Integrand size = 92, antiderivative size = 33 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=\frac {2 \left (-e^3-e^4+\frac {1}{2} \left (3+\frac {4}{1+e^x}\right )+x\right )}{\log ^2(x)} \] Output:
2*(2/(1+exp(x))+3/2-exp(4)+x-exp(3))/ln(x)^2
Time = 0.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=-\frac {-7+2 e^3+2 e^4-3 e^x+2 e^{3+x}+2 e^{4+x}-2 x-2 e^x x}{\left (1+e^x\right ) \log ^2(x)} \] Input:
Integrate[(-14 + 4*E^3 + 4*E^4 + E^x*(-20 + 8*E^3 + 8*E^4 - 8*x) + E^(2*x) *(-6 + 4*E^3 + 4*E^4 - 4*x) - 4*x + (2*x + 2*E^(2*x)*x)*Log[x])/((x + 2*E^ x*x + E^(2*x)*x)*Log[x]^3),x]
Output:
-((-7 + 2*E^3 + 2*E^4 - 3*E^x + 2*E^(3 + x) + 2*E^(4 + x) - 2*x - 2*E^x*x) /((1 + E^x)*Log[x]^2))
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.
\(\displaystyle \int \frac {e^x \left (-8 x+8 e^4+8 e^3-20\right )+e^{2 x} \left (-4 x+4 e^4+4 e^3-6\right )-4 x+\left (2 e^{2 x} x+2 x\right ) \log (x)+4 e^4+4 e^3-14}{\left (2 e^x x+e^{2 x} x+x\right ) \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (-8 x+8 e^4+8 e^3-20\right )+e^{2 x} \left (-4 x+4 e^4+4 e^3-6\right )-4 x+\left (2 e^{2 x} x+2 x\right ) \log (x)-14 \left (1-\frac {2}{7} e^3 (1+e)\right )}{\left (e^x+1\right )^2 x \log ^3(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {4 (x \log (x)+2)}{\left (e^x+1\right ) x \log ^3(x)}+\frac {2 \left (-2 x+x \log (x)-3 \left (1-\frac {2}{3} e^3 (1+e)\right )\right )}{x \log ^3(x)}+\frac {4}{\left (e^x+1\right )^2 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {1}{\left (1+e^x\right ) x \log ^3(x)}dx+4 \int \frac {1}{\left (1+e^x\right )^2 \log ^2(x)}dx-4 \int \frac {1}{\left (1+e^x\right ) \log ^2(x)}dx+\frac {2 x}{\log ^2(x)}+\frac {3-2 e^3-2 e^4}{\log ^2(x)}\) |
Input:
Int[(-14 + 4*E^3 + 4*E^4 + E^x*(-20 + 8*E^3 + 8*E^4 - 8*x) + E^(2*x)*(-6 + 4*E^3 + 4*E^4 - 4*x) - 4*x + (2*x + 2*E^(2*x)*x)*Log[x])/((x + 2*E^x*x + E^(2*x)*x)*Log[x]^3),x]
Output:
$Aborted
Time = 4.41 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42
method | result | size |
risch | \(-\frac {2 \,{\mathrm e}^{4+x}+2 \,{\mathrm e}^{4}+2 \,{\mathrm e}^{3+x}+2 \,{\mathrm e}^{3}-2 \,{\mathrm e}^{x} x -2 x -3 \,{\mathrm e}^{x}-7}{\left ({\mathrm e}^{x}+1\right ) \ln \left (x \right )^{2}}\) | \(47\) |
parallelrisch | \(-\frac {-7+2 \,{\mathrm e}^{x} {\mathrm e}^{3}+2 \,{\mathrm e}^{4} {\mathrm e}^{x}-2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{3}+2 \,{\mathrm e}^{4}-2 x -3 \,{\mathrm e}^{x}}{\left ({\mathrm e}^{x}+1\right ) \ln \left (x \right )^{2}}\) | \(47\) |
