Integrand size = 107, antiderivative size = 34 \[ \int \frac {\left (2 e^5 x+e^x \left (-e^5 x+x^3\right )\right ) \log ^3(x)+\left (-4 x^2+2 e^x x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} \left (16-8 e^x+\left (-32+16 e^x\right ) \log (x)\right )}{\left (-2 x+e^x x\right ) \log ^3(x)} \, dx=x \left (-e^5+\frac {e^{4 \left (-2+\frac {1}{\log (x)}\right )^2}}{x}+x \log \left (2-e^x\right )\right ) \]
Time = 1.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {\left (2 e^5 x+e^x \left (-e^5 x+x^3\right )\right ) \log ^3(x)+\left (-4 x^2+2 e^x x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} \left (16-8 e^x+\left (-32+16 e^x\right ) \log (x)\right )}{\left (-2 x+e^x x\right ) \log ^3(x)} \, dx=e^{16+\frac {4}{\log ^2(x)}-\frac {16}{\log (x)}}-e^5 x+x^2 \log \left (2-e^x\right ) \]
Integrate[((2*E^5*x + E^x*(-(E^5*x) + x^3))*Log[x]^3 + (-4*x^2 + 2*E^x*x^2 )*Log[2 - E^x]*Log[x]^3 + E^((4 - 16*Log[x] + 16*Log[x]^2)/Log[x]^2)*(16 - 8*E^x + (-32 + 16*E^x)*Log[x]))/((-2*x + E^x*x)*Log[x]^3),x]
Time = 1.58 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {\left (e^x \left (x^3-e^5 x\right )+2 e^5 x\right ) \log ^3(x)+\left (2 e^x x^2-4 x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {16 \log ^2(x)-16 \log (x)+4}{\log ^2(x)}} \left (-8 e^x+\left (16 e^x-32\right ) \log (x)+16\right )}{\left (e^x x-2 x\right ) \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {8 e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}} (2 \log (x)-1)}{x \log ^3(x)}-\frac {-e^x x^2+e^{x+5}-2 e^x x \log \left (2-e^x\right )+4 x \log \left (2-e^x\right )-2 e^5}{e^x-2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2 \log \left (2-e^x\right )-e^5 x+e^{\frac {4 (1-2 \log (x))^2}{\log ^2(x)}}\) |
Int[((2*E^5*x + E^x*(-(E^5*x) + x^3))*Log[x]^3 + (-4*x^2 + 2*E^x*x^2)*Log[ 2 - E^x]*Log[x]^3 + E^((4 - 16*Log[x] + 16*Log[x]^2)/Log[x]^2)*(16 - 8*E^x + (-32 + 16*E^x)*Log[x]))/((-2*x + E^x*x)*Log[x]^3),x]
3.4.43.3.1 Defintions of rubi rules used
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97
\[\ln \left (-{\mathrm e}^{x}+2\right ) x^{2}-x \,{\mathrm e}^{5}+{\mathrm e}^{\frac {4 \left (2 \ln \left (x \right )-1\right )^{2}}{\ln \left (x \right )^{2}}}\]
int(((2*exp(x)*x^2-4*x^2)*ln(x)^3*ln(-exp(x)+2)+((16*exp(x)-32)*ln(x)-8*ex p(x)+16)*exp((16*ln(x)^2-16*ln(x)+4)/ln(x)^2)+((-x*exp(5)+x^3)*exp(x)+2*x* exp(5))*ln(x)^3)/(exp(x)*x-2*x)/ln(x)^3,x)
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {\left (2 e^5 x+e^x \left (-e^5 x+x^3\right )\right ) \log ^3(x)+\left (-4 x^2+2 e^x x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} \left (16-8 e^x+\left (-32+16 e^x\right ) \log (x)\right )}{\left (-2 x+e^x x\right ) \log ^3(x)} \, dx=x^{2} \log \left (-e^{x} + 2\right ) - x e^{5} + e^{\left (\frac {4 \, {\left (4 \, \log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 1\right )}}{\log \left (x\right )^{2}}\right )} \]
integrate(((2*exp(x)*x^2-4*x^2)*log(x)^3*log(-exp(x)+2)+((16*exp(x)-32)*lo g(x)-8*exp(x)+16)*exp((16*log(x)^2-16*log(x)+4)/log(x)^2)+((-x*exp(5)+x^3) *exp(x)+2*x*exp(5))*log(x)^3)/(exp(x)*x-2*x)/log(x)^3,x, algorithm=\
