Integrand size = 57, antiderivative size = 29 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=\log \left (e^2+\frac {4+7 e^{-x}+x-\frac {2+x}{x}+\log (x)}{e}\right ) \] Output:
ln(exp(2)+(ln(x)+x-(2+x)/x+4+7/exp(x))/exp(1))
Time = 0.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=-x-\log (x)+\log \left (-2 e^x+7 x+3 e^x x+e^{3+x} x+e^x x^2+e^x x \log (x)\right ) \] Input:
Integrate[(-7*x^2 + E^x*(2 + x + x^2))/(7*x^2 + E^x*(-2*x + 3*x^2 + E^3*x^ 2 + x^3) + E^x*x^2*Log[x]),x]
Output:
-x - Log[x] + Log[-2*E^x + 7*x + 3*E^x*x + E^(3 + x)*x + E^x*x^2 + E^x*x*L og[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.
\(\displaystyle \int \frac {e^x \left (x^2+x+2\right )-7 x^2}{7 x^2+e^x x^2 \log (x)+e^x \left (x^3+e^3 x^2+3 x^2-2 x\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-x^2-x-2}{x \left (-x^2-3 \left (1+\frac {e^3}{3}\right ) x-x \log (x)+2\right )}+\frac {7 \left (-x^3-4 \left (1+\frac {e^3}{4}\right ) x^2-x^2 \log (x)+x-2\right )}{\left (-x^2-3 \left (1+\frac {e^3}{3}\right ) x-x \log (x)+2\right ) \left (-e^x x^2-3 \left (1+\frac {e^3}{3}\right ) e^x x-7 x+2 e^x-e^x x \log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x^2+\log (x) x+3 \left (1+\frac {e^3}{3}\right ) x-2}dx+2 \int \frac {1}{x \left (x^2+\log (x) x+3 \left (1+\frac {e^3}{3}\right ) x-2\right )}dx+\int \frac {x}{x^2+\log (x) x+3 \left (1+\frac {e^3}{3}\right ) x-2}dx+7 \int \frac {x}{\left (-x^2-\log (x) x-3 \left (1+\frac {e^3}{3}\right ) x+2\right ) \left (-e^x x^2-e^x \log (x) x-3 e^x \left (1+\frac {e^3}{3}\right ) x-7 x+2 e^x\right )}dx-7 \left (4+e^3\right ) \int \frac {x^2}{\left (-x^2-\log (x) x-3 \left (1+\frac {e^3}{3}\right ) x+2\right ) \left (-e^x x^2-e^x \log (x) x-3 e^x \left (1+\frac {e^3}{3}\right ) x-7 x+2 e^x\right )}dx+14 \int \frac {1}{\left (x^2+\log (x) x+3 \left (1+\frac {e^3}{3}\right ) x-2\right ) \left (-e^x x^2-e^x \log (x) x-3 e^x \left (1+\frac {e^3}{3}\right ) x-7 x+2 e^x\right )}dx+7 \int \frac {x^2 \log (x)}{\left (x^2+\log (x) x+3 \left (1+\frac {e^3}{3}\right ) x-2\right ) \left (-e^x x^2-e^x \log (x) x-3 e^x \left (1+\frac {e^3}{3}\right ) x-7 x+2 e^x\right )}dx+7 \int \frac {x^3}{\left (x^2+\log (x) x+3 \left (1+\frac {e^3}{3}\right ) x-2\right ) \left (-e^x x^2-e^x \log (x) x-3 e^x \left (1+\frac {e^3}{3}\right ) x-7 x+2 e^x\right )}dx\) |
Input:
Int[(-7*x^2 + E^x*(2 + x + x^2))/(7*x^2 + E^x*(-2*x + 3*x^2 + E^3*x^2 + x^ 3) + E^x*x^2*Log[x]),x]
Output:
$Aborted
Time = 0.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\ln \left (\frac {x^{2}+3 x +x \,{\mathrm e}^{3}-2+x \ln \left (x \right )+7 x \,{\mathrm e}^{-x}}{x}\right )\) | \(29\) |
norman | \(-\ln \left (x \right )-x +\ln \left ({\mathrm e}^{x} {\mathrm e} \,{\mathrm e}^{2} x +x \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{x} x^{2}+3 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+7 x \right )\) | \(43\) |
parallelrisch | \(-\ln \left (x \right )-x +\ln \left ({\mathrm e}^{x} {\mathrm e} \,{\mathrm e}^{2} x +x \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{x} x^{2}+3 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+7 x \right )\) | \(43\) |
Input:
