Integrand size = 73, antiderivative size = 30 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {\left (-10+\frac {x}{2}+\log (x)\right ) \left (e^{e^{4 x^2}}-x+3 \log (x)\right )}{x} \] Output:
(1/2*x-10+ln(x))*(exp(exp(4*x^2))-x+3*ln(x))/x
Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {e^{e^{4 x^2}} (-20+x)-x^2+\left (-60+2 e^{e^{4 x^2}}+x\right ) \log (x)+6 \log ^2(x)}{2 x} \] Input:
Integrate[(-60 + x - x^2 + 72*Log[x] - 6*Log[x]^2 + E^E^(4*x^2)*(22 + E^(4 *x^2)*(-160*x^2 + 8*x^3) + (-2 + 16*E^(4*x^2)*x^2)*Log[x]))/(2*x^2),x]
Output:
(E^E^(4*x^2)*(-20 + x) - x^2 + (-60 + 2*E^E^(4*x^2) + x)*Log[x] + 6*Log[x] ^2)/(2*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.
\(\displaystyle \int \frac {-x^2+e^{e^{4 x^2}} \left (\left (16 e^{4 x^2} x^2-2\right ) \log (x)+e^{4 x^2} \left (8 x^3-160 x^2\right )+22\right )+x-6 \log ^2(x)+72 \log (x)-60}{2 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {x^2-x+6 \log ^2(x)-72 \log (x)-2 e^{e^{4 x^2}} \left (-4 e^{4 x^2} \left (20 x^2-x^3\right )-\left (1-8 e^{4 x^2} x^2\right ) \log (x)+11\right )+60}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {x^2-x+6 \log ^2(x)-72 \log (x)-2 e^{e^{4 x^2}} \left (-4 e^{4 x^2} \left (20 x^2-x^3\right )-\left (1-8 e^{4 x^2} x^2\right ) \log (x)+11\right )+60}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {x^2-x-22 e^{e^{4 x^2}}+6 \log ^2(x)+2 e^{e^{4 x^2}} \log (x)-72 \log (x)+60}{x^2}-8 e^{4 x^2+e^{4 x^2}} (x+2 \log (x)-20)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-160 \int e^{4 x^2+e^{4 x^2}}dx+22 \int \frac {e^{e^{4 x^2}}}{x^2}dx-16 \int \frac {\int e^{4 x^2+e^{4 x^2}}dx}{x}dx+2 \int \frac {\int \frac {e^{e^{4 x^2}}}{x^2}dx}{x}dx+16 \log (x) \int e^{4 x^2+e^{4 x^2}}dx-2 \log (x) \int \frac {e^{e^{4 x^2}}}{x^2}dx+e^{e^{4 x^2}}-x+\frac {6 \log ^2(x)}{x}-\frac {60 \log (x)}{x}+\log (x)\right )\) |
Input:
Int[(-60 + x - x^2 + 72*Log[x] - 6*Log[x]^2 + E^E^(4*x^2)*(22 + E^(4*x^2)* (-160*x^2 + 8*x^3) + (-2 + 16*E^(4*x^2)*x^2)*Log[x]))/(2*x^2),x]
Output:
$Aborted
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47
method | result | size |
risch | \(\frac {3 \ln \left (x \right )^{2}}{x}-\frac {30 \ln \left (x \right )}{x}+\frac {\ln \left (x \right )}{2}-\frac {x}{2}+\frac {\left (x -20+2 \ln \left (x \right )\right ) {\mathrm e}^{{\mathrm e}^{4 x^{2}}}}{2 x}\) | \(44\) |
parallelrisch | \(\frac {6 \ln \left (x \right )^{2}+2 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{4 x^{2}}}+x \ln \left (x \right )+x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}}}-x^{2}-60 \ln \left (x \right )-20 \,{\mathrm e}^{{\mathrm e}^{4 x^{2}}}}{2 x}\) | \(55\) |
Input:
int(1/2*(((16*x^2*exp(4*x^2)-2)*ln(x)+(8*x^3-160*x^2)*exp(4*x^2)+22)*exp(e xp(4*x^2))-6*ln(x)^2+72*ln(x)-x^2+x-60)/x^2,x,method=_RETURNVERBOSE)
Output:
3*ln(x)^2/x-30*ln(x)/x+1/2*ln(x)-1/2*x+1/2/x*(x-20+2*ln(x))*exp(exp(4*x^2) )
Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=-\frac {x^{2} - {\left (x + 2 \, \log \left (x\right ) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} - {\left (x - 60\right )} \log \left (x\right ) - 6 \, \log \left (x\right )^{2}}{2 \, x} \] Input:
integrate(1/2*(((16*x^2*exp(4*x^2)-2)*log(x)+(8*x^3-160*x^2)*exp(4*x^2)+22 )*exp(exp(4*x^2))-6*log(x)^2+72*log(x)-x^2+x-60)/x^2,x, algorithm="fricas" )
Output:
-1/2*(x^2 - (x + 2*log(x) - 20)*e^(e^(4*x^2)) - (x - 60)*log(x) - 6*log(x) ^2)/x
Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=- \frac {x}{2} + \frac {\log {\left (x \right )}}{2} + \frac {\left (x + 2 \log {\left (x \right )} - 20\right ) e^{e^{4 x^{2}}}}{2 x} + \frac {3 \log {\left (x \right )}^{2}}{x} - \frac {30 \log {\left (x \right )}}{x} \] Input:
integrate(1/2*(((16*x**2*exp(4*x**2)-2)*ln(x)+(8*x**3-160*x**2)*exp(4*x**2 )+22)*exp(exp(4*x**2))-6*ln(x)**2+72*ln(x)-x**2+x-60)/x**2,x)
Output:
-x/2 + log(x)/2 + (x + 2*log(x) - 20)*exp(exp(4*x**2))/(2*x) + 3*log(x)**2 /x - 30*log(x)/x
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=-\frac {1}{2} \, x + \frac {{\left (x + 2 \, \log \left (x\right ) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \left (x\right )^{2} - 60 \, \log \left (x\right ) - 60}{2 \, x} + \frac {30}{x} + \frac {1}{2} \, \log \left (x\right ) \] Input:
integrate(1/2*(((16*x^2*exp(4*x^2)-2)*log(x)+(8*x^3-160*x^2)*exp(4*x^2)+22 )*exp(exp(4*x^2))-6*log(x)^2+72*log(x)-x^2+x-60)/x^2,x, algorithm="maxima" )
Output:
-1/2*x + 1/2*((x + 2*log(x) - 20)*e^(e^(4*x^2)) + 6*log(x)^2 - 60*log(x) - 60)/x + 30/x + 1/2*log(x)
\[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\int { -\frac {x^{2} - 2 \, {\left (4 \, {\left (x^{3} - 20 \, x^{2}\right )} e^{\left (4 \, x^{2}\right )} + {\left (8 \, x^{2} e^{\left (4 \, x^{2}\right )} - 1\right )} \log \left (x\right ) + 11\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \left (x\right )^{2} - x - 72 \, \log \left (x\right ) + 60}{2 \, x^{2}} \,d x } \] Input:
integrate(1/2*(((16*x^2*exp(4*x^2)-2)*log(x)+(8*x^3-160*x^2)*exp(4*x^2)+22 )*exp(exp(4*x^2))-6*log(x)^2+72*log(x)-x^2+x-60)/x^2,x, algorithm="giac")
Output:
integrate(-1/2*(x^2 - 2*(4*(x^3 - 20*x^2)*e^(4*x^2) + (8*x^2*e^(4*x^2) - 1 )*log(x) + 11)*e^(e^(4*x^2)) + 6*log(x)^2 - x - 72*log(x) + 60)/x^2, x)
Time = 3.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {\ln \left (x\right )}{2}-\frac {x}{2}-\frac {30\,\ln \left (x\right )}{x}+\frac {3\,{\ln \left (x\right )}^2}{x}+\frac {{\mathrm {e}}^{{\mathrm {e}}^{4\,x^2}}\,\left (\frac {x}{2}+\ln \left (x\right )-10\right )}{x} \] Input:
int((x/2 + 36*log(x) + (exp(exp(4*x^2))*(log(x)*(16*x^2*exp(4*x^2) - 2) - exp(4*x^2)*(160*x^2 - 8*x^3) + 22))/2 - 3*log(x)^2 - x^2/2 - 30)/x^2,x)
Output:
log(x)/2 - x/2 - (30*log(x))/x + (3*log(x)^2)/x + (exp(exp(4*x^2))*(x/2 + log(x) - 10))/x
Time = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {2 e^{e^{4 x^{2}}} \mathrm {log}\left (x \right )+e^{e^{4 x^{2}}} x -20 e^{e^{4 x^{2}}}+6 \mathrm {log}\left (x \right )^{2}+\mathrm {log}\left (x \right ) x -60 \,\mathrm {log}\left (x \right )-x^{2}}{2 x} \] Input:
int(1/2*(((16*x^2*exp(4*x^2)-2)*log(x)+(8*x^3-160*x^2)*exp(4*x^2)+22)*exp( exp(4*x^2))-6*log(x)^2+72*log(x)-x^2+x-60)/x^2,x)
Output:
(2*e**(e**(4*x**2))*log(x) + e**(e**(4*x**2))*x - 20*e**(e**(4*x**2)) + 6* log(x)**2 + log(x)*x - 60*log(x) - x**2)/(2*x)