Integrand size = 79, antiderivative size = 30 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {x^2}{2}-\log \left (e^x+4 e^{3 x+\frac {30 \log (x)}{x}}\right ) \] Output:
1/2*x^2-ln(exp(x)+4*exp(30*ln(x)/x+3*x))
Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=-x+\frac {x^2}{2}-\log \left (1+4 e^{2 x} x^{30/x}\right ) \] Input:
Integrate[(E^x*(-x^2 + x^3) + E^((3*x^2 + 30*Log[x])/x)*(-120 - 12*x^2 + 4 *x^3 + 120*Log[x]))/(E^x*x^2 + 4*E^((3*x^2 + 30*Log[x])/x)*x^2),x]
Output:
-x + x^2/2 - Log[1 + 4*E^(2*x)*x^(30/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.
\(\displaystyle \int \frac {e^x \left (x^3-x^2\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (4 x^3-12 x^2+120 \log (x)-120\right )}{e^x x^2+4 x^2 e^{\frac {3 x^2+30 \log (x)}{x}}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x} \left (e^x \left (x^3-x^2\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (4 x^3-12 x^2+120 \log (x)-120\right )\right )}{x^2 \left (4 e^{2 x} x^{30/x}+1\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {120 e^{2 x} x^{\frac {30}{x}-2}}{4 e^{2 x} x^{30/x}+1}+\frac {4 e^{2 x} x^{\frac {30}{x}+1}}{4 e^{2 x} x^{30/x}+1}-\frac {12 e^{2 x} x^{30/x}}{4 e^{2 x} x^{30/x}+1}+\frac {x}{4 e^{2 x} x^{30/x}+1}-\frac {1}{4 e^{2 x} x^{30/x}+1}+\frac {120 e^{2 x} x^{\frac {30}{x}-2} \log (x)}{4 e^{2 x} x^{30/x}+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {1}{4 e^{2 x} x^{30/x}+1}dx+\int \frac {x}{4 e^{2 x} x^{30/x}+1}dx-120 \int \frac {e^{2 x} x^{\frac {30}{x}-2}}{4 e^{2 x} x^{30/x}+1}dx+4 \int \frac {e^{2 x} x^{1+\frac {30}{x}}}{4 e^{2 x} x^{30/x}+1}dx+120 \log (x) \int \frac {e^{2 x} x^{\frac {30}{x}-2}}{4 e^{2 x} x^{30/x}+1}dx-120 \int \frac {\int \frac {e^{2 x}}{x^2 \left (x^{-30/x}+4 e^{2 x}\right )}dx}{x}dx-3 x\) |
Input:
Int[(E^x*(-x^2 + x^3) + E^((3*x^2 + 30*Log[x])/x)*(-120 - 12*x^2 + 4*x^3 + 120*Log[x]))/(E^x*x^2 + 4*E^((3*x^2 + 30*Log[x])/x)*x^2),x]
Output:
$Aborted
Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(\frac {x^{2}}{2}-\ln \left ({\mathrm e}^{x}+4 \,{\mathrm e}^{\frac {30 \ln \left (x \right )+3 x^{2}}{x}}\right )\) | \(29\) |
risch | \(-\frac {30 \ln \left (x \right )}{x}-3 x +\frac {x^{2}}{2}+\frac {30 \ln \left (x \right )+3 x^{2}}{x}-\ln \left (\frac {{\mathrm e}^{x}}{4}+x^{\frac {30}{x}} {\mathrm e}^{3 x}\right )\) | \(51\) |
Input:
int(((120*ln(x)+4*x^3-12*x^2-120)*exp((30*ln(x)+3*x^2)/x)+(x^3-x^2)*exp(x) )/(4*x^2*exp((30*ln(x)+3*x^2)/x)+exp(x)*x^2),x,method=_RETURNVERBOSE)
Output:
1/2*x^2-ln(exp(x)+4*exp(3*(x^2+10*ln(x))/x))
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {1}{2} \, x^{2} - \log \left (e^{x} + 4 \, e^{\left (\frac {3 \, {\left (x^{2} + 10 \, \log \left (x\right )\right )}}{x}\right )}\right ) \] Input:
integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2 )*exp(x))/(4*x^2*exp((30*log(x)+3*x^2)/x)+exp(x)*x^2),x, algorithm="fricas ")
Output:
1/2*x^2 - log(e^x + 4*e^(3*(x^2 + 10*log(x))/x))
