3.12.67 \(\int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} (50 x^2-20 x^3+2 x^4+(-100 x^2+40 x^3-4 x^4) \log (2)+(50 x^2-20 x^3+2 x^4) \log ^2(2)))}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} (-2 x^2+4 x^3)} \, dx\) [1167]

3.12.67.1 Optimal result

Integrand size = 200, antiderivative size = 36 \[ \int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} \left (50 x^2-20 x^3+2 x^4+\left (-100 x^2+40 x^3-4 x^4\right ) \log (2)+\left (50 x^2-20 x^3+2 x^4\right ) \log ^2(2)\right )\right )}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (-2 x^2+4 x^3\right )} \, dx=5+\frac {\left (-1+\frac {5}{x}\right )^2 x}{1-e^{e^{(-x+x \log (2))^2}}-2 x} \]

output

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

3.12.67.2 Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} \left (50 x^2-20 x^3+2 x^4+\left (-100 x^2+40 x^3-4 x^4\right ) \log (2)+\left (50 x^2-20 x^3+2 x^4\right ) \log ^2(2)\right )\right )}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (-2 x^2+4 x^3\right )} \, dx=-\frac {(-5+x)^2}{x \left (-1+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}+2 x\right )} \]

input

Integrate[(-25 + 100*x - 19*x^2 + E^E^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2)* 
(25 - x^2 + E^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2)*(50*x^2 - 20*x^3 + 2*x^4 
 + (-100*x^2 + 40*x^3 - 4*x^4)*Log[2] + (50*x^2 - 20*x^3 + 2*x^4)*Log[2]^2 
)))/(x^2 + E^(2*E^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2))*x^2 - 4*x^3 + 4*x^4 
 + E^E^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2)*(-2*x^2 + 4*x^3)),x]
 

output

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

3.12.67.3 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.

\(\displaystyle \int \frac {-19 x^2+e^{e^{x^2+x^2 \log ^2(2)-2 x^2 \log (2)}} \left (-x^2+e^{x^2+x^2 \log ^2(2)-2 x^2 \log (2)} \left (2 x^4-20 x^3+50 x^2+\left (2 x^4-20 x^3+50 x^2\right ) \log ^2(2)+\left (-4 x^4+40 x^3-100 x^2\right ) \log (2)\right )+25\right )+100 x-25}{4 x^4-4 x^3+x^2+x^2 e^{2 e^{x^2+x^2 \log ^2(2)-2 x^2 \log (2)}}+\left (4 x^3-2 x^2\right ) e^{e^{x^2+x^2 \log ^2(2)-2 x^2 \log (2)}}} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-19 x^2+e^{e^{x^2+x^2 \log ^2(2)-2 x^2 \log (2)}} \left (-x^2+e^{x^2+x^2 \log ^2(2)-2 x^2 \log (2)} \left (2 x^4-20 x^3+50 x^2+\left (2 x^4-20 x^3+50 x^2\right ) \log ^2(2)+\left (-4 x^4+40 x^3-100 x^2\right ) \log (2)\right )+25\right )+100 x-25}{x^2 \left (-e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-2 x+1\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2^{1-2 x^2} (x-5)^2 (\log (2)-1)^2 \exp \left (x^2 \left (1+\log ^2(2)\right )+2^{-2 x^2} e^{x^2 \left (1+\log ^2(2)\right )}\right )}{\left (e^{2^{-2 x^2} e^{x^2 \left (1+\log ^2(2)\right )}}+2 x-1\right )^2}-\frac {e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}}{\left (e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}+2 x-1\right )^2}+\frac {100}{x \left (e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}+2 x-1\right )^2}+\frac {25 e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}}{x^2 \left (e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}+2 x-1\right )^2}-\frac {25}{x^2 \left (e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}+2 x-1\right )^2}-\frac {19}{\left (e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}+2 x-1\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 25 (1-\log (2))^2 \int \frac {2^{1-2 x^2} \exp \left (\left (1+\log ^2(2)\right ) x^2+2^{-2 x^2} e^{x^2 \left (1+\log ^2(2)\right )}\right )}{\left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx-5 (1-\log (2))^2 \int \frac {2^{2-2 x^2} \exp \left (\left (1+\log ^2(2)\right ) x^2+2^{-2 x^2} e^{x^2 \left (1+\log ^2(2)\right )}\right ) x}{\left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx+(1-\log (2))^2 \int \frac {2^{1-2 x^2} \exp \left (\left (1+\log ^2(2)\right ) x^2+2^{-2 x^2} e^{x^2 \left (1+\log ^2(2)\right )}\right ) x^2}{\left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx-19 \int \frac {1}{\left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx-\int \frac {e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}}{\left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx-25 \int \frac {1}{x^2 \left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx+25 \int \frac {e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}}{x^2 \left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx+100 \int \frac {1}{x \left (2 x+e^{4^{-x^2} e^{x^2 \left (1+\log ^2(2)\right )}}-1\right )^2}dx\)

