Integrand size = 82, antiderivative size = 29 \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=\log (2)+\log \left (x+5 \left (3+x+\frac {e^{\frac {(5+x)^2}{x}}}{-5+\log (2)}\right )\right ) \] Output:
ln(2)+ln(6*x+15+5*exp((5+x)^2/x)/(ln(2)-5))
Time = 1.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=\log \left (75-5 e^{10+\frac {25}{x}+x}+30 x-15 \log (2)-x \log (64)\right ) \] Input:
Integrate[(-30*x^2 + E^((25 + 10*x + x^2)/x)*(-125 + 5*x^2) + 6*x^2*Log[2] )/(-75*x^2 + 5*E^((25 + 10*x + x^2)/x)*x^2 - 30*x^3 + (15*x^2 + 6*x^3)*Log [2]),x]
Output:
Log[75 - 5*E^(10 + 25/x + x) + 30*x - 15*Log[2] - x*Log[64]]
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.
\(\displaystyle \int \frac {-30 x^2+e^{\frac {x^2+10 x+25}{x}} \left (5 x^2-125\right )+6 x^2 \log (2)}{-30 x^3+5 e^{\frac {x^2+10 x+25}{x}} x^2-75 x^2+\left (6 x^3+15 x^2\right ) \log (2)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{\frac {x^2+10 x+25}{x}} \left (5 x^2-125\right )+x^2 (6 \log (2)-30)}{-30 x^3+5 e^{\frac {x^2+10 x+25}{x}} x^2-75 x^2+\left (6 x^3+15 x^2\right ) \log (2)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2-25}{x^2}+\frac {3 \left (-2 x^3-3 x^2+50 x+125\right ) (\log (2)-5)}{x^2 \left (5 e^{\frac {(x+5)^2}{x}}-30 x \left (1-\frac {\log (2)}{5}\right )-75 \left (1-\frac {\log (2)}{5}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -375 (5-\log (2)) \int \frac {1}{x^2 \left (-30 \left (1-\frac {\log (2)}{5}\right ) x+5 e^{\frac {(x+5)^2}{x}}-75 \left (1-\frac {\log (2)}{5}\right )\right )}dx-150 (5-\log (2)) \int \frac {1}{x \left (-30 \left (1-\frac {\log (2)}{5}\right ) x+5 e^{\frac {(x+5)^2}{x}}-75 \left (1-\frac {\log (2)}{5}\right )\right )}dx-9 (5-\log (2)) \int \frac {1}{30 \left (1-\frac {\log (2)}{5}\right ) x-5 e^{\frac {(x+5)^2}{x}}+75 \left (1-\frac {\log (2)}{5}\right )}dx-6 (5-\log (2)) \int \frac {x}{30 \left (1-\frac {\log (2)}{5}\right ) x-5 e^{\frac {(x+5)^2}{x}}+75 \left (1-\frac {\log (2)}{5}\right )}dx+x+\frac {25}{x}\) |
Input:
Int[(-30*x^2 + E^((25 + 10*x + x^2)/x)*(-125 + 5*x^2) + 6*x^2*Log[2])/(-75 *x^2 + 5*E^((25 + 10*x + x^2)/x)*x^2 - 30*x^3 + (15*x^2 + 6*x^3)*Log[2]),x ]
Output:
$Aborted
Time = 0.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07
method | result | size |
norman | \(\ln \left (6 x \ln \left (2\right )+5 \,{\mathrm e}^{\frac {x^{2}+10 x +25}{x}}+15 \ln \left (2\right )-30 x -75\right )\) | \(31\) |
parallelrisch | \(\ln \left (\frac {6 x \ln \left (2\right )+5 \,{\mathrm e}^{\frac {x^{2}+10 x +25}{x}}+15 \ln \left (2\right )-30 x -75}{6 \ln \left (2\right )-30}\right )\) | \(39\) |
risch | \(x +\frac {25}{x}-\frac {x^{2}+10 x +25}{x}+\ln \left (\frac {6 x \ln \left (2\right )}{5}+3 \ln \left (2\right )-6 x +{\mathrm e}^{\frac {\left (5+x \right )^{2}}{x}}-15\right )\) | \(46\) |
Input:
int(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*ln(2)-30*x^2)/(5*x^2*exp((x^2+ 10*x+25)/x)+(6*x^3+15*x^2)*ln(2)-30*x^3-75*x^2),x,method=_RETURNVERBOSE)
Output:
ln(6*x*ln(2)+5*exp((x^2+10*x+25)/x)+15*ln(2)-30*x-75)
