Integrand size = 59, antiderivative size = 26 \[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx=e^{2-\frac {1}{20} e^{-1+x^2} (1-2 x)+x}-3 \log (x) \] Output:
exp(2-1/20*(1-2*x)*exp(x^2-1))*exp(x)-3*ln(x)
Time = 0.57 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx=e^{2+x+\frac {1}{20} e^{-1+x^2} (-1+2 x)}-3 \log (x) \] Input:
Integrate[(-30 + E^((40 + E^(-1 + x^2)*(-1 + 2*x))/20)*(10*E^x*x + E^(-1 + x + x^2)*(x - x^2 + 2*x^3)))/(10*x),x]
Output:
E^(2 + x + (E^(-1 + x^2)*(-1 + 2*x))/20) - 3*Log[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.
\(\displaystyle \int \frac {e^{\frac {1}{20} \left (e^{x^2-1} (2 x-1)+40\right )} \left (e^{x^2+x-1} \left (2 x^3-x^2+x\right )+10 e^x x\right )-30}{10 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int -\frac {30-e^{\frac {1}{20} \left (40-e^{x^2-1} (1-2 x)\right )} \left (10 e^x x+e^{x^2+x-1} \left (2 x^3-x^2+x\right )\right )}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{10} \int \frac {30-e^{\frac {1}{20} \left (40-e^{x^2-1} (1-2 x)\right )} \left (10 e^x x+e^{x^2+x-1} \left (2 x^3-x^2+x\right )\right )}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{10} \int \left (-\frac {10 \left (e^{x+\frac {1}{20} e^{x^2-1} (2 x-1)+2} x-3\right )}{x}-e^{x^2+x+\frac {1}{20} e^{x^2-1} (2 x-1)+1} \left (2 x^2-x+1\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} \left (10 \int e^{x+\frac {1}{20} e^{x^2-1} (2 x-1)+2}dx+\int e^{x^2+x+\frac {1}{20} e^{x^2-1} (2 x-1)+1}dx-\int e^{x^2+x+\frac {1}{20} e^{x^2-1} (2 x-1)+1} xdx+2 \int e^{x^2+x+\frac {1}{20} e^{x^2-1} (2 x-1)+1} x^2dx-30 \log (x)\right )\) |
Input:
Int[(-30 + E^((40 + E^(-1 + x^2)*(-1 + 2*x))/20)*(10*E^x*x + E^(-1 + x + x ^2)*(x - x^2 + 2*x^3)))/(10*x),x]
Output:
$Aborted
Time = 0.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-3 \ln \left (x \right )+{\mathrm e}^{x} {\mathrm e}^{\frac {\left (-1+2 x \right ) {\mathrm e}^{x^{2}-1}}{20}+2}\) | \(25\) |
parts | \(-3 \ln \left (x \right )+{\mathrm e}^{x} {\mathrm e}^{\frac {\left (-1+2 x \right ) {\mathrm e}^{x^{2}-1}}{20}+2}\) | \(25\) |
risch | \(-3 \ln \left (x \right )+{\mathrm e}^{x +\frac {{\mathrm e}^{\left (-1+x \right ) \left (1+x \right )} x}{10}-\frac {{\mathrm e}^{\left (-1+x \right ) \left (1+x \right )}}{20}+2}\) | \(31\) |
Input:
int(1/10*(((2*x^3-x^2+x)*exp(x)*exp(x^2-1)+10*exp(x)*x)*exp(1/20*(-1+2*x)* exp(x^2-1)+2)-30)/x,x,method=_RETURNVERBOSE)
Output:
-3*ln(x)+exp(x)*exp(1/20*(-1+2*x)*exp(x^2-1)+2)
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx=-{\left (3 \, e^{\left (x^{2} - 1\right )} \log \left (x\right ) - e^{\left (x^{2} + \frac {1}{20} \, {\left (2 \, x - 1\right )} e^{\left (x^{2} - 1\right )} + x + 1\right )}\right )} e^{\left (-x^{2} + 1\right )} \] Input:
