Integrand size = 133, antiderivative size = 25 \[ \int \frac {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx=e^{10}-2 x-\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \]
Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx=-2 x-\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \]
Integrate[((-10 - 2*E^(E^x/2))*Log[5 + E^(E^x/2)] + Log[5 + E^(E^x/2)]^x^2 *(-1/2*(E^(E^x/2 + x)*x^2) + (-10*x - 2*E^(E^x/2)*x)*Log[5 + E^(E^x/2)]*Lo g[Log[5 + E^(E^x/2)]]))/((5 + E^(E^x/2))*Log[5 + E^(E^x/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.
| \(\displaystyle \int \frac {\left (\left (-2 e^{\frac {e^x}{2}} x-10 x\right ) \log \left (e^{\frac {e^x}{2}}+5\right ) \log \left (\log \left (e^{\frac {e^x}{2}}+5\right )\right )-\frac {1}{2} e^{x+\frac {e^x}{2}} x^2\right ) \log ^{x^2}\left (e^{\frac {e^x}{2}}+5\right )+\left (-2 e^{\frac {e^x}{2}}-10\right ) \log \left (e^{\frac {e^x}{2}}+5\right )}{\left (e^{\frac {e^x}{2}}+5\right ) \log \left (e^{\frac {e^x}{2}}+5\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {x \left (e^{x+\frac {e^x}{2}} x+4 e^{\frac {e^x}{2}} \log \left (e^{\frac {e^x}{2}}+5\right ) \log \left (\log \left (e^{\frac {e^x}{2}}+5\right )\right )+20 \log \left (e^{\frac {e^x}{2}}+5\right ) \log \left (\log \left (e^{\frac {e^x}{2}}+5\right )\right )\right ) \log ^{x^2-1}\left (e^{\frac {e^x}{2}}+5\right )}{2 \left (e^{\frac {e^x}{2}}+5\right )}-2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{2} \int \frac {e^{\frac {1}{2} \left (2 x+e^x\right )} x^2 \log ^{x^2-1}\left (5+e^{\frac {e^x}{2}}\right )}{5+e^{\frac {e^x}{2}}}dx-2 \int x \log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )dx-2 x\) |
Int[((-10 - 2*E^(E^x/2))*Log[5 + E^(E^x/2)] + Log[5 + E^(E^x/2)]^x^2*(-1/2 *(E^(E^x/2 + x)*x^2) + (-10*x - 2*E^(E^x/2)*x)*Log[5 + E^(E^x/2)]*Log[Log[ 5 + E^(E^x/2)]]))/((5 + E^(E^x/2))*Log[5 + E^(E^x/2)]),x]
3.16.70.3.1 Defintions of rubi rules used
Time = 5.43 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76
| method | result | size |
| risch | \(-2 x -\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x}}{2}}+5\right )^{x^{2}}\) | \(19\) |
| parallelrisch | \(-2 x -{\mathrm e}^{x^{2} \ln \left (\ln \left ({\mathrm e}^{{\mathrm e}^{x -\ln \left (2\right )}}+5\right )\right )}\) | \(24\) |
int((((-2*x*exp(exp(x-ln(2)))-10*x)*ln(exp(exp(x-ln(2)))+5)*ln(ln(exp(exp( x-ln(2)))+5))-x^2*exp(x-ln(2))*exp(exp(x-ln(2))))*exp(x^2*ln(ln(exp(exp(x- ln(2)))+5)))+(-2*exp(exp(x-ln(2)))-10)*ln(exp(exp(x-ln(2)))+5))/(exp(exp(x -ln(2)))+5)/ln(exp(exp(x-ln(2)))+5),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx=-2 \, x - \log \left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right )^{\left (x^{2}\right )} \]
integrate((((-2*x*exp(exp(x-log(2)))-10*x)*log(exp(exp(x-log(2)))+5)*log(l og(exp(exp(x-log(2)))+5))-x^2*exp(x-log(2))*exp(exp(x-log(2))))*exp(x^2*lo g(log(exp(exp(x-log(2)))+5)))+(-2*exp(exp(x-log(2)))-10)*log(exp(exp(x-log (2)))+5))/(exp(exp(x-log(2)))+5)/log(exp(exp(x-log(2)))+5),x, algorithm=\
