\(\int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} (-1+e^{e^4} (-2 x+2 x^2))}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx\) [1783]

Optimal result

Integrand size = 75, antiderivative size = 31 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{e^{48 \log ^2(5)} \left (2+\frac {e^{-e^4}}{(1-x) x}\right )} \] Output:

exp((2+1/x/(1-x)/exp(exp(4)))*exp(48*ln(5)^2))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{e^{48 \log ^2(5)} \left (2+\frac {e^{-e^4}}{x-x^2}\right )} \] Input:

Integrate[(E^(-E^4 + (E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/ 
(-x + x^2) + 48*Log[5]^2)*(-1 + 2*x))/(x^2 - 2*x^3 + x^4),x]
 

Output:

E^(E^(48*Log[5]^2)*(2 + 1/(E^E^4*(x - x^2))))
 

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.

\(\displaystyle \int \frac {(2 x-1) \exp \left (\frac {\left (e^{e^4} \left (2 x^2-2 x\right )-1\right ) e^{48 \log ^2(5)-e^4}}{x^2-x}-e^4+48 \log ^2(5)\right )}{x^4-2 x^3+x^2} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {(2 x-1) \exp \left (\frac {\left (e^{e^4} \left (2 x^2-2 x\right )-1\right ) e^{48 \log ^2(5)-e^4}}{x^2-x}-e^4+48 \log ^2(5)\right )}{x^2 \left (x^2-2 x+1\right )}dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 4 \int -\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}+48 \log ^2(5)-e^4\right ) (1-2 x)}{4 (1-x)^2 x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle -\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}+48 \log ^2(5)-e^4\right ) (1-2 x)}{(1-x)^2 x^2}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle -\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right ) (1-2 x)}{(1-x)^2 x^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\int \left (\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{x^2}-\frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{(x-1)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{(x-1)^2}dx-\int \frac {\exp \left (\frac {e^{-e^4+48 \log ^2(5)} \left (2 e^{e^4} \left (x-x^2\right )+1\right )}{x-x^2}-e^4 \left (1-\frac {48 \log ^2(5)}{e^4}\right )\right )}{x^2}dx\)

Input:

Int[(E^(-E^4 + (E^(-E^4 + 48*Log[5]^2)*(-1 + E^E^4*(-2*x + 2*x^2)))/(-x + 
x^2) + 48*Log[5]^2)*(-1 + 2*x))/(x^2 - 2*x^3 + x^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x \left (-1+x \right )}}\)	\(38\)
gosper	\({\mathrm e}^{\frac {\left (2 x^{2} {\mathrm e}^{{\mathrm e}^{4}}-2 x \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x \left (-1+x \right )}}\)	\(39\)
risch	\({\mathrm e}^{\frac {\left (2 x^{2} {\mathrm e}^{{\mathrm e}^{4}}-2 x \,{\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}-{\mathrm e}^{4}}}{x \left (-1+x \right )}}\)	\(39\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x^{2}-x}}-x \,{\mathrm e}^{\frac {\left (\left (2 x^{2}-2 x \right ) {\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{48 \ln \left (5\right )^{2}} {\mathrm e}^{-{\mathrm e}^{4}}}{x^{2}-x}}}{x \left (-1+x \right )}\)	\(94\)

Input:

int((-1+2*x)*exp(48*ln(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*ln(5)^ 
2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x,method=_RETURNVERBOS 
E)
 

Output:

exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*ln(5)^2)/x/(-1+x)/exp(exp(4)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (26) = 52\).

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.32 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{\left (-48 \, \log \left (5\right )^{2} + \frac {48 \, {\left (x^{2} - x\right )} \log \left (5\right )^{2} - {\left (x^{2} - x\right )} e^{4} + {\left (2 \, {\left (x^{2} - x\right )} e^{\left (e^{4}\right )} - 1\right )} e^{\left (48 \, \log \left (5\right )^{2} - e^{4}\right )}}{x^{2} - x} + e^{4}\right )} \] Input:

integrate((-1+2*x)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48 
*log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x, algorithm=" 
fricas")
 

Output:

e^(-48*log(5)^2 + (48*(x^2 - x)*log(5)^2 - (x^2 - x)*e^4 + (2*(x^2 - x)*e^ 
(e^4) - 1)*e^(48*log(5)^2 - e^4))/(x^2 - x) + e^4)
 

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{\frac {\left (\left (2 x^{2} - 2 x\right ) e^{e^{4}} - 1\right ) e^{48 \log {\left (5 \right )}^{2}}}{\left (x^{2} - x\right ) e^{e^{4}}}} \] Input:

integrate((-1+2*x)*exp(48*ln(5)**2)*exp(((2*x**2-2*x)*exp(exp(4))-1)*exp(4 
8*ln(5)**2)/(x**2-x)/exp(exp(4)))/(x**4-2*x**3+x**2)/exp(exp(4)),x)
 

Output:

exp(((2*x**2 - 2*x)*exp(exp(4)) - 1)*exp(-exp(4))*exp(48*log(5)**2)/(x**2 
- x))
 

Maxima [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=e^{\left (-\frac {e^{\left (48 \, \log \left (5\right )^{2}\right )}}{x e^{\left (e^{4}\right )} - e^{\left (e^{4}\right )}} + \frac {e^{\left (48 \, \log \left (5\right )^{2} - e^{4}\right )}}{x} + 2 \, e^{\left (48 \, \log \left (5\right )^{2}\right )}\right )} \] Input:

integrate((-1+2*x)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48 
*log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x, algorithm=" 
maxima")
 