int(((x^2+x+2)*exp(x)-7*x^2)/(x^2*exp(x)*ln(x)+(x^2*exp(1)*exp(2)+x^3+3*x^ 2-2*x)*exp(x)+7*x^2),x,method=_RETURNVERBOSE)
Output:
ln((x^2+3*x+x*exp(3)-2+x*ln(x)+7*x*exp(-x))/x)
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=\log \left (\frac {{\left (x e^{x} \log \left (x\right ) + {\left (x^{2} + x e^{3} + 3 \, x - 2\right )} e^{x} + 7 \, x\right )} e^{\left (-x\right )}}{x}\right ) \] Input:
integrate(((x^2+x+2)*exp(x)-7*x^2)/(x^2*exp(x)*log(x)+(x^2*exp(1)*exp(2)+x ^3+3*x^2-2*x)*exp(x)+7*x^2),x, algorithm="fricas")
Output:
log((x*e^x*log(x) + (x^2 + x*e^3 + 3*x - 2)*e^x + 7*x)*e^(-x)/x)
Time = 1.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=- x + \log {\left (\frac {7 x}{x^{2} + x \log {\left (x \right )} + 3 x + x e^{3} - 2} + e^{x} \right )} + \log {\left (\log {\left (x \right )} + \frac {x^{2} + 3 x + x e^{3} - 2}{x} \right )} \] Input:
integrate(((x**2+x+2)*exp(x)-7*x**2)/(x**2*exp(x)*ln(x)+(x**2*exp(1)*exp(2 )+x**3+3*x**2-2*x)*exp(x)+7*x**2),x)
Output:
-x + log(7*x/(x**2 + x*log(x) + 3*x + x*exp(3) - 2) + exp(x)) + log(log(x) + (x**2 + 3*x + x*exp(3) - 2)/x)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (26) = 52\).
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.24 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=-x + \log \left (\frac {{\left (x^{2} + x {\left (e^{3} + 3\right )} + x \log \left (x\right ) - 2\right )} e^{x} + 7 \, x}{x^{2} + x {\left (e^{3} + 3\right )} + x \log \left (x\right ) - 2}\right ) + \log \left (\frac {x^{2} + x {\left (e^{3} + 3\right )} + x \log \left (x\right ) - 2}{x}\right ) \] Input:
integrate(((x^2+x+2)*exp(x)-7*x^2)/(x^2*exp(x)*log(x)+(x^2*exp(1)*exp(2)+x ^3+3*x^2-2*x)*exp(x)+7*x^2),x, algorithm="maxima")
Output:
-x + log(((x^2 + x*(e^3 + 3) + x*log(x) - 2)*e^x + 7*x)/(x^2 + x*(e^3 + 3) + x*log(x) - 2)) + log((x^2 + x*(e^3 + 3) + x*log(x) - 2)/x)
Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=-x + \log \left (x^{2} e^{x} + x e^{x} \log \left (x\right ) + x e^{\left (x + 3\right )} + 3 \, x e^{x} + 7 \, x - 2 \, e^{x}\right ) - \log \left (x\right ) \] Input:
integrate(((x^2+x+2)*exp(x)-7*x^2)/(x^2*exp(x)*log(x)+(x^2*exp(1)*exp(2)+x ^3+3*x^2-2*x)*exp(x)+7*x^2),x, algorithm="giac")
Output:
-x + log(x^2*e^x + x*e^x*log(x) + x*e^(x + 3) + 3*x*e^x + 7*x - 2*e^x) - l og(x)
Time = 4.63 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=\ln \left (x+7\,{\mathrm {e}}^{-x}+{\mathrm {e}}^3+\ln \left (x\right )-\frac {2}{x}+3\right ) \] Input:
int(-(7*x^2 - exp(x)*(x + x^2 + 2))/(exp(x)*(x^2*exp(3) - 2*x + 3*x^2 + x^ 3) + 7*x^2 + x^2*exp(x)*log(x)),x)
Output:
log(x + 7*exp(-x) + exp(3) + log(x) - 2/x + 3)
Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {-7 x^2+e^x \left (2+x+x^2\right )}{7 x^2+e^x \left (-2 x+3 x^2+e^3 x^2+x^3\right )+e^x x^2 \log (x)} \, dx=\mathrm {log}\left (e^{x} \mathrm {log}\left (x \right ) x +e^{x} e^{3} x +e^{x} x^{2}+3 e^{x} x -2 e^{x}+7 x \right )-\mathrm {log}\left (x \right )-x \] Input:
int(((x^2+x+2)*exp(x)-7*x^2)/(x^2*exp(x)*log(x)+(x^2*exp(1)*exp(2)+x^3+3*x ^2-2*x)*exp(x)+7*x^2),x)
Output:
log(e**x*log(x)*x + e**x*e**3*x + e**x*x**2 + 3*e**x*x - 2*e**x + 7*x) - l og(x) - x