input

Int[(-25 + 100*x - 19*x^2 + E^E^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2)*(25 - 
x^2 + E^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2)*(50*x^2 - 20*x^3 + 2*x^4 + (-1 
00*x^2 + 40*x^3 - 4*x^4)*Log[2] + (50*x^2 - 20*x^3 + 2*x^4)*Log[2]^2)))/(x 
^2 + E^(2*E^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2))*x^2 - 4*x^3 + 4*x^4 + E^E 
^(x^2 - 2*x^2*Log[2] + x^2*Log[2]^2)*(-2*x^2 + 4*x^3)),x]
 

output

$Aborted
 

3.12.67.3.1 Defintions of rubi rules used

rule 2009

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

rule 7292

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

rule 7293

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

3.12.67.4 Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(\frac {-x^{2}+10 x -25}{x \left (-1+2 x +{\mathrm e}^{{\mathrm e}^{x^{2} \left (\ln \left (2\right )^{2}-2 \ln \left (2\right )+1\right )}}\right )}\)	\(38\)
risch	\(-\frac {x^{2}-10 x +25}{x \left (-1+{\mathrm e}^{\left (\frac {1}{4}\right )^{x^{2}} {\mathrm e}^{x^{2} \left (\ln \left (2\right )^{2}+1\right )}}+2 x \right )}\)	\(39\)
norman	\(\frac {-25+\frac {19 x}{2}+\frac {{\mathrm e}^{{\mathrm e}^{x^{2} \ln \left (2\right )^{2}-2 x^{2} \ln \left (2\right )+x^{2}}} x}{2}}{x \left (-1+{\mathrm e}^{{\mathrm e}^{x^{2} \ln \left (2\right )^{2}-2 x^{2} \ln \left (2\right )+x^{2}}}+2 x \right )}\)	\(62\)

input

int(((((2*x^4-20*x^3+50*x^2)*ln(2)^2+(-4*x^4+40*x^3-100*x^2)*ln(2)+2*x^4-2 
0*x^3+50*x^2)*exp(x^2*ln(2)^2-2*x^2*ln(2)+x^2)-x^2+25)*exp(exp(x^2*ln(2)^2 
-2*x^2*ln(2)+x^2))-19*x^2+100*x-25)/(x^2*exp(exp(x^2*ln(2)^2-2*x^2*ln(2)+x 
^2))^2+(4*x^3-2*x^2)*exp(exp(x^2*ln(2)^2-2*x^2*ln(2)+x^2))+4*x^4-4*x^3+x^2 
),x,method=_RETURNVERBOSE)
 

output

(-x^2+10*x-25)/x/(-1+2*x+exp(exp(x^2*(ln(2)^2-2*ln(2)+1))))
 

