Integrand size = 69, antiderivative size = 24 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=2 e^{\frac {(5-x)^2 \left (1+5 x+\log \left (x^2\right )\right )}{e^4}} \] Output:
2*exp((1+ln(x^2)+5*x)*(5-x)^2/exp(4))
Time = 0.98 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=2 e^{\frac {(-5+x)^2 (1+5 x)}{e^4}} \left (x^2\right )^{\frac {(-5+x)^2}{e^4}} \] Input:
Integrate[(E^(-4 + (25 + 115*x - 49*x^2 + 5*x^3 + (25 - 10*x + x^2)*Log[x^ 2])/E^4)*(100 + 190*x - 192*x^2 + 30*x^3 + (-20*x + 4*x^2)*Log[x^2]))/x,x]
Output:
2*E^(((-5 + x)^2*(1 + 5*x))/E^4)*(x^2)^((-5 + x)^2/E^4)
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.
\(\displaystyle \int \frac {\left (30 x^3-192 x^2+\left (4 x^2-20 x\right ) \log \left (x^2\right )+190 x+100\right ) \exp \left (\frac {5 x^3-49 x^2+\left (x^2-10 x+25\right ) \log \left (x^2\right )+115 x+25}{e^4}-4\right )}{x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 (5-x) \left (-15 x^2-2 x \log \left (x^2\right )+21 x+10\right ) \exp \left (\frac {5 x^3-49 x^2+\left (x^2-10 x+25\right ) \log \left (x^2\right )+115 x+25}{e^4}-4\right )}{x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {\exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right ) (5-x) \left (-15 x^2-2 \log \left (x^2\right ) x+21 x+10\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {\exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right ) \left (15 x^3-96 x^2+95 x+50\right )}{x}+2 \exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right ) (x-5) \log \left (x^2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (95 \int \exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right )dx+50 \int \frac {\exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right )}{x}dx-96 \int \exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right ) xdx+15 \int \exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right ) x^2dx-10 \int \exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right ) \log \left (x^2\right )dx+2 \int \exp \left (\frac {5 x^3-49 x^2+115 x+\left (x^2-10 x+25\right ) \log \left (x^2\right )+25}{e^4}-4\right ) x \log \left (x^2\right )dx\right )\) |
Input:
Int[(E^(-4 + (25 + 115*x - 49*x^2 + 5*x^3 + (25 - 10*x + x^2)*Log[x^2])/E^ 4)*(100 + 190*x - 192*x^2 + 30*x^3 + (-20*x + 4*x^2)*Log[x^2]))/x,x]
Output:
$Aborted
Time = 0.51 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
risch | \(2 \,{\mathrm e}^{\left (-5+x \right )^{2} \left (1+\ln \left (x^{2}\right )+5 x \right ) {\mathrm e}^{-4}}\) | \(21\) |
default | \(2 \,{\mathrm e}^{\left (\left (x^{2}-10 x +25\right ) \ln \left (x^{2}\right )+5 x^{3}-49 x^{2}+115 x +25\right ) {\mathrm e}^{-4}}\) | \(37\) |
norman | \(2 \,{\mathrm e}^{\left (\left (x^{2}-10 x +25\right ) \ln \left (x^{2}\right )+5 x^{3}-49 x^{2}+115 x +25\right ) {\mathrm e}^{-4}}\) | \(37\) |
parallelrisch | \(2 \,{\mathrm e}^{\left (\left (x^{2}-10 x +25\right ) \ln \left (x^{2}\right )+5 x^{3}-49 x^{2}+115 x +25\right ) {\mathrm e}^{-4}}\) | \(37\) |
Input:
int(((4*x^2-20*x)*ln(x^2)+30*x^3-192*x^2+190*x+100)*exp(((x^2-10*x+25)*ln( x^2)+5*x^3-49*x^2+115*x+25)/exp(4))/x/exp(4),x,method=_RETURNVERBOSE)
Output:
2*exp((-5+x)^2*(1+ln(x^2)+5*x)*exp(-4))
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=2 \, e^{\left ({\left (5 \, x^{3} - 49 \, x^{2} + {\left (x^{2} - 10 \, x + 25\right )} \log \left (x^{2}\right ) + 115 \, x - 4 \, e^{4} + 25\right )} e^{\left (-4\right )} + 4\right )} \] Input:
integrate(((4*x^2-20*x)*log(x^2)+30*x^3-192*x^2+190*x+100)*exp(((x^2-10*x+ 25)*log(x^2)+5*x^3-49*x^2+115*x+25)/exp(4))/x/exp(4),x, algorithm="fricas" )
Output:
2*e^((5*x^3 - 49*x^2 + (x^2 - 10*x + 25)*log(x^2) + 115*x - 4*e^4 + 25)*e^ (-4) + 4)
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=2 e^{\frac {5 x^{3} - 49 x^{2} + 115 x + \left (x^{2} - 10 x + 25\right ) \log {\left (x^{2} \right )} + 25}{e^{4}}} \] Input:
integrate(((4*x**2-20*x)*ln(x**2)+30*x**3-192*x**2+190*x+100)*exp(((x**2-1 0*x+25)*ln(x**2)+5*x**3-49*x**2+115*x+25)/exp(4))/x/exp(4),x)
Output:
2*exp((5*x**3 - 49*x**2 + 115*x + (x**2 - 10*x + 25)*log(x**2) + 25)*exp(- 4))
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).