Time = 0.98 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2 e^5 x+e^x \left (-e^5 x+x^3\right )\right ) \log ^3(x)+\left (-4 x^2+2 e^x x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} \left (16-8 e^x+\left (-32+16 e^x\right ) \log (x)\right )}{\left (-2 x+e^x x\right ) \log ^3(x)} \, dx=x^{2} \log {\left (2 - e^{x} \right )} - x e^{5} + e^{\frac {16 \log {\left (x \right )}^{2} - 16 \log {\left (x \right )} + 4}{\log {\left (x \right )}^{2}}} \]
integrate(((2*exp(x)*x**2-4*x**2)*ln(x)**3*ln(-exp(x)+2)+((16*exp(x)-32)*l n(x)-8*exp(x)+16)*exp((16*ln(x)**2-16*ln(x)+4)/ln(x)**2)+((-x*exp(5)+x**3) *exp(x)+2*x*exp(5))*ln(x)**3)/(exp(x)*x-2*x)/ln(x)**3,x)
Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {\left (2 e^5 x+e^x \left (-e^5 x+x^3\right )\right ) \log ^3(x)+\left (-4 x^2+2 e^x x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} \left (16-8 e^x+\left (-32+16 e^x\right ) \log (x)\right )}{\left (-2 x+e^x x\right ) \log ^3(x)} \, dx={\left (x^{2} e^{\frac {16}{\log \left (x\right )}} \log \left (-e^{x} + 2\right ) - x e^{\left (\frac {16}{\log \left (x\right )} + 5\right )} + e^{\left (\frac {4}{\log \left (x\right )^{2}} + 16\right )}\right )} e^{\left (-\frac {16}{\log \left (x\right )}\right )} \]
integrate(((2*exp(x)*x^2-4*x^2)*log(x)^3*log(-exp(x)+2)+((16*exp(x)-32)*lo g(x)-8*exp(x)+16)*exp((16*log(x)^2-16*log(x)+4)/log(x)^2)+((-x*exp(5)+x^3) *exp(x)+2*x*exp(5))*log(x)^3)/(exp(x)*x-2*x)/log(x)^3,x, algorithm=\
(x^2*e^(16/log(x))*log(-e^x + 2) - x*e^(16/log(x) + 5) + e^(4/log(x)^2 + 1 6))*e^(-16/log(x))
\[ \int \frac {\left (2 e^5 x+e^x \left (-e^5 x+x^3\right )\right ) \log ^3(x)+\left (-4 x^2+2 e^x x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} \left (16-8 e^x+\left (-32+16 e^x\right ) \log (x)\right )}{\left (-2 x+e^x x\right ) \log ^3(x)} \, dx=\int { \frac {2 \, {\left (x^{2} e^{x} - 2 \, x^{2}\right )} \log \left (x\right )^{3} \log \left (-e^{x} + 2\right ) + {\left (2 \, x e^{5} + {\left (x^{3} - x e^{5}\right )} e^{x}\right )} \log \left (x\right )^{3} + 8 \, {\left (2 \, {\left (e^{x} - 2\right )} \log \left (x\right ) - e^{x} + 2\right )} e^{\left (\frac {4 \, {\left (4 \, \log \left (x\right )^{2} - 4 \, \log \left (x\right ) + 1\right )}}{\log \left (x\right )^{2}}\right )}}{{\left (x e^{x} - 2 \, x\right )} \log \left (x\right )^{3}} \,d x } \]
integrate(((2*exp(x)*x^2-4*x^2)*log(x)^3*log(-exp(x)+2)+((16*exp(x)-32)*lo g(x)-8*exp(x)+16)*exp((16*log(x)^2-16*log(x)+4)/log(x)^2)+((-x*exp(5)+x^3) *exp(x)+2*x*exp(5))*log(x)^3)/(exp(x)*x-2*x)/log(x)^3,x, algorithm=\
Time = 8.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2 e^5 x+e^x \left (-e^5 x+x^3\right )\right ) \log ^3(x)+\left (-4 x^2+2 e^x x^2\right ) \log \left (2-e^x\right ) \log ^3(x)+e^{\frac {4-16 \log (x)+16 \log ^2(x)}{\log ^2(x)}} \left (16-8 e^x+\left (-32+16 e^x\right ) \log (x)\right )}{\left (-2 x+e^x x\right ) \log ^3(x)} \, dx=x^2\,\ln \left (2-{\mathrm {e}}^x\right )-x\,{\mathrm {e}}^5+{\mathrm {e}}^{\frac {4}{{\ln \left (x\right )}^2}}\,{\mathrm {e}}^{-\frac {16}{\ln \left (x\right )}}\,{\mathrm {e}}^{16} \]