Integrand size = 148, antiderivative size = 30 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=5 \left (3+x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right )\right ) \]
Time = 0.60 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=5 x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right ) \]
Integrate[(25*x^2 + 20*x^3 + E^(4*x)*(25 + 20*x) + E^(2*x)*(-50*x - 40*x^2 ) + E^E^(2/(E^(2*x) - x))*(5*E^(4*x) - 10*E^(2*x)*x + 5*x^2 + E^(2/(E^(2*x ) - x))*(10*x - 20*E^(2*x)*x)) + (5*E^(4*x) - 10*E^(2*x)*x + 5*x^2)*Log[x] )/(E^(4*x) - 2*E^(2*x)*x + x^2),x]
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.
\(\displaystyle \int \frac {20 x^3+25 x^2+e^{2 x} \left (-40 x^2-50 x\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 x^2-10 e^{2 x} x+5 e^{4 x}+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 x^2-10 e^{2 x} x+5 e^{4 x}\right ) \log (x)+e^{4 x} (20 x+25)}{x^2-2 e^{2 x} x+e^{4 x}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {20 x^3+25 x^2+e^{2 x} \left (-40 x^2-50 x\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 x^2-10 e^{2 x} x+5 e^{4 x}+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 x^2-10 e^{2 x} x+5 e^{4 x}\right ) \log (x)+e^{4 x} (20 x+25)}{\left (e^{2 x}-x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {10 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} (2 x-1) x}{\left (e^{2 x}-x\right )^2}-\frac {20 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x}+5 \left (4 x+e^{e^{\frac {2}{e^{2 x}-x}}}+\log (x)+5\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x^2}{\left (e^{2 x}-x\right )^2}dx+5 \int e^{e^{\frac {2}{e^{2 x}-x}}}dx+10 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{\left (e^{2 x}-x\right )^2}dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x}dx+10 x^2+20 x+5 x \log (x)\) |
Int[(25*x^2 + 20*x^3 + E^(4*x)*(25 + 20*x) + E^(2*x)*(-50*x - 40*x^2) + E^ E^(2/(E^(2*x) - x))*(5*E^(4*x) - 10*E^(2*x)*x + 5*x^2 + E^(2/(E^(2*x) - x) )*(10*x - 20*E^(2*x)*x)) + (5*E^(4*x) - 10*E^(2*x)*x + 5*x^2)*Log[x])/(E^( 4*x) - 2*E^(2*x)*x + x^2),x]
3.15.64.3.1 Defintions of rubi rules used
Time = 52.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
risch | \(10 x^{2}+5 x \ln \left (x \right )+5 x \,{\mathrm e}^{{\mathrm e}^{-\frac {2}{x -{\mathrm e}^{2 x}}}}+20 x\) | \(32\) |
parallelrisch | \(10 x^{2}+5 x \ln \left (x \right )+5 x \,{\mathrm e}^{{\mathrm e}^{\frac {2}{{\mathrm e}^{2 x}-x}}}+20 x\) | \(32\) |
int((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5 *x^2)*exp(exp(2/(exp(x)^2-x)))+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*ln(x)+(20* x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2*x*exp(x) ^2+x^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \left (x\right ) + 20 \, x \]
integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp( x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log( x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2*x *exp(x)^2+x^2),x, algorithm=\
Timed out. \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\text {Timed out} \]
integrate((((-20*x*exp(x)**2+10*x)*exp(2/(exp(x)**2-x))+5*exp(x)**4-10*x*e xp(x)**2+5*x**2)*exp(exp(2/(exp(x)**2-x)))+(5*exp(x)**4-10*x*exp(x)**2+5*x **2)*ln(x)+(20*x+25)*exp(x)**4+(-40*x**2-50*x)*exp(x)**2+20*x**3+25*x**2)/ (exp(x)**4-2*x*exp(x)**2+x**2),x)
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \left (x\right ) + 20 \, x \]
integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp( x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log( x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2*x *exp(x)^2+x^2),x, algorithm=\
\[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\int { \frac {5 \, {\left (4 \, x^{3} + 5 \, x^{2} + {\left (4 \, x + 5\right )} e^{\left (4 \, x\right )} - 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} + {\left (x^{2} - 2 \, x e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x e^{\left (2 \, x\right )} - x\right )} e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )} + e^{\left (4 \, x\right )}\right )} e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + {\left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )}\right )} \log \left (x\right )\right )}}{x^{2} - 2 \, x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )}} \,d x } \]
integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp( x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log( x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2*x *exp(x)^2+x^2),x, algorithm=\
integrate(5*(4*x^3 + 5*x^2 + (4*x + 5)*e^(4*x) - 2*(4*x^2 + 5*x)*e^(2*x) + (x^2 - 2*x*e^(2*x) - 2*(2*x*e^(2*x) - x)*e^(-2/(x - e^(2*x))) + e^(4*x))* e^(e^(-2/(x - e^(2*x)))) + (x^2 - 2*x*e^(2*x) + e^(4*x))*log(x))/(x^2 - 2* x*e^(2*x) + e^(4*x)), x)
Time = 8.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=20\,x+5\,x\,\ln \left (x\right )+5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2}{x-{\mathrm {e}}^{2\,x}}}}+10\,x^2 \]