Integrand size = 76, antiderivative size = 27 \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=\log \left (e^{-1+x-\frac {1+x}{2 x}}-\frac {x}{5}+3 x^2\right ) \] Output:
ln(exp(x-1/2*(1+x)/x-1)-1/5*x+3*x^2)
Time = 0.59 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=\frac {1}{2} \left (-\frac {1}{x}+2 \log \left (5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )\right ) \] Input:
Integrate[(-2*x^2 + 60*x^3 + E^((-1 - 3*x + 2*x^2)/(2*x))*(5 + 10*x^2))/(1 0*E^((-1 - 3*x + 2*x^2)/(2*x))*x^2 - 2*x^3 + 30*x^4),x]
Output:
(-x^(-1) + 2*Log[5*E^x - E^(3/2 + 1/(2*x))*x + 15*E^(3/2 + 1/(2*x))*x^2])/ 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.
\(\displaystyle \int \frac {60 x^3-2 x^2+e^{\frac {2 x^2-3 x-1}{2 x}} \left (10 x^2+5\right )}{30 x^4-2 x^3+10 e^{\frac {2 x^2-3 x-1}{2 x}} x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2+1}{2 x^2}-\frac {e^{\frac {1}{2 x}+\frac {3}{2}} \left (30 x^3-62 x^2+17 x-1\right )}{2 x \left (15 e^{\frac {1}{2 x}+\frac {3}{2}} x^2-e^{\frac {1}{2 x}+\frac {3}{2}} x+5 e^x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\text {Subst}\left (\int \frac {\exp \left (\frac {45 e^{\frac {3}{2}+\frac {1}{2 x}} x^2+x+e^{\frac {3}{2}+\frac {1}{2 x}}+15 e^x}{10 \left (3 e^{\frac {3}{2}+\frac {1}{2 x}} x^2+e^x\right )}\right )}{x}dx,x,\frac {15 e^{\frac {1}{2 x}+\frac {3}{2}} x^2-e^{\frac {1}{2 x}+\frac {3}{2}} x+5 e^x}{x}\right )}{10 \left (3 e^{\frac {1}{2 x}+\frac {3}{2}} x^2+e^x\right )}-\frac {17}{2} \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}}}{15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2-e^{\frac {3}{2}+\frac {1}{2 x}} x+5 e^x}dx+31 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x}{15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2-e^{\frac {3}{2}+\frac {1}{2 x}} x+5 e^x}dx-15 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x^2}{15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2-e^{\frac {3}{2}+\frac {1}{2 x}} x+5 e^x}dx+x-\frac {1}{2 x}\) |
Input:
Int[(-2*x^2 + 60*x^3 + E^((-1 - 3*x + 2*x^2)/(2*x))*(5 + 10*x^2))/(10*E^(( -1 - 3*x + 2*x^2)/(2*x))*x^2 - 2*x^3 + 30*x^4),x]
Output:
$Aborted
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\ln \left (x^{2}-\frac {x}{15}+\frac {{\mathrm e}^{\frac {2 x^{2}-3 x -1}{2 x}}}{3}\right )\) | \(27\) |
norman | \(\ln \left (15 x^{2}+5 \,{\mathrm e}^{\frac {2 x^{2}-3 x -1}{2 x}}-x \right )\) | \(29\) |
risch | \(x -\frac {1}{2 x}-\frac {2 x^{2}-3 x -1}{2 x}+\ln \left (3 x^{2}-\frac {x}{5}+{\mathrm e}^{\frac {2 x^{2}-3 x -1}{2 x}}\right )\) | \(49\) |
Input:
int(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1/2*(2* x^2-3*x-1)/x)+30*x^4-2*x^3),x,method=_RETURNVERBOSE)
Output:
ln(x^2-1/15*x+1/3*exp(1/2*(2*x^2-3*x-1)/x))
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=\log \left (15 \, x^{2} - x + 5 \, e^{\left (\frac {2 \, x^{2} - 3 \, x - 1}{2 \, x}\right )}\right ) \] Input:
integrate(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1 /2*(2*x^2-3*x-1)/x)+30*x^4-2*x^3),x, algorithm="fricas")
Output:
log(15*x^2 - x + 5*e^(1/2*(2*x^2 - 3*x - 1)/x))
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=\log {\left (3 x^{2} - \frac {x}{5} + e^{\frac {x^{2} - \frac {3 x}{2} - \frac {1}{2}}{x}} \right )} \] Input:
integrate(((10*x**2+5)*exp(1/2*(2*x**2-3*x-1)/x)+60*x**3-2*x**2)/(10*x**2* exp(1/2*(2*x**2-3*x-1)/x)+30*x**4-2*x**3),x)
Output:
log(3*x**2 - x/5 + exp((x**2 - 3*x/2 - 1/2)/x))
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=-\frac {1}{2 \, x} + \log \left (15 \, x - 1\right ) + \log \left (x\right ) + \log \left (\frac {{\left (15 \, x^{2} e^{\frac {3}{2}} - x e^{\frac {3}{2}}\right )} e^{\left (\frac {1}{2 \, x}\right )} + 5 \, e^{x}}{15 \, x^{2} e^{\frac {3}{2}} - x e^{\frac {3}{2}}}\right ) \] Input:
integrate(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1 /2*(2*x^2-3*x-1)/x)+30*x^4-2*x^3),x, algorithm="maxima")
Output:
-1/2/x + log(15*x - 1) + log(x) + log(((15*x^2*e^(3/2) - x*e^(3/2))*e^(1/2 /x) + 5*e^x)/(15*x^2*e^(3/2) - x*e^(3/2)))
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=\log \left (15 \, x^{2} - x + 5 \, e^{\left (\frac {2 \, x^{2} - 3 \, x - 1}{2 \, x}\right )}\right ) \] Input:
integrate(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1 /2*(2*x^2-3*x-1)/x)+30*x^4-2*x^3),x, algorithm="giac")
Output:
log(15*x^2 - x + 5*e^(1/2*(2*x^2 - 3*x - 1)/x))
Timed out. \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=\int \frac {60\,x^3-2\,x^2+{\mathrm {e}}^{-\frac {-x^2+\frac {3\,x}{2}+\frac {1}{2}}{x}}\,\left (10\,x^2+5\right )}{10\,x^2\,{\mathrm {e}}^{-\frac {-x^2+\frac {3\,x}{2}+\frac {1}{2}}{x}}-2\,x^3+30\,x^4} \,d x \] Input:
int((60*x^3 - 2*x^2 + exp(-((3*x)/2 - x^2 + 1/2)/x)*(10*x^2 + 5))/(10*x^2* exp(-((3*x)/2 - x^2 + 1/2)/x) - 2*x^3 + 30*x^4),x)
Output:
int((60*x^3 - 2*x^2 + exp(-((3*x)/2 - x^2 + 1/2)/x)*(10*x^2 + 5))/(10*x^2* exp(-((3*x)/2 - x^2 + 1/2)/x) - 2*x^3 + 30*x^4), x)
Time = 2.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx=\frac {2 \,\mathrm {log}\left (\frac {15 e^{\frac {1}{2 x}} e^{2} x^{2}-e^{\frac {1}{2 x}} e^{2} x +5 e^{x +\frac {1}{2}}}{x^{2}}\right ) x +4 \,\mathrm {log}\left (2 x \right ) x -1}{2 x} \] Input:
int(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1/2*(2* x^2-3*x-1)/x)+30*x^4-2*x^3),x)
Output:
(2*log((15*e**(1/(2*x))*e**2*x**2 - e**(1/(2*x))*e**2*x + 5*e**((2*x + 1)/ 2))/x**2)*x + 4*log(2*x)*x - 1)/(2*x)