3.12.67.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} \left (50 x^2-20 x^3+2 x^4+\left (-100 x^2+40 x^3-4 x^4\right ) \log (2)+\left (50 x^2-20 x^3+2 x^4\right ) \log ^2(2)\right )\right )}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (-2 x^2+4 x^3\right )} \, dx=-\frac {x^{2} - 10 \, x + 25}{2 \, x^{2} + x e^{\left (e^{\left (x^{2} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (2\right ) + x^{2}\right )}\right )} - x} \]

input

integrate(((((2*x^4-20*x^3+50*x^2)*log(2)^2+(-4*x^4+40*x^3-100*x^2)*log(2) 
+2*x^4-20*x^3+50*x^2)*exp(x^2*log(2)^2-2*x^2*log(2)+x^2)-x^2+25)*exp(exp(x 
^2*log(2)^2-2*x^2*log(2)+x^2))-19*x^2+100*x-25)/(x^2*exp(exp(x^2*log(2)^2- 
2*x^2*log(2)+x^2))^2+(4*x^3-2*x^2)*exp(exp(x^2*log(2)^2-2*x^2*log(2)+x^2)) 
+4*x^4-4*x^3+x^2),x, algorithm=\
 

output

-(x^2 - 10*x + 25)/(2*x^2 + x*e^(e^(x^2*log(2)^2 - 2*x^2*log(2) + x^2)) - 
x)
 

3.12.67.6 Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} \left (50 x^2-20 x^3+2 x^4+\left (-100 x^2+40 x^3-4 x^4\right ) \log (2)+\left (50 x^2-20 x^3+2 x^4\right ) \log ^2(2)\right )\right )}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (-2 x^2+4 x^3\right )} \, dx=\frac {- x^{2} + 10 x - 25}{2 x^{2} + x e^{e^{- 2 x^{2} \log {\left (2 \right )} + x^{2} \log {\left (2 \right )}^{2} + x^{2}}} - x} \]

input

integrate(((((2*x**4-20*x**3+50*x**2)*ln(2)**2+(-4*x**4+40*x**3-100*x**2)* 
ln(2)+2*x**4-20*x**3+50*x**2)*exp(x**2*ln(2)**2-2*x**2*ln(2)+x**2)-x**2+25 
)*exp(exp(x**2*ln(2)**2-2*x**2*ln(2)+x**2))-19*x**2+100*x-25)/(x**2*exp(ex 
p(x**2*ln(2)**2-2*x**2*ln(2)+x**2))**2+(4*x**3-2*x**2)*exp(exp(x**2*ln(2)* 
*2-2*x**2*ln(2)+x**2))+4*x**4-4*x**3+x**2),x)
 

output

(-x**2 + 10*x - 25)/(2*x**2 + x*exp(exp(-2*x**2*log(2) + x**2*log(2)**2 + 
x**2)) - x)
 

3.12.67.7 Maxima [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} \left (50 x^2-20 x^3+2 x^4+\left (-100 x^2+40 x^3-4 x^4\right ) \log (2)+\left (50 x^2-20 x^3+2 x^4\right ) \log ^2(2)\right )\right )}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (-2 x^2+4 x^3\right )} \, dx=-\frac {x^{2} - 10 \, x + 25}{2 \, x^{2} + x e^{\left (e^{\left (x^{2} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (2\right ) + x^{2}\right )}\right )} - x} \]

input

integrate(((((2*x^4-20*x^3+50*x^2)*log(2)^2+(-4*x^4+40*x^3-100*x^2)*log(2) 
+2*x^4-20*x^3+50*x^2)*exp(x^2*log(2)^2-2*x^2*log(2)+x^2)-x^2+25)*exp(exp(x 
^2*log(2)^2-2*x^2*log(2)+x^2))-19*x^2+100*x-25)/(x^2*exp(exp(x^2*log(2)^2- 
2*x^2*log(2)+x^2))^2+(4*x^3-2*x^2)*exp(exp(x^2*log(2)^2-2*x^2*log(2)+x^2)) 
+4*x^4-4*x^3+x^2),x, algorithm=\
 

output

-(x^2 - 10*x + 25)/(2*x^2 + x*e^(e^(x^2*log(2)^2 - 2*x^2*log(2) + x^2)) - 
x)
 