Input:
int(((2*x*exp(x)^2+2*x)*ln(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8*exp(4) +8*exp(3)-8*x-20)*exp(x)+4*exp(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*exp(x)*x+ x)/ln(x)^3,x,method=_RETURNVERBOSE)
Output:
-(2*exp(4+x)+2*exp(4)+2*exp(3+x)+2*exp(3)-2*exp(x)*x-2*x-3*exp(x)-7)/(exp( x)+1)/ln(x)^2
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=\frac {{\left (2 \, x - 2 \, e^{4} - 2 \, e^{3} + 3\right )} e^{x} + 2 \, x - 2 \, e^{4} - 2 \, e^{3} + 7}{{\left (e^{x} + 1\right )} \log \left (x\right )^{2}} \] Input:
integrate(((2*x*exp(x)^2+2*x)*log(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8 *exp(4)+8*exp(3)-8*x-20)*exp(x)+4*exp(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*ex p(x)*x+x)/log(x)^3,x, algorithm="fricas")
Output:
((2*x - 2*e^4 - 2*e^3 + 3)*e^x + 2*x - 2*e^4 - 2*e^3 + 7)/((e^x + 1)*log(x )^2)
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=\frac {2 x - 2 e^{4} - 2 e^{3} + 3}{\log {\left (x \right )}^{2}} + \frac {4}{e^{x} \log {\left (x \right )}^{2} + \log {\left (x \right )}^{2}} \] Input:
integrate(((2*x*exp(x)**2+2*x)*ln(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)**2+( 8*exp(4)+8*exp(3)-8*x-20)*exp(x)+4*exp(4)+4*exp(3)-4*x-14)/(x*exp(x)**2+2* exp(x)*x+x)/ln(x)**3,x)
Output:
(2*x - 2*exp(4) - 2*exp(3) + 3)/log(x)**2 + 4/(exp(x)*log(x)**2 + log(x)** 2)
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=\frac {{\left (2 \, x - 2 \, e^{4} - 2 \, e^{3} + 3\right )} e^{x} + 2 \, x - 2 \, e^{4} - 2 \, e^{3} + 7}{e^{x} \log \left (x\right )^{2} + \log \left (x\right )^{2}} \] Input:
integrate(((2*x*exp(x)^2+2*x)*log(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8 *exp(4)+8*exp(3)-8*x-20)*exp(x)+4*exp(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*ex p(x)*x+x)/log(x)^3,x, algorithm="maxima")
Output:
((2*x - 2*e^4 - 2*e^3 + 3)*e^x + 2*x - 2*e^4 - 2*e^3 + 7)/(e^x*log(x)^2 + log(x)^2)
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=\frac {2 \, x e^{x} + 2 \, x - 2 \, e^{4} - 2 \, e^{3} - 2 \, e^{\left (x + 4\right )} - 2 \, e^{\left (x + 3\right )} + 3 \, e^{x} + 7}{e^{x} \log \left (x\right )^{2} + \log \left (x\right )^{2}} \] Input:
integrate(((2*x*exp(x)^2+2*x)*log(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8 *exp(4)+8*exp(3)-8*x-20)*exp(x)+4*exp(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*ex p(x)*x+x)/log(x)^3,x, algorithm="giac")
Output:
(2*x*e^x + 2*x - 2*e^4 - 2*e^3 - 2*e^(x + 4) - 2*e^(x + 3) + 3*e^x + 7)/(e ^x*log(x)^2 + log(x)^2)
