right )} \]

input

int((exp(exp(x^2) + exp(log(x)/x))*(250*x - 500*x^3*exp(x^2) + exp(log(x)/ 
x)*(250*log(x) - 250)))/(125*exp(3*exp(x^2) + 3*exp(log(x)/x)) - 1050*x*ex 
p(2*exp(x^2) + 2*exp(log(x)/x)) + 2940*x^2*exp(exp(x^2) + exp(log(x)/x)) - 
 2744*x^3),x)
 

output

-(25*x^5*(x - 2*x^3*exp(x^2) - x^(1/x) + x^(1/x)*log(x)))/((25*exp(2*exp(x 
^2) + 2*x^(1/x)) - 140*x*exp(exp(x^2) + x^(1/x)) + 196*x^2)*(2*x^6*exp(x^2 
) - x^4 + x^(1/x)*x^3 - x^(1/x)*x^3*log(x)))