Integrand size = 97, antiderivative size = 32 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=x \left (2 x+\frac {2+e^{-x} (1-x)}{4 e^2 (x+\log (x))}\right ) \]
Time = 5.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {1}{4} x \left (8 x+\frac {e^{-2-x} \left (1+2 e^x-x\right )}{x+\log (x)}\right ) \]
Integrate[(-1 + x - 2*x^2 + x^3 + E^x*(-2 + 16*E^2*x^3) + (1 - 3*x + x^2 + E^x*(2 + 32*E^2*x^2))*Log[x] + 16*E^(2 + x)*x*Log[x]^2)/(4*E^(2 + x)*x^2 + 8*E^(2 + x)*x*Log[x] + 4*E^(2 + x)*Log[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.
\(\displaystyle \int \frac {x^3+e^x \left (16 e^2 x^3-2\right )-2 x^2+\left (x^2+e^x \left (32 e^2 x^2+2\right )-3 x+1\right ) \log (x)+x+16 e^{x+2} x \log ^2(x)-1}{4 e^{x+2} x^2+4 e^{x+2} \log ^2(x)+8 e^{x+2} x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x-2} \left (x^3+e^x \left (16 e^2 x^3-2\right )-2 x^2+\left (x^2+e^x \left (32 e^2 x^2+2\right )-3 x+1\right ) \log (x)+x+16 e^{x+2} x \log ^2(x)-1\right )}{4 (x+\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {e^{-x-2} \left (-x^3+2 x^2-16 e^{x+2} \log ^2(x) x-x+2 e^x \left (1-8 e^2 x^3\right )-\left (x^2-3 x+2 e^x \left (16 e^2 x^2+1\right )+1\right ) \log (x)+1\right )}{(x+\log (x))^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {e^{-x-2} \left (-x^3+2 x^2-16 e^{x+2} \log ^2(x) x-x+2 e^x \left (1-8 e^2 x^3\right )-\left (x^2-3 x+2 e^x \left (16 e^2 x^2+1\right )+1\right ) \log (x)+1\right )}{(x+\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (-\frac {e^{-x-2} x^3}{(x+\log (x))^2}+\frac {2 e^{-x-2} x^2}{(x+\log (x))^2}-\frac {e^{-x-2} \log (x) x^2}{(x+\log (x))^2}-\frac {e^{-x-2} x}{(x+\log (x))^2}+\frac {3 e^{-x-2} \log (x) x}{(x+\log (x))^2}-\frac {2 \left (8 e^2 x^3+16 e^2 \log (x) x^2+8 e^2 \log ^2(x) x+\log (x)-1\right )}{e^2 (x+\log (x))^2}+\frac {e^{-x-2}}{(x+\log (x))^2}-\frac {e^{-x-2} \log (x)}{(x+\log (x))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\int \frac {e^{-x-2} x^2}{(x+\log (x))^2}dx+\int \frac {e^{-x-2} x^2}{x+\log (x)}dx-\frac {2 \int \frac {1}{(x+\log (x))^2}dx}{e^2}-\int \frac {e^{-x-2}}{(x+\log (x))^2}dx-\frac {2 \int \frac {x}{(x+\log (x))^2}dx}{e^2}+\frac {2 \int \frac {1}{x+\log (x)}dx}{e^2}+\int \frac {e^{-x-2}}{x+\log (x)}dx-3 \int \frac {e^{-x-2} x}{x+\log (x)}dx+8 x^2\right )\) |
Int[(-1 + x - 2*x^2 + x^3 + E^x*(-2 + 16*E^2*x^3) + (1 - 3*x + x^2 + E^x*( 2 + 32*E^2*x^2))*Log[x] + 16*E^(2 + x)*x*Log[x]^2)/(4*E^(2 + x)*x^2 + 8*E^ (2 + x)*x*Log[x] + 4*E^(2 + x)*Log[x]^2),x]
3.15.21.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91
method | result | size |
risch | \(2 x^{2}-\frac {x \left (x -2 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2-x}}{4 \left (x +\ln \left (x \right )\right )}\) | \(29\) |
parallelrisch | \(\frac {\left (8 x^{3} {\mathrm e}^{2} {\mathrm e}^{x}+8 x^{2} {\mathrm e}^{2} {\mathrm e}^{x} \ln \left (x \right )-x^{2}+2 \,{\mathrm e}^{x} x +x \right ) {\mathrm e}^{-x} {\mathrm e}^{-2}}{4 x +4 \ln \left (x \right )}\) | \(49\) |
int((16*x*exp(2)*exp(x)*ln(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1)*ln(x) +(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*ln(x)^2+8*x*exp( 2)*exp(x)*ln(x)+4*x^2*exp(2)*exp(x)),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (27) = 54\).
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {8 \, x^{2} e^{\left (x + 4\right )} \log \left (x\right ) - {\left (x^{2} - x\right )} e^{2} + 2 \, {\left (4 \, x^{3} e^{2} + x\right )} e^{\left (x + 2\right )}}{4 \, {\left (x e^{\left (x + 4\right )} + e^{\left (x + 4\right )} \log \left (x\right )\right )}} \]
integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1 )*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2 +8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm=\
1/4*(8*x^2*e^(x + 4)*log(x) - (x^2 - x)*e^2 + 2*(4*x^3*e^2 + x)*e^(x + 2)) /(x*e^(x + 4) + e^(x + 4)*log(x))
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=2 x^{2} + \frac {x}{2 x e^{2} + 2 e^{2} \log {\left (x \right )}} + \frac {\left (- x^{2} + x\right ) e^{- x}}{4 x e^{2} + 4 e^{2} \log {\left (x \right )}} \]
integrate((16*x*exp(2)*exp(x)*ln(x)**2+((32*x**2*exp(2)+2)*exp(x)+x**2-3*x +1)*ln(x)+(16*x**3*exp(2)-2)*exp(x)+x**3-2*x**2+x-1)/(4*exp(2)*exp(x)*ln(x )**2+8*x*exp(2)*exp(x)*ln(x)+4*x**2*exp(2)*exp(x)),x)
2*x**2 + x/(2*x*exp(2) + 2*exp(2)*log(x)) + (-x**2 + x)*exp(-x)/(4*x*