Integrand size = 93, antiderivative size = 25 \[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\frac {e^{1+5 e^x} (x+\log (x))}{x \left (1+x+x^2\right )} \] Output:
exp(5*exp(x)+1)/(x^2+x+1)/x*(x+ln(x))
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\frac {e^{1+5 e^x} (x+\log (x))}{x \left (1+x+x^2\right )} \] Input:
Integrate[(E^(1 + 5*E^x)*(1 + x - 2*x^3 + E^x*(5*x^2 + 5*x^3 + 5*x^4) + (- 1 - 2*x - 3*x^2 + E^x*(5*x + 5*x^2 + 5*x^3))*Log[x]))/(x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6),x]
Output:
(E^(1 + 5*E^x)*(x + Log[x]))/(x*(1 + x + x^2))
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.
| \(\displaystyle \int \frac {e^{5 e^x+1} \left (-2 x^3+\left (-3 x^2+e^x \left (5 x^3+5 x^2+5 x\right )-2 x-1\right ) \log (x)+e^x \left (5 x^4+5 x^3+5 x^2\right )+x+1\right )}{x^6+2 x^5+3 x^4+2 x^3+x^2} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{5 e^x+1} \left (-2 x^3+\left (-3 x^2+e^x \left (5 x^3+5 x^2+5 x\right )-2 x-1\right ) \log (x)+e^x \left (5 x^4+5 x^3+5 x^2\right )+x+1\right )}{x^2 \left (x^4+2 x^3+3 x^2+2 x+1\right )}dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {4 i e^{5 e^x+1} \left (-2 x^3+\left (-3 x^2+e^x \left (5 x^3+5 x^2+5 x\right )-2 x-1\right ) \log (x)+e^x \left (5 x^4+5 x^3+5 x^2\right )+x+1\right )}{3 \sqrt {3} x^2 \left (2 x+i \sqrt {3}+1\right )}+\frac {4 i e^{5 e^x+1} \left (-2 x^3+\left (-3 x^2+e^x \left (5 x^3+5 x^2+5 x\right )-2 x-1\right ) \log (x)+e^x \left (5 x^4+5 x^3+5 x^2\right )+x+1\right )}{3 \sqrt {3} \left (-2 x+i \sqrt {3}-1\right ) x^2}-\frac {4 e^{5 e^x+1} \left (-2 x^3+\left (-3 x^2+e^x \left (5 x^3+5 x^2+5 x\right )-2 x-1\right ) \log (x)+e^x \left (5 x^4+5 x^3+5 x^2\right )+x+1\right )}{3 \left (-2 x+i \sqrt {3}-1\right )^2 x^2}-\frac {4 e^{5 e^x+1} \left (-2 x^3+\left (-3 x^2+e^x \left (5 x^3+5 x^2+5 x\right )-2 x-1\right ) \log (x)+e^x \left (5 x^4+5 x^3+5 x^2\right )+x+1\right )}{3 x^2 \left (2 x+i \sqrt {3}+1\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{5 e^x+1} \left (5 e^x x^4+\left (5 e^x-2\right ) x^3+5 e^x x^2+\left (5 e^x x^3+\left (5 e^x-3\right ) x^2+\left (5 e^x-2\right ) x-1\right ) \log (x)+x+1\right )}{x^2 \left (x^2+x+1\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {2 e^{5 e^x+1} x}{\left (x^2+x+1\right )^2}+\frac {e^{5 e^x+1}}{\left (x^2+x+1\right )^2 x}+\frac {e^{5 e^x+1}}{\left (x^2+x+1\right )^2 x^2}-\frac {3 e^{5 e^x+1} \log (x)}{\left (x^2+x+1\right )^2}-\frac {2 e^{5 e^x+1} \log (x)}{\left (x^2+x+1\right )^2 x}+\frac {5 e^{x+5 e^x+1} (x+\log (x))}{\left (x^2+x+1\right ) x}-\frac {e^{5 