integrate(1/10*(((2*x^3-x^2+x)*exp(x)*exp(x^2-1)+10*exp(x)*x)*exp(1/20*(-1 +2*x)*exp(x^2-1)+2)-30)/x,x, algorithm="fricas")
Output:
-(3*e^(x^2 - 1)*log(x) - e^(x^2 + 1/20*(2*x - 1)*e^(x^2 - 1) + x + 1))*e^( -x^2 + 1)
Time = 27.71 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx=e^{x} e^{\left (\frac {x}{10} - \frac {1}{20}\right ) e^{x^{2} - 1} + 2} - 3 \log {\left (x \right )} \] Input:
integrate(1/10*(((2*x**3-x**2+x)*exp(x)*exp(x**2-1)+10*exp(x)*x)*exp(1/20* (-1+2*x)*exp(x**2-1)+2)-30)/x,x)
Output:
exp(x)*exp((x/10 - 1/20)*exp(x**2 - 1) + 2) - 3*log(x)
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx=e^{\left (\frac {1}{10} \, x e^{\left (x^{2} - 1\right )} + x - \frac {1}{20} \, e^{\left (x^{2} - 1\right )} + 2\right )} - 3 \, \log \left (x\right ) \] Input:
integrate(1/10*(((2*x^3-x^2+x)*exp(x)*exp(x^2-1)+10*exp(x)*x)*exp(1/20*(-1 +2*x)*exp(x^2-1)+2)-30)/x,x, algorithm="maxima")
Output:
e^(1/10*x*e^(x^2 - 1) + x - 1/20*e^(x^2 - 1) + 2) - 3*log(x)
\[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx=\int { \frac {{\left ({\left (2 \, x^{3} - x^{2} + x\right )} e^{\left (x^{2} + x - 1\right )} + 10 \, x e^{x}\right )} e^{\left (\frac {1}{20} \, {\left (2 \, x - 1\right )} e^{\left (x^{2} - 1\right )} + 2\right )} - 30}{10 \, x} \,d x } \] Input:
integrate(1/10*(((2*x^3-x^2+x)*exp(x)*exp(x^2-1)+10*exp(x)*x)*exp(1/20*(-1 +2*x)*exp(x^2-1)+2)-30)/x,x, algorithm="giac")
Output:
integrate(1/10*(((2*x^3 - x^2 + x)*e^(x^2 + x - 1) + 10*x*e^x)*e^(1/20*(2* x - 1)*e^(x^2 - 1) + 2) - 30)/x, x)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx={\mathrm {e}}^{x-\frac {{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-1}}{20}+\frac {x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-1}}{10}+2}-3\,\ln \left (x\right ) \] Input:
int(((exp((exp(x^2 - 1)*(2*x - 1))/20 + 2)*(10*x*exp(x) + exp(x^2 - 1)*exp (x)*(x - x^2 + 2*x^3)))/10 - 3)/x,x)
Output:
exp(x - (exp(x^2)*exp(-1))/20 + (x*exp(x^2)*exp(-1))/10 + 2) - 3*log(x)
Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {-30+e^{\frac {1}{20} \left (40+e^{-1+x^2} (-1+2 x)\right )} \left (10 e^x x+e^{-1+x+x^2} \left (x-x^2+2 x^3\right )\right )}{10 x} \, dx=\frac {-3 e^{\frac {e^{x^{2}}}{20 e}} \mathrm {log}\left (x \right )+e^{\frac {e^{x^{2}} x +10 e x}{10 e}} e^{2}}{e^{\frac {e^{x^{2}}}{20 e}}} \] Input:
int(1/10*(((2*x^3-x^2+x)*exp(x)*exp(x^2-1)+10*exp(x)*x)*exp(1/20*(-1+2*x)* exp(x^2-1)+2)-30)/x,x)
Output:
( - 3*e**(e**(x**2)/(20*e))*log(x) + e**((e**(x**2)*x + 10*e*x)/(10*e))*e* *2)/e**(e**(x**2)/(20*e))