Time = 2.90 (sec) , antiderivative size = 213, normalized size of antiderivative = 6.45 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=x-\frac {2\,x^2+2\,x}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1}+\frac {2\,x+{\mathrm {e}}^{2\,x}\,\left (\frac {2\,x^2}{3}-2\,x+\frac {2}{3}\right )-{\mathrm {e}}^x\,\left (\frac {8\,x^2}{3}-\frac {4}{3}\right )+\frac {2\,x^2}{3}+\frac {2}{3}}{3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x+1}+\frac {\frac {4\,x^2}{3}-\frac {2}{3}}{{\mathrm {e}}^x+1}+\frac {\frac {2\,x-2\,{\mathrm {e}}^{x+3}-2\,{\mathrm {e}}^{x+4}-2\,{\mathrm {e}}^3-2\,{\mathrm {e}}^4+3\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^x+7}{{\mathrm {e}}^x+1}-\frac {x\,\ln \left (x\right )\,\left ({\mathrm {e}}^{2\,x}+1\right )}{{\left ({\mathrm {e}}^x+1\right )}^2}}{{\ln \left (x\right )}^2}+\frac {\frac {x\,\left ({\mathrm {e}}^{2\,x}+1\right )}{{\left ({\mathrm {e}}^x+1\right )}^2}-\frac {x\,\ln \left (x\right )\,\left ({\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^{2\,x}-2\,x\,{\mathrm {e}}^x+1\right )}{{\left ({\mathrm {e}}^x+1\right )}^3}}{\ln \left (x\right )} \] Input:
int(-(4*x - 4*exp(3) - 4*exp(4) + exp(2*x)*(4*x - 4*exp(3) - 4*exp(4) + 6) - log(x)*(2*x + 2*x*exp(2*x)) + exp(x)*(8*x - 8*exp(3) - 8*exp(4) + 20) + 14)/(log(x)^3*(x + x*exp(2*x) + 2*x*exp(x))),x)
Output:
x - (2*x + 2*x^2)/(exp(2*x) + 2*exp(x) + 1) + (2*x + exp(2*x)*((2*x^2)/3 - 2*x + 2/3) - exp(x)*((8*x^2)/3 - 4/3) + (2*x^2)/3 + 2/3)/(3*exp(2*x) + ex p(3*x) + 3*exp(x) + 1) + ((4*x^2)/3 - 2/3)/(exp(x) + 1) + ((2*x - 2*exp(x + 3) - 2*exp(x + 4) - 2*exp(3) - 2*exp(4) + 3*exp(x) + 2*x*exp(x) + 7)/(ex p(x) + 1) - (x*log(x)*(exp(2*x) + 1))/(exp(x) + 1)^2)/log(x)^2 + ((x*(exp( 2*x) + 1))/(exp(x) + 1)^2 - (x*log(x)*(exp(2*x) + exp(3*x) + exp(x) + 2*x* exp(2*x) - 2*x*exp(x) + 1))/(exp(x) + 1)^3)/log(x)
Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {-14+4 e^3+4 e^4+e^x \left (-20+8 e^3+8 e^4-8 x\right )+e^{2 x} \left (-6+4 e^3+4 e^4-4 x\right )-4 x+\left (2 x+2 e^{2 x} x\right ) \log (x)}{\left (x+2 e^x x+e^{2 x} x\right ) \log ^3(x)} \, dx=\frac {-2 e^{x} e^{4}-2 e^{x} e^{3}+2 e^{x} x +3 e^{x}-2 e^{4}-2 e^{3}+2 x +7}{\mathrm {log}\left (x \right )^{2} \left (e^{x}+1\right )} \] Input:
int(((2*x*exp(x)^2+2*x)*log(x)+(4*exp(4)+4*exp(3)-4*x-6)*exp(x)^2+(8*exp(4 )+8*exp(3)-8*x-20)*exp(x)+4*exp(4)+4*exp(3)-4*x-14)/(x*exp(x)^2+2*exp(x)*x +x)/log(x)^3,x)
Output:
( - 2*e**x*e**4 - 2*e**x*e**3 + 2*e**x*x + 3*e**x - 2*e**4 - 2*e**3 + 2*x + 7)/(log(x)**2*(e**x + 1))