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=\log \left (3 \, {\left (2 \, x + 5\right )} \log \left (2\right ) - 30 \, x + 5 \, e^{\left (\frac {x^{2} + 10 \, x + 25}{x}\right )} - 75\right ) \] Input:
integrate(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*log(2)-30*x^2)/(5*x^2*ex p((x^2+10*x+25)/x)+(6*x^3+15*x^2)*log(2)-30*x^3-75*x^2),x, algorithm="fric as")
Output:
log(3*(2*x + 5)*log(2) - 30*x + 5*e^((x^2 + 10*x + 25)/x) - 75)
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=\log {\left (- 6 x + \frac {6 x \log {\left (2 \right )}}{5} + e^{\frac {x^{2} + 10 x + 25}{x}} - 15 + 3 \log {\left (2 \right )} \right )} \] Input:
integrate(((5*x**2-125)*exp((x**2+10*x+25)/x)+6*x**2*ln(2)-30*x**2)/(5*x** 2*exp((x**2+10*x+25)/x)+(6*x**3+15*x**2)*ln(2)-30*x**3-75*x**2),x)
Output:
log(-6*x + 6*x*log(2)/5 + exp((x**2 + 10*x + 25)/x) - 15 + 3*log(2))
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=x + \log \left (\frac {1}{5} \, {\left (6 \, x {\left (\log \left (2\right ) - 5\right )} + 5 \, e^{\left (x + \frac {25}{x} + 10\right )} + 15 \, \log \left (2\right ) - 75\right )} e^{\left (-x - 10\right )}\right ) \] Input:
integrate(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*log(2)-30*x^2)/(5*x^2*ex p((x^2+10*x+25)/x)+(6*x^3+15*x^2)*log(2)-30*x^3-75*x^2),x, algorithm="maxi ma")
Output:
x + log(1/5*(6*x*(log(2) - 5) + 5*e^(x + 25/x + 10) + 15*log(2) - 75)*e^(- x - 10))
Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=\log \left (6 \, x \log \left (2\right ) - 30 \, x + 5 \, e^{\left (\frac {x^{2} + 10 \, x + 25}{x}\right )} + 15 \, \log \left (2\right ) - 75\right ) \] Input:
integrate(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*log(2)-30*x^2)/(5*x^2*ex p((x^2+10*x+25)/x)+(6*x^3+15*x^2)*log(2)-30*x^3-75*x^2),x, algorithm="giac ")
Output:
log(6*x*log(2) - 30*x + 5*e^((x^2 + 10*x + 25)/x) + 15*log(2) - 75)
Timed out. \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=\int \frac {{\mathrm {e}}^{\frac {x^2+10\,x+25}{x}}\,\left (5\,x^2-125\right )+6\,x^2\,\ln \left (2\right )-30\,x^2}{\ln \left (2\right )\,\left (6\,x^3+15\,x^2\right )-75\,x^2-30\,x^3+5\,x^2\,{\mathrm {e}}^{\frac {x^2+10\,x+25}{x}}} \,d x \] Input:
int((exp((10*x + x^2 + 25)/x)*(5*x^2 - 125) + 6*x^2*log(2) - 30*x^2)/(log( 2)*(15*x^2 + 6*x^3) - 75*x^2 - 30*x^3 + 5*x^2*exp((10*x + x^2 + 25)/x)),x)
Output:
int((exp((10*x + x^2 + 25)/x)*(5*x^2 - 125) + 6*x^2*log(2) - 30*x^2)/(log( 2)*(15*x^2 + 6*x^3) - 75*x^2 - 30*x^3 + 5*x^2*exp((10*x + x^2 + 25)/x)), x )
Time = 1.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-30 x^2+e^{\frac {25+10 x+x^2}{x}} \left (-125+5 x^2\right )+6 x^2 \log (2)}{-75 x^2+5 e^{\frac {25+10 x+x^2}{x}} x^2-30 x^3+\left (15 x^2+6 x^3\right ) \log (2)} \, dx=\mathrm {log}\left (5 e^{\frac {x^{2}+25}{x}} e^{10}+6 \,\mathrm {log}\left (2\right ) x +15 \,\mathrm {log}\left (2\right )-30 x -75\right ) \] Input:
int(((5*x^2-125)*exp((x^2+10*x+25)/x)+6*x^2*log(2)-30*x^2)/(5*x^2*exp((x^2 +10*x+25)/x)+(6*x^3+15*x^2)*log(2)-30*x^3-75*x^2),x)
Output:
log(5*e**((x**2 + 25)/x)*e**10 + 6*log(2)*x + 15*log(2) - 30*x - 75)