Integrand size = 86, antiderivative size = 28 \[ \int \frac {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx=\frac {3-x+5 e^{-x} x-\log (2)}{-4+x^2 \log (x)} \]
Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(28)=56\).
Time = 5.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.07 \[ \int \frac {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx=\frac {e^{-x} \left (5 x \left (8+x^2\right )-e^x \left (8 x+x^3+x^2 (-3+\log (2))+4 (-6+\log (4))\right )\right )}{\left (8+x^2\right ) \left (-4+x^2 \log (x)\right )} \]
Integrate[(-20 + 20*x - 5*x^2 + E^x*(4 - 3*x + x^2 + x*Log[2]) + (-5*x^2 - 5*x^3 + E^x*(-6*x + x^2 + 2*x*Log[2]))*Log[x])/(16*E^x - 8*E^x*x^2*Log[x] + E^x*x^4*Log[x]^2),x]
(5*x*(8 + x^2) - E^x*(8*x + x^3 + x^2*(-3 + Log[2]) + 4*(-6 + Log[4])))/(E ^x*(8 + x^2)*(-4 + x^2*Log[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.
\(\displaystyle \int \frac {-5 x^2+e^x \left (x^2-3 x+x \log (2)+4\right )+\left (-5 x^3-5 x^2+e^x \left (x^2-6 x+2 x \log (2)\right )\right ) \log (x)+20 x-20}{e^x x^4 \log ^2(x)-8 e^x x^2 \log (x)+16 e^x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x} \left (-5 x^2+e^x \left (x^2-3 x+x \log (2)+4\right )+\left (-5 x^3-5 x^2+e^x \left (x^2-6 x+2 x \log (2)\right )\right ) \log (x)+20 x-20\right )}{\left (4-x^2 \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {5 e^{-x} x^2}{\left (x^2 \log (x)-4\right )^2}-\frac {5 e^{-x} x^2 \log (x)}{\left (x^2 \log (x)-4\right )^2}+\frac {20 e^{-x} x}{\left (x^2 \log (x)-4\right )^2}+\frac {x^2+x^2 \log (x)-6 x \left (1-\frac {\log (2)}{3}\right ) \log (x)-3 x \left (1-\frac {\log (2)}{3}\right )+4}{\left (4-x^2 \log (x)\right )^2}-\frac {20 e^{-x}}{\left (x^2 \log (x)-4\right )^2}-\frac {5 e^{-x} x^3 \log (x)}{\left (x^2 \log (x)-4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \int \frac {1}{\left (x^2 \log (x)-4\right )^2}dx-40 \int \frac {e^{-x}}{\left (x^2 \log (x)-4\right )^2}dx-4 (6-\log (4)) \int \frac {1}{x \left (x^2 \log (x)-4\right )^2}dx-(3-\log (2)) \int \frac {x}{\left (x^2 \log (x)-4\right )^2}dx+\int \frac {x^2}{\left (x^2 \log (x)-4\right )^2}dx-5 \int \frac {e^{-x} x^2}{\left (x^2 \log (x)-4\right )^2}dx+\int \frac {1}{x^2 \log (x)-4}dx-5 \int \frac {e^{-x}}{x^2 \log (x)-4}dx-(6-\log (4)) \int \frac {1}{x \left (x^2 \log (x)-4\right )}dx-5 \int \frac {e^{-x} x}{x^2 \log (x)-4}dx\) |
Int[(-20 + 20*x - 5*x^2 + E^x*(4 - 3*x + x^2 + x*Log[2]) + (-5*x^2 - 5*x^3 + E^x*(-6*x + x^2 + 2*x*Log[2]))*Log[x])/(16*E^x - 8*E^x*x^2*Log[x] + E^x *x^4*Log[x]^2),x]
3.24.57.3.1 Defintions of rubi rules used
Time = 1.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
risch | \(-\frac {\left ({\mathrm e}^{x} \ln \left (2\right )+{\mathrm e}^{x} x -5 x -3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x^{2} \ln \left (x \right )-4}\) | \(34\) |