3.12.67.8 Giac [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} \left (50 x^2-20 x^3+2 x^4+\left (-100 x^2+40 x^3-4 x^4\right ) \log (2)+\left (50 x^2-20 x^3+2 x^4\right ) \log ^2(2)\right )\right )}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (-2 x^2+4 x^3\right )} \, dx=-\frac {x^{2} - 10 \, x + 25}{2 \, x^{2} + x e^{\left (e^{\left (x^{2} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (2\right ) + x^{2}\right )}\right )} - x} \]

input

integrate(((((2*x^4-20*x^3+50*x^2)*log(2)^2+(-4*x^4+40*x^3-100*x^2)*log(2) 
+2*x^4-20*x^3+50*x^2)*exp(x^2*log(2)^2-2*x^2*log(2)+x^2)-x^2+25)*exp(exp(x 
^2*log(2)^2-2*x^2*log(2)+x^2))-19*x^2+100*x-25)/(x^2*exp(exp(x^2*log(2)^2- 
2*x^2*log(2)+x^2))^2+(4*x^3-2*x^2)*exp(exp(x^2*log(2)^2-2*x^2*log(2)+x^2)) 
+4*x^4-4*x^3+x^2),x, algorithm=\
 

output

-(x^2 - 10*x + 25)/(2*x^2 + x*e^(e^(x^2*log(2)^2 - 2*x^2*log(2) + x^2)) - 
x)
 

3.12.67.9 Mupad [F(-1)]

Timed out. \[ \int \frac {-25+100 x-19 x^2+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (25-x^2+e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)} \left (50 x^2-20 x^3+2 x^4+\left (-100 x^2+40 x^3-4 x^4\right ) \log (2)+\left (50 x^2-20 x^3+2 x^4\right ) \log ^2(2)\right )\right )}{x^2+e^{2 e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} x^2-4 x^3+4 x^4+e^{e^{x^2-2 x^2 \log (2)+x^2 \log ^2(2)}} \left (-2 x^2+4 x^3\right )} \, dx=\int \frac {100\,x+{\mathrm {e}}^{{\mathrm {e}}^{x^2\,{\ln \left (2\right )}^2-2\,x^2\,\ln \left (2\right )+x^2}}\,\left ({\mathrm {e}}^{x^2\,{\ln \left (2\right )}^2-2\,x^2\,\ln \left (2\right )+x^2}\,\left ({\ln \left (2\right )}^2\,\left (2\,x^4-20\,x^3+50\,x^2\right )-\ln \left (2\right )\,\left (4\,x^4-40\,x^3+100\,x^2\right )+50\,x^2-20\,x^3+2\,x^4\right )-x^2+25\right )-19\,x^2-25}{x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2\,{\ln \left (2\right )}^2-2\,x^2\,\ln \left (2\right )+x^2}}-{\mathrm {e}}^{{\mathrm {e}}^{x^2\,{\ln \left (2\right )}^2-2\,x^2\,\ln \left (2\right )+x^2}}\,\left (2\,x^2-4\,x^3\right )+x^2-4\,x^3+4\,x^4} \,d x \]

input

int((100*x + exp(exp(x^2*log(2)^2 - 2*x^2*log(2) + x^2))*(exp(x^2*log(2)^2 
 - 2*x^2*log(2) + x^2)*(log(2)^2*(50*x^2 - 20*x^3 + 2*x^4) - log(2)*(100*x 
^2 - 40*x^3 + 4*x^4) + 50*x^2 - 20*x^3 + 2*x^4) - x^2 + 25) - 19*x^2 - 25) 
/(x^2*exp(2*exp(x^2*log(2)^2 - 2*x^2*log(2) + x^2)) - exp(exp(x^2*log(2)^2 
 - 2*x^2*log(2) + x^2))*(2*x^2 - 4*x^3) + x^2 - 4*x^3 + 4*x^4),x)
 

output

int((100*x + exp(exp(x^2*log(2)^2 - 2*x^2*log(2) + x^2))*(exp(x^2*log(2)^2 
 - 2*x^2*log(2) + x^2)*(log(2)^2*(50*x^2 - 20*x^3 + 2*x^4) - log(2)*(100*x 
^2 - 40*x^3 + 4*x^4) + 50*x^2 - 20*x^3 + 2*x^4) - x^2 + 25) - 19*x^2 - 25) 
/(x^2*exp(2*exp(x^2*log(2)^2 - 2*x^2*log(2) + x^2)) - exp(exp(x^2*log(2)^2 
 - 2*x^2*log(2) + x^2))*(2*x^2 - 4*x^3) + x^2 - 4*x^3 + 4*x^4), x)