Timed out. \[ \int \frac {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx=\text {Timed out} \]
integrate((((-2*x*exp(exp(x-ln(2)))-10*x)*ln(exp(exp(x-ln(2)))+5)*ln(ln(ex p(exp(x-ln(2)))+5))-x**2*exp(x-ln(2))*exp(exp(x-ln(2))))*exp(x**2*ln(ln(ex p(exp(x-ln(2)))+5)))+(-2*exp(exp(x-ln(2)))-10)*ln(exp(exp(x-ln(2)))+5))/(e xp(exp(x-ln(2)))+5)/ln(exp(exp(x-ln(2)))+5),x)
Time = 0.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx=-2 \, x - \log \left (e^{\left (\frac {1}{2} \, e^{x}\right )} + 5\right )^{\left (x^{2}\right )} \]
integrate((((-2*x*exp(exp(x-log(2)))-10*x)*log(exp(exp(x-log(2)))+5)*log(l og(exp(exp(x-log(2)))+5))-x^2*exp(x-log(2))*exp(exp(x-log(2))))*exp(x^2*lo g(log(exp(exp(x-log(2)))+5)))+(-2*exp(exp(x-log(2)))-10)*log(exp(exp(x-log (2)))+5))/(exp(exp(x-log(2)))+5)/log(exp(exp(x-log(2)))+5),x, algorithm=\
\[ \int \frac {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx=\int { -\frac {{\left (x^{2} e^{\left (x + e^{\left (x - \log \left (2\right )\right )} - \log \left (2\right )\right )} + 2 \, {\left (x e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5 \, x\right )} \log \left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right ) \log \left (\log \left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right )\right )\right )} \log \left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right )^{\left (x^{2}\right )} + 2 \, {\left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right )} \log \left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right )}{{\left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right )} \log \left (e^{\left (e^{\left (x - \log \left (2\right )\right )}\right )} + 5\right )} \,d x } \]
integrate((((-2*x*exp(exp(x-log(2)))-10*x)*log(exp(exp(x-log(2)))+5)*log(l og(exp(exp(x-log(2)))+5))-x^2*exp(x-log(2))*exp(exp(x-log(2))))*exp(x^2*lo g(log(exp(exp(x-log(2)))+5)))+(-2*exp(exp(x-log(2)))-10)*log(exp(exp(x-log (2)))+5))/(exp(exp(x-log(2)))+5)/log(exp(exp(x-log(2)))+5),x, algorithm=\
integrate(-((x^2*e^(x + e^(x - log(2)) - log(2)) + 2*(x*e^(e^(x - log(2))) + 5*x)*log(e^(e^(x - log(2))) + 5)*log(log(e^(e^(x - log(2))) + 5)))*log( e^(e^(x - log(2))) + 5)^(x^2) + 2*(e^(e^(x - log(2))) + 5)*log(e^(e^(x - l og(2))) + 5))/((e^(e^(x - log(2))) + 5)*log(e^(e^(x - log(2))) + 5)), x)
Time = 8.88 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx=-2\,x-{\ln \left ({\mathrm {e}}^{\frac {{\mathrm {e}}^x}{2}}+5\right )}^{x^2} \]
int(-(exp(x^2*log(log(exp(exp(x - log(2))) + 5)))*(x^2*exp(x - log(2))*exp (exp(x - log(2))) + log(exp(exp(x - log(2))) + 5)*log(log(exp(exp(x - log( 2))) + 5))*(10*x + 2*x*exp(exp(x - log(2))))) + log(exp(exp(x - log(2))) + 5)*(2*exp(exp(x - log(2))) + 10))/(log(exp(exp(x - log(2))) + 5)*(exp(exp (x - log(2))) + 5)),x)