Output:

e^(-e^(48*log(5)^2)/(x*e^(e^4) - e^(e^4)) + e^(48*log(5)^2 - e^4)/x + 2*e^ 
(48*log(5)^2))
 

Giac [F]

\[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=\int { \frac {{\left (2 \, x - 1\right )} e^{\left (48 \, \log \left (5\right )^{2} + \frac {{\left (2 \, {\left (x^{2} - x\right )} e^{\left (e^{4}\right )} - 1\right )} e^{\left (48 \, \log \left (5\right )^{2} - e^{4}\right )}}{x^{2} - x} - e^{4}\right )}}{x^{4} - 2 \, x^{3} + x^{2}} \,d x } \] Input:

integrate((-1+2*x)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48 
*log(5)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x, algorithm=" 
giac")
 

Output:

integrate((2*x - 1)*e^(48*log(5)^2 + (2*(x^2 - x)*e^(e^4) - 1)*e^(48*log(5 
)^2 - e^4)/(x^2 - x) - e^4)/(x^4 - 2*x^3 + x^2), x)
 

Mupad [B] (verification not implemented)

Time = 5.48 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.13 \[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx={\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{48\,{\ln \left (5\right )}^2}}{x-x^2}}\,{\mathrm {e}}^{-\frac {2\,x^2\,{\mathrm {e}}^{48\,{\ln \left (5\right )}^2}}{x-x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^{48\,{\ln \left (5\right )}^2}}{x-x^2}} \] Input:

int((exp(-exp(4))*exp(48*log(5)^2)*exp((exp(-exp(4))*exp(48*log(5)^2)*(exp 
(exp(4))*(2*x - 2*x^2) + 1))/(x - x^2))*(2*x - 1))/(x^2 - 2*x^3 + x^4),x)
 

Output:

exp((2*x*exp(48*log(5)^2))/(x - x^2))*exp(-(2*x^2*exp(48*log(5)^2))/(x - x 
^2))*exp((exp(-exp(4))*exp(48*log(5)^2))/(x - x^2))
 

Reduce [F]

\[ \int \frac {e^{-e^4+\frac {e^{-e^4+48 \log ^2(5)} \left (-1+e^{e^4} \left (-2 x+2 x^2\right )\right )}{-x+x^2}+48 \log ^2(5)} (-1+2 x)}{x^2-2 x^3+x^4} \, dx=\text {too large to display} \] Input:

int((-1+2*x)*exp(48*log(5)^2)*exp(((2*x^2-2*x)*exp(exp(4))-1)*exp(48*log(5 
)^2)/(x^2-x)/exp(exp(4)))/(x^4-2*x^3+x^2)/exp(exp(4)),x)
 

Output:

(e**(2*e**(48*log(5)**2) + 48*log(5)**2)*(64*e**((e**(48*log(5)**2) + 288* 
e**(e**4)*log(5)**2*x**2 - 288*e**(e**4)*log(5)**2*x + e**(e**4)*e**4*x**2 
 - e**(e**4)*e**4*x)/(e**(e**4)*x**2 - e**(e**4)*x))*int(1/(16*e**((e**(48 
*log(5)**2) + 192*e**(e**4)*log(5)**2*x**2 - 192*e**(e**4)*log(5)**2*x)/(e 
**(e**4)*x**2 - e**(e**4)*x))*x**4 - 32*e**((e**(48*log(5)**2) + 192*e**(e 
**4)*log(5)**2*x**2 - 192*e**(e**4)*log(5)**2*x)/(e**(e**4)*x**2 - e**(e** 
4)*x))*x**3 + 16*e**((e**(48*log(5)**2) + 192*e**(e**4)*log(5)**2*x**2 - 1 
92*e**(e**4)*log(5)**2*x)/(e**(e**4)*x**2 - e**(e**4)*x))*x**2 + 32*e**((e 
**(48*log(5)**2) + 144*e**(e**4)*log(5)**2*x**2 - 144*e**(e**4)*log(5)**2* 
x + e**(e**4)*e**4*x**2 - e**(e**4)*e**4*x)/(e**(e**4)*x**2 - e**(e**4)*x) 
)*x**4 - 64*e**((e**(48*log(5)**2) + 144*e**(e**4)*log(5)**2*x**2 - 144*e* 
*(e**4)*log(5)**2*x + e**(e**4)*e**4*x**2 - e**(e**4)*e**4*x)/(e**(e**4)*x 
**2 - e**(e**4)*x))*x**3 + 32*e**((e**(48*log(5)**2) + 144*e**(e**4)*log(5 
)**2*x**2 - 144*e**(e**4)*log(5)**2*x + e**(e**4)*e**4*x**2 - e**(e**4)*e* 
*4*x)/(e**(e**4)*x**2 - e**(e**4)*x))*x**2 + 24*e**((e**(48*log(5)**2) + 9 
6*e**(e**4)*log(5)**2*x**2 - 96*e**(e**4)*log(5)**2*x + 2*e**(e**4)*e**4*x 
**2 - 2*e**(e**4)*e**4*x)/(e**(e**4)*x**2 - e**(e**4)*x))*x**4 - 48*e**((e 
**(48*log(5)**2) + 96*e**(e**4)*log(5)**2*x**2 - 96*e**(e**4)*log(5)**2*x 
+ 2*e**(e**4)*e**4*x**2 - 2*e**(e**4)*e**4*x)/(e**(e**4)*x**2 - e**(e**4)* 
x))*x**3 + 24*e**((e**(48*log(5)**2) + 96*e**(e**4)*log(5)**2*x**2 - 96...