Integrand size = 150, antiderivative size = 27 \[ \int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx=\frac {1}{5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)} \]
Timed out. \[ \int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx=\text {\$Aborted} \]
Integrate[(-x^3 + E^(-2 + x + (E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*((2 - x) *Log[2] + E^(5/x)*(5*x - x^3)*Log[2]))/(25*x^3 + 10*x^4 + x^5 + E^((E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*(10*x^3 + 2*x^4)*Log[2] + E^((2*E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*x^3*Log[2]^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.
| \(\displaystyle \int \frac {e^{\frac {e^{x-2} \left (e^{5/x} x^2+1\right )}{x^2}+x-2} \left (e^{5/x} \left (5 x-x^3\right ) \log (2)+(2-x) \log (2)\right )-x^3}{x^5+10 x^4+25 x^3+e^{\frac {2 e^{x-2} \left (e^{5/x} x^2+1\right )}{x^2}} x^3 \log ^2(2)+e^{\frac {e^{x-2} \left (e^{5/x} x^2+1\right )}{x^2}} \left (2 x^4+10 x^3\right ) \log (2)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{\frac {e^{x-2} \left (e^{5/x} x^2+1\right )}{x^2}+x-2} \left (e^{5/x} \left (5 x-x^3\right ) \log (2)+(2-x) \log (2)\right )-x^3}{x^3 \left (e^{e^{x-2} \left (\frac {1}{x^2}+e^{5/x}\right )} \log (2)+x+5\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {1}{\left (e^{e^{x-2} \left (\frac {1}{x^2}+e^{5/x}\right )} \log (2)+x+5\right )^2}-\frac {e^{e^{x-2} \left (\frac {1}{x^2}+e^{5/x}\right )+x-2} \left (e^{5/x} x^3-5 e^{5/x} x+x-2\right ) \log (2)}{x^3 \left (e^{e^{x-2} \left (\frac {1}{x^2}+e^{5/x}\right )} \log (2)+x+5\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (-\frac {1}{\left (e^{e^{x-2} \left (\frac {1}{x^2}+e^{5/x}\right )} \log (2)+x+5\right )^2}-\frac {e^{e^{x-2} \left (\frac {1}{x^2}+e^{5/x}\right )+x-2} \left (e^{5/x} x^3-5 e^{5/x} x+x-2\right ) \log (2)}{x^3 \left (e^{e^{x-2} \left (\frac {1}{x^2}+e^{5/x}\right )} \log (2)+x+5\right )^2}\right )dx\) |
Int[(-x^3 + E^(-2 + x + (E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*((2 - x)*Log[2 ] + E^(5/x)*(5*x - x^3)*Log[2]))/(25*x^3 + 10*x^4 + x^5 + E^((E^(-2 + x)*( 1 + E^(5/x)*x^2))/x^2)*(10*x^3 + 2*x^4)*Log[2] + E^((2*E^(-2 + x)*(1 + E^( 5/x)*x^2))/x^2)*x^3*Log[2]^2),x]
3.3.100.3.1 Defintions of rubi rules used
Time = 28.83 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
| method | result | size |
| risch | \(\frac {1}{\ln \left (2\right ) {\mathrm e}^{\frac {\left (x^{2} {\mathrm e}^{\frac {5}{x}}+1\right ) {\mathrm e}^{-2+x}}{x^{2}}}+x +5}\) | \(30\) |
| parallelrisch | \(\frac {1}{\ln \left (2\right ) {\mathrm e}^{\frac {\left (x^{2} {\mathrm e}^{\frac {5}{x}}+1\right ) {\mathrm e}^{-2+x}}{x^{2}}}+x +5}\) | \(30\) |
int((((-x^3+5*x)*ln(2)*exp(5/x)+(2-x)*ln(2))*exp(-2+x)*exp((x^2*exp(5/x)+1 )*exp(-2+x)/x^2)-x^3)/(x^3*ln(2)^2*exp((x^2*exp(5/x)+1)*exp(-2+x)/x^2)^2+( 2*x^4+10*x^3)*ln(2)*exp((x^2*exp(5/x)+1)*exp(-2+x)/x^2)+x^5+10*x^4+25*x^3) ,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx=\frac {e^{\left (x - 2\right )}}{{\left (x + 5\right )} e^{\left (x - 2\right )} + e^{\left (\frac {x^{3} - 2 \, x^{2} + {\left (x^{2} e^{\frac {5}{x}} + 1\right )} e^{\left (x - 2\right )}}{x^{2}}\right )} \log \left (2\right )} \]
integrate((((-x^3+5*x)*log(2)*exp(5/x)+(2-x)*log(2))*exp(-2+x)*exp((x^2*ex p(5/x)+1)*exp(-2+x)/x^2)-x^3)/(x^3*log(2)^2*exp((x^2*exp(5/x)+1)*exp(-2+x) /x^2)^2+(2*x^4+10*x^3)*log(2)*exp((x^2*exp(5/x)+1)*exp(-2+x)/x^2)+x^5+10*x ^4+25*x^3),x, algorithm=\
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx=\frac {1}{x + e^{\frac {\left (x^{2} e^{\frac {5}{x}} + 1\right ) e^{x - 2}}{x^{2}}} \log {\left (2 \right )} + 5} \]
integrate((((-x**3+5*x)*ln(2)*exp(5/x)+(2-x)*ln(2))*exp(-2+x)*exp((x**2*ex p(5/x)+1)*exp(-2+x)/x**2)-x**3)/(x**3*ln(2)**2*exp((x**2*exp(5/x)+1)*exp(- 2+x)/x**2)**2+(2*x**4+10*x**3)*ln(2)*exp((x**2*exp(5/x)+1)*exp(-2+x)/x**2) +x**5+10*x**4+25*x**3),x)
Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx=\frac {1}{e^{\left (\frac {e^{\left (x - 2\right )}}{x^{2}} + e^{\left (x + \frac {5}{x} - 2\right )}\right )} \log \left (2\right ) + x + 5} \]
integrate((((-x^3+5*x)*log(2)*exp(5/x)+(2-x)*log(2))*exp(-2+x)*exp((x^2*ex p(5/x)+1)*exp(-2+x)/x^2)-x^3)/(x^3*log(2)^2*exp((x^2*exp(5/x)+1)*exp(-2+x) /x^2)^2+(2*x^4+10*x^3)*log(2)*exp((x^2*exp(5/x)+1)*exp(-2+x)/x^2)+x^5+10*x ^4+25*x^3),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (24) = 48\).