parallelrisch | \(-\frac {\left ({\mathrm e}^{x} \ln \left (2\right )+{\mathrm e}^{x} x -5 x -3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x^{2} \ln \left (x \right )-4}\) | \(34\) |
int((((2*x*ln(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*ln(x)+(x*ln(2)+x^2-3*x+4)*ex p(x)-5*x^2+20*x-20)/(x^4*exp(x)*ln(x)^2-8*x^2*exp(x)*ln(x)+16*exp(x)),x,me thod=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx=-\frac {{\left (x + \log \left (2\right ) - 3\right )} e^{x} - 5 \, x}{x^{2} e^{x} \log \left (x\right ) - 4 \, e^{x}} \]
integrate((((2*x*log(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*log(x)+(x*log(2)+x^2- 3*x+4)*exp(x)-5*x^2+20*x-20)/(x^4*exp(x)*log(x)^2-8*x^2*exp(x)*log(x)+16*e xp(x)),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx=\frac {5 x e^{- x}}{x^{2} \log {\left (x \right )} - 4} + \frac {- x - \log {\left (2 \right )} + 3}{x^{2} \log {\left (x \right )} - 4} \]
integrate((((2*x*ln(2)+x**2-6*x)*exp(x)-5*x**3-5*x**2)*ln(x)+(x*ln(2)+x**2 -3*x+4)*exp(x)-5*x**2+20*x-20)/(x**4*exp(x)*ln(x)**2-8*x**2*exp(x)*ln(x)+1 6*exp(x)),x)
Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx=\frac {5 \, x e^{\left (-x\right )} - x - \log \left (2\right ) + 3}{x^{2} \log \left (x\right ) - 4} \]
integrate((((2*x*log(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*log(x)+(x*log(2)+x^2- 3*x+4)*exp(x)-5*x^2+20*x-20)/(x^4*exp(x)*log(x)^2-8*x^2*exp(x)*log(x)+16*e xp(x)),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx=\frac {5 \, x e^{\left (-x\right )} - x - \log \left (2\right ) + 3}{x^{2} \log \left (x\right ) - 4} \]
integrate((((2*x*log(2)+x^2-6*x)*exp(x)-5*x^3-5*x^2)*log(x)+(x*log(2)+x^2- 3*x+4)*exp(x)-5*x^2+20*x-20)/(x^4*exp(x)*log(x)^2-8*x^2*exp(x)*log(x)+16*e xp(x)),x, algorithm=\
Time = 14.76 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.07 \[ \int \frac {-20+20 x-5 x^2+e^x \left (4-3 x+x^2+x \log (2)\right )+\left (-5 x^2-5 x^3+e^x \left (-6 x+x^2+2 x \log (2)\right )\right ) \log (x)}{16 e^x-8 e^x x^2 \log (x)+e^x x^4 \log ^2(x)} \, dx=-\frac {x^2\,{\mathrm {e}}^{-x}\,\left (8\,{\mathrm {e}}^x-40\right )-x\,{\mathrm {e}}^{-x}\,\left (24\,{\mathrm {e}}^x-8\,{\mathrm {e}}^x\,\ln \left (2\right )\right )+x^4\,{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^x-5\right )-x^3\,{\mathrm {e}}^{-x}\,\left (3\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \left (2\right )\right )}{\left (x^3+8\,x\right )\,\left (x^2\,\ln \left (x\right )-4\right )} \]
int(-(log(x)*(5*x^2 - exp(x)*(2*x*log(2) - 6*x + x^2) + 5*x^3) - exp(x)*(x *log(2) - 3*x + x^2 + 4) - 20*x + 5*x^2 + 20)/(16*exp(x) + x^4*exp(x)*log( x)^2 - 8*x^2*exp(x)*log(x)),x)