e^x+1} \log (x)}{\left (x^2+x+1\right )^2 x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{3} \left (1-i \sqrt {3}\right ) \log (x) \int \frac {e^{1+5 e^x}}{\left (-2 x+i \sqrt {3}-1\right )^2}dx+\frac {4}{3} \log (x) \int \frac {e^{1+5 e^x}}{\left (-2 x+i \sqrt {3}-1\right )^2}dx-\frac {4}{3} \left (1-i \sqrt {3}\right ) \int \frac {e^{1+5 e^x}}{\left (-2 x+i \sqrt {3}-1\right )^2}dx+\frac {4}{3} \int \frac {e^{1+5 e^x}}{\left (-2 x+i \sqrt {3}-1\right )^2}dx-\frac {2 i \log (x) \int \frac {e^{1+5 e^x}}{-2 x+i \sqrt {3}-1}dx}{\sqrt {3}}+\frac {10 i \int \frac {e^{x+5 e^x+1}}{-2 x+i \sqrt {3}-1}dx}{\sqrt {3}}-\log (x) \int \frac {e^{1+5 e^x}}{x^2}dx+\int \frac {e^{1+5 e^x}}{x^2}dx-\int \frac {e^{1+5 e^x}}{x}dx+5 \log (x) \int \frac {e^{x+5 e^x+1}}{x}dx+\frac {2}{3} \left (3-i \sqrt {3}\right ) \log (x) \int \frac {e^{1+5 e^x}}{2 x-i \sqrt {3}+1}dx-2 \log (x) \int \frac {e^{1+5 e^x}}{2 x-i \sqrt {3}+1}dx-\frac {1}{3} \left (3-i \sqrt {3}\right ) \int \frac {e^{1+5 e^x}}{2 x-i \sqrt {3}+1}dx+2 \int \frac {e^{1+5 e^x}}{2 x-i \sqrt {3}+1}dx-\frac {5}{3} \left (3-i \sqrt {3}\right ) \log (x) \int \frac {e^{x+5 e^x+1}}{2 x-i \sqrt {3}+1}dx+\frac {2}{3} \left (1+i \sqrt {3}\right ) \log (x) \int \frac {e^{1+5 e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx+\frac {4}{3} \log (x) \int \frac {e^{1+5 e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx-\frac {4}{3} \left (1+i \sqrt {3}\right ) \int \frac {e^{1+5 e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx+\frac {4}{3} \int \frac {e^{1+5 e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx+\frac {2}{3} \left (3+i \sqrt {3}\right ) \log (x) \int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx-\frac {2 i \log (x) \int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx}{\sqrt {3}}-2 \log (x) \int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx-\frac {1}{3} \left (3+i \sqrt {3}\right ) \int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx+2 \int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx-\frac {5}{3} \left (3+i \sqrt {3}\right ) \log (x) \int \frac {e^{x+5 e^x+1}}{2 x+i \sqrt {3}+1}dx+\frac {10 i \int \frac {e^{x+5 e^x+1}}{2 x+i \sqrt {3}+1}dx}{\sqrt {3}}+\frac {2 i \int \frac {\int \frac {e^{1+5 e^x}}{-2 x+i \sqrt {3}-1}dx}{x}dx}{\sqrt {3}}+\frac {4}{3} \left (1+i \sqrt {3}\right ) \int \frac {\int -\frac {e^{1+5 e^x}}{\left (2 i x+\sqrt {3}+i\right )^2}dx}{x}dx+\frac {2}{3} \left (1-i \sqrt {3}\right ) \int \frac {\int -\frac {e^{1+5 e^x}}{\left (2 i x+\sqrt {3}+i\right )^2}dx}{x}dx-4 \int \frac {\int -\frac {e^{1+5 e^x}}{\left (2 i x+\sqrt {3}+i\right )^2}dx}{x}dx+\int \frac {\int \frac {e^{1+5 e^x}}{x^2}dx}{x}dx-5 \int \frac {\int \frac {e^{x+5 e^x+1}}{x}dx}{x}dx-\frac {2}{3} \left (3-i \sqrt {3}\right ) \int \frac {\int \frac {e^{1+5 e^x}}{2 x-i \sqrt {3}+1}dx}{x}dx+2 \int \frac {\int \frac {e^{1+5 e^x}}{2 x-i \sqrt {3}+1}dx}{x}dx+\frac {5}{3} \left (3-i \sqrt {3}\right ) \int \frac {\int \frac {e^{x+5 e^x+1}}{2 x-i \sqrt {3}+1}dx}{x}dx+\frac {2}{3} \left (1+i \sqrt {3}\right ) \int \frac {\int \frac {e^{1+5 e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx}{x}dx+\frac {4}{3} \left (1-i \sqrt {3}\right ) \int \frac {\int \frac {e^{1+5 e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx}{x}dx-4 \int \frac {\int \frac {e^{1+5 e^x}}{\left (2 x+i \sqrt {3}+1\right )^2}dx}{x}dx-\frac {2}{9} \left (9+5 i \sqrt {3}\right ) \int \frac {\int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx}{x}dx+\frac {2}{9} \left (9-i \sqrt {3}\right ) \int \frac {\int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx}{x}dx+\frac {4 i \int \frac {\int \frac {e^{1+5 e^x}}{2 x+i \sqrt {3}+1}dx}{x}dx}{\sqrt {3}}+\frac {5}{3} \left (3+i \sqrt {3}\right ) \int \frac {\int \frac {e^{x+5 e^x+1}}{2 x+i \sqrt {3}+1}dx}{x}dx\) |
Input:
Int[(E^(1 + 5*E^x)*(1 + x - 2*x^3 + E^x*(5*x^2 + 5*x^3 + 5*x^4) + (-1 - 2* x - 3*x^2 + E^x*(5*x + 5*x^2 + 5*x^3))*Log[x]))/(x^2 + 2*x^3 + 3*x^4 + 2*x ^5 + x^6),x]
Output:
$Aborted
Time = 19.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(\frac {{\mathrm e}^{5 \,{\mathrm e}^{x}+1} \left (x +\ln \left (x \right )\right )}{\left (x^{2}+x +1\right ) x}\) | \(24\) |
| parallelrisch | \(\frac {{\mathrm e}^{5 \,{\mathrm e}^{x}+1} x +\ln \left (x \right ) {\mathrm e}^{5 \,{\mathrm e}^{x}+1}}{x \left (x^{2}+x +1\right )}\) | \(33\) |
Input:
int((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*ln(x)+(5*x^4+5*x^3+5*x^2)*exp( x)-2*x^3+x+1)*exp(5*exp(x)+1)/(x^6+2*x^5+3*x^4+2*x^3+x^2),x,method=_RETURN VERBOSE)
Output:
exp(5*exp(x)+1)/(x^2+x+1)/x*(x+ln(x))
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\frac {{\left (x + \log \left (x\right )\right )} e^{\left (5 \, e^{x} + 1\right )}}{x^{3} + x^{2} + x} \] Input:
integrate((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*log(x)+(5*x^4+5*x^3+5*x^ 2)*exp(x)-2*x^3+x+1)*exp(5*exp(x)+1)/(x^6+2*x^5+3*x^4+2*x^3+x^2),x, algori thm="fricas")
Output:
(x + log(x))*e^(5*e^x + 1)/(x^3 + x^2 + x)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\frac {\left (x + \log {\left (x \right )}\right ) e^{5 e^{x} + 1}}{x^{3} + x^{2} + x} \] Input:
integrate((((5*x**3+5*x**2+5*x)*exp(x)-3*x**2-2*x-1)*ln(x)+(5*x**4+5*x**3+ 5*x**2)*exp(x)-2*x**3+x+1)*exp(5*exp(x)+1)/(x**6+2*x**5+3*x**4+2*x**3+x**2 ),x)
Output:
(x + log(x))*exp(5*exp(x) + 1)/(x**3 + x**2 + x)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\frac {{\left (x e + e \log \left (x\right )\right )} e^{\left (5 \, e^{x}\right )}}{x^{3} + x^{2} + x} \] Input:
integrate((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*log(x)+(5*x^4+5*x^3+5*x^ 2)*exp(x)-2*x^3+x+1)*exp(5*exp(x)+1)/(x^6+2*x^5+3*x^4+2*x^3+x^2),x, algori thm="maxima")
Output:
(x*e + e*log(x))*e^(5*e^x)/(x^3 + x^2 + x)
\[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\int { -\frac {{\left (2 \, x^{3} - 5 \, {\left (x^{4} + x^{3} + x^{2}\right )} e^{x} + {\left (3 \, x^{2} - 5 \, {\left (x^{3} + x^{2} + x\right )} e^{x} + 2 \, x + 1\right )} \log \left (x\right ) - x - 1\right )} e^{\left (5 \, e^{x} + 1\right )}}{x^{6} + 2 \, x^{5} + 3 \, x^{4} + 2 \, x^{3} + x^{2}} \,d x } \] Input:
integrate((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*log(x)+(5*x^4+5*x^3+5*x^ 2)*exp(x)-2*x^3+x+1)*exp(5*exp(x)+1)/(x^6+2*x^5+3*x^4+2*x^3+x^2),x, algori thm="giac")
Output:
integrate(-(2*x^3 - 5*(x^4 + x^3 + x^2)*e^x + (3*x^2 - 5*(x^3 + x^2 + x)*e ^x + 2*x + 1)*log(x) - x - 1)*e^(5*e^x + 1)/(x^6 + 2*x^5 + 3*x^4 + 2*x^3 + x^2), x)
Timed out. \[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\int \frac {{\mathrm {e}}^{5\,{\mathrm {e}}^x+1}\,\left (x+{\mathrm {e}}^x\,\left (5\,x^4+5\,x^3+5\,x^2\right )-\ln \left (x\right )\,\left (2\,x+3\,x^2-{\mathrm {e}}^x\,\left (5\,x^3+5\,x^2+5\,x\right )+1\right )-2\,x^3+1\right )}{x^6+2\,x^5+3\,x^4+2\,x^3+x^2} \,d x \] Input:
int((exp(5*exp(x) + 1)*(x + exp(x)*(5*x^2 + 5*x^3 + 5*x^4) - log(x)*(2*x + 3*x^2 - exp(x)*(5*x + 5*x^2 + 5*x^3) + 1) - 2*x^3 + 1))/(x^2 + 2*x^3 + 3* x^4 + 2*x^5 + x^6),x)
Output:
int((exp(5*exp(x) + 1)*(x + exp(x)*(5*x^2 + 5*x^3 + 5*x^4) - log(x)*(2*x + 3*x^2 - exp(x)*(5*x + 5*x^2 + 5*x^3) + 1) - 2*x^3 + 1))/(x^2 + 2*x^3 + 3* x^4 + 2*x^5 + x^6), x)
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{1+5 e^x} \left (1+x-2 x^3+e^x \left (5 x^2+5 x^3+5 x^4\right )+\left (-1-2 x-3 x^2+e^x \left (5 x+5 x^2+5 x^3\right )\right ) \log (x)\right )}{x^2+2 x^3+3 x^4+2 x^5+x^6} \, dx=\frac {e^{5 e^{x}} e \left (\mathrm {log}\left (x \right )+x \right )}{x \left (x^{2}+x +1\right )} \] Input:
int((((5*x^3+5*x^2+5*x)*exp(x)-3*x^2-2*x-1)*log(x)+(5*x^4+5*x^3+5*x^2)*exp (x)-2*x^3+x+1)*exp(5*exp(x)+1)/(x^6+2*x^5+3*x^4+2*x^3+x^2),x)
Output:
(e**(5*e**x)*e*(log(x) + x))/(x*(x**2 + x + 1))