Time = 1.06 (sec) , antiderivative size = 564, normalized size of antiderivative = 20.89 \[ \int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx=\frac {x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - x^{3} e^{2} + 5 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{2} e^{x} - 5 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 3 \, x e^{x} - 25 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - 10 \, e^{x}}{x^{5} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \left (2\right ) - x^{4} e^{2} + 10 \, x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 5 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \left (2\right ) - x^{3} e^{\left (\frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \left (2\right ) - 5 \, x^{3} e^{2} + x^{3} e^{x} + 20 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{2} e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \left (2\right ) - 5 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \left (2\right ) + 8 \, x^{2} e^{x} - 50 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 3 \, x e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \left (2\right ) - 25 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \left (2\right ) + 5 \, x e^{x} - 125 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - 10 \, e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \left (2\right ) - 50 \, e^{x}} \]
integrate((((-x^3+5*x)*log(2)*exp(5/x)+(2-x)*log(2))*exp(-2+x)*exp((x^2*ex p(5/x)+1)*exp(-2+x)/x^2)-x^3)/(x^3*log(2)^2*exp((x^2*exp(5/x)+1)*exp(-2+x) /x^2)^2+(2*x^4+10*x^3)*log(2)*exp((x^2*exp(5/x)+1)*exp(-2+x)/x^2)+x^5+10*x ^4+25*x^3),x, algorithm=\
(x^4*e^((x^2 - 2*x + 5)/x + 2) - x^3*e^2 + 5*x^3*e^((x^2 - 2*x + 5)/x + 2) + x^2*e^x - 5*x^2*e^((x^2 - 2*x + 5)/x + 2) + 3*x*e^x - 25*x*e^((x^2 - 2* x + 5)/x + 2) - 10*e^x)/(x^5*e^((x^2 - 2*x + 5)/x + 2) + x^4*e^((x^2 - 2*x + 5)/x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log(2) - x^4*e^ 2 + 10*x^4*e^((x^2 - 2*x + 5)/x + 2) + 5*x^3*e^((x^2 - 2*x + 5)/x + (x^2*e ^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log(2) - x^3*e^((x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log(2) - 5*x^3*e^2 + x^3*e^x + 20*x^3*e^ ((x^2 - 2*x + 5)/x + 2) + x^2*e^(x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2 ))/x^2)*log(2) - 5*x^2*e^((x^2 - 2*x + 5)/x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log(2) + 8*x^2*e^x - 50*x^2*e^((x^2 - 2*x + 5)/x + 2) + 3*x*e^(x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2)*log(2) - 25*x*e ^((x^2 - 2*x + 5)/x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log (2) + 5*x*e^x - 125*x*e^((x^2 - 2*x + 5)/x + 2) - 10*e^(x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2)*log(2) - 50*e^x)
Time = 7.76 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.44 \[ \int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx=\frac {x\,\left (\ln \left (2\right )-{\mathrm {e}}^{5/x}\,\ln \left (32\right )\right )-\ln \left (4\right )+x^3\,{\mathrm {e}}^{5/x}\,\ln \left (2\right )}{{\ln \left (2\right )}^2\,\left ({\mathrm {e}}^{{\mathrm {e}}^{-2}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^x+\frac {{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x^2}}+\frac {x+5}{\ln \left (2\right )}\right )\,\left (x-5\,x\,{\mathrm {e}}^{5/x}+x^3\,{\mathrm {e}}^{5/x}-2\right )} \]
int(-(x^3 + exp((exp(x - 2)*(x^2*exp(5/x) + 1))/x^2)*exp(x - 2)*(log(2)*(x - 2) - exp(5/x)*log(2)*(5*x - x^3)))/(25*x^3 + 10*x^4 + x^5 + x^3*exp((2* exp(x - 2)*(x^2*exp(5/x) + 1))/x^2)*log(2)^2 + exp((exp(x - 2)*(x^2*exp(5/ x) + 1))/x^2)*log(2)*(10*x^3 + 2*x^4)),x)