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {x^{2}}{2} - \log {\left (\frac {e^{x}}{4} + e^{\frac {3 x^{2} + 30 \log {\left (x \right )}}{x}} \right )} \] Input:
integrate(((120*ln(x)+4*x**3-12*x**2-120)*exp((30*ln(x)+3*x**2)/x)+(x**3-x **2)*exp(x))/(4*x**2*exp((30*ln(x)+3*x**2)/x)+exp(x)*x**2),x)
Output:
x**2/2 - log(exp(x)/4 + exp((3*x**2 + 30*log(x))/x))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {1}{2} \, x^{2} - 3 \, x - \log \left (\frac {1}{4} \, {\left (4 \, e^{\left (2 \, x + \frac {30 \, \log \left (x\right )}{x}\right )} + 1\right )} e^{\left (-2 \, x\right )}\right ) \] Input:
integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2 )*exp(x))/(4*x^2*exp((30*log(x)+3*x^2)/x)+exp(x)*x^2),x, algorithm="maxima ")
Output:
1/2*x^2 - 3*x - log(1/4*(4*e^(2*x + 30*log(x)/x) + 1)*e^(-2*x))
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {1}{2} \, x^{2} - \log \left (e^{x} + 4 \, e^{\left (\frac {3 \, {\left (x^{2} + 10 \, \log \left (x\right )\right )}}{x}\right )}\right ) \] Input:
integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2 )*exp(x))/(4*x^2*exp((30*log(x)+3*x^2)/x)+exp(x)*x^2),x, algorithm="giac")
Output:
1/2*x^2 - log(e^x + 4*e^(3*(x^2 + 10*log(x))/x))
Time = 3.79 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {x^2}{2}-\ln \left (\frac {{\mathrm {e}}^x}{4}+x^{30/x}\,{\mathrm {e}}^{3\,x}\right ) \] Input:
int(-(exp(x)*(x^2 - x^3) - exp((30*log(x) + 3*x^2)/x)*(120*log(x) - 12*x^2 + 4*x^3 - 120))/(x^2*exp(x) + 4*x^2*exp((30*log(x) + 3*x^2)/x)),x)
Output:
x^2/2 - log(exp(x)/4 + x^(30/x)*exp(3*x))
\[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=-120 \left (\int \frac {e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}}}{4 e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}} x^{2}+x^{2}}d x \right )-12 \left (\int \frac {e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}}}{4 e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}}+1}d x \right )+120 \left (\int \frac {e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}} \mathrm {log}\left (x \right )}{4 e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}} x^{2}+x^{2}}d x \right )+4 \left (\int \frac {e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}} x}{4 e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}}+1}d x \right )+\int \frac {x}{4 e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}}+1}d x -\left (\int \frac {1}{4 e^{\frac {30 \,\mathrm {log}\left (x \right )+2 x^{2}}{x}}+1}d x \right ) \] Input:
int(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2)*exp( x))/(4*x^2*exp((30*log(x)+3*x^2)/x)+exp(x)*x^2),x)
Output:
- 120*int(e**((30*log(x) + 2*x**2)/x)/(4*e**((30*log(x) + 2*x**2)/x)*x**2 + x**2),x) - 12*int(e**((30*log(x) + 2*x**2)/x)/(4*e**((30*log(x) + 2*x** 2)/x) + 1),x) + 120*int((e**((30*log(x) + 2*x**2)/x)*log(x))/(4*e**((30*lo g(x) + 2*x**2)/x)*x**2 + x**2),x) + 4*int((e**((30*log(x) + 2*x**2)/x)*x)/ (4*e**((30*log(x) + 2*x**2)/x) + 1),x) + int(x/(4*e**((30*log(x) + 2*x**2) /x) + 1),x) - int(1/(4*e**((30*log(x) + 2*x**2)/x) + 1),x)