Time = 0.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=2 \, e^{\left (5 \, x^{3} e^{\left (-4\right )} + 2 \, x^{2} e^{\left (-4\right )} \log \left (x\right ) - 49 \, x^{2} e^{\left (-4\right )} - 20 \, x e^{\left (-4\right )} \log \left (x\right ) + 115 \, x e^{\left (-4\right )} + 50 \, e^{\left (-4\right )} \log \left (x\right ) + 25 \, e^{\left (-4\right )}\right )} \] Input:
integrate(((4*x^2-20*x)*log(x^2)+30*x^3-192*x^2+190*x+100)*exp(((x^2-10*x+ 25)*log(x^2)+5*x^3-49*x^2+115*x+25)/exp(4))/x/exp(4),x, algorithm="maxima" )
Output:
2*e^(5*x^3*e^(-4) + 2*x^2*e^(-4)*log(x) - 49*x^2*e^(-4) - 20*x*e^(-4)*log( x) + 115*x*e^(-4) + 50*e^(-4)*log(x) + 25*e^(-4))
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (20) = 40\).
Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.25 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=2 \, e^{\left (5 \, x^{3} e^{\left (-4\right )} + x^{2} e^{\left (-4\right )} \log \left (x^{2}\right ) - 49 \, x^{2} e^{\left (-4\right )} - 10 \, x e^{\left (-4\right )} \log \left (x^{2}\right ) + 115 \, x e^{\left (-4\right )} + 25 \, e^{\left (-4\right )} \log \left (x^{2}\right ) + 25 \, e^{\left (-4\right )}\right )} \] Input:
integrate(((4*x^2-20*x)*log(x^2)+30*x^3-192*x^2+190*x+100)*exp(((x^2-10*x+ 25)*log(x^2)+5*x^3-49*x^2+115*x+25)/exp(4))/x/exp(4),x, algorithm="giac")
Output:
2*e^(5*x^3*e^(-4) + x^2*e^(-4)*log(x^2) - 49*x^2*e^(-4) - 10*x*e^(-4)*log( x^2) + 115*x*e^(-4) + 25*e^(-4)*log(x^2) + 25*e^(-4))
Time = 8.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=2\,{\mathrm {e}}^{5\,x^3\,{\mathrm {e}}^{-4}}\,{\mathrm {e}}^{-49\,x^2\,{\mathrm {e}}^{-4}}\,{\mathrm {e}}^{25\,{\mathrm {e}}^{-4}}\,{\mathrm {e}}^{115\,x\,{\mathrm {e}}^{-4}}\,{\left (x^2\right )}^{{\mathrm {e}}^{-4}\,\left (x^2-10\,x+25\right )} \] Input:
int((exp(-4)*exp(exp(-4)*(115*x + log(x^2)*(x^2 - 10*x + 25) - 49*x^2 + 5* x^3 + 25))*(190*x - log(x^2)*(20*x - 4*x^2) - 192*x^2 + 30*x^3 + 100))/x,x )
Output:
2*exp(5*x^3*exp(-4))*exp(-49*x^2*exp(-4))*exp(25*exp(-4))*exp(115*x*exp(-4 ))*(x^2)^(exp(-4)*(x^2 - 10*x + 25))
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {e^{-4+\frac {25+115 x-49 x^2+5 x^3+\left (25-10 x+x^2\right ) \log \left (x^2\right )}{e^4}} \left (100+190 x-192 x^2+30 x^3+\left (-20 x+4 x^2\right ) \log \left (x^2\right )\right )}{x} \, dx=\frac {2 e^{\frac {\mathrm {log}\left (x^{2}\right ) x^{2}+25 \,\mathrm {log}\left (x^{2}\right )+5 x^{3}+115 x +25}{e^{4}}}}{e^{\frac {10 \,\mathrm {log}\left (x^{2}\right ) x +49 x^{2}}{e^{4}}}} \] Input:
int(((4*x^2-20*x)*log(x^2)+30*x^3-192*x^2+190*x+100)*exp(((x^2-10*x+25)*lo g(x^2)+5*x^3-49*x^2+115*x+25)/exp(4))/x/exp(4),x)
Output:
(2*e**((log(x**2)*x**2 + 25*log(x**2) + 5*x**3 + 115*x + 25)/e**4))/e**((1 0*log(x**2)*x + 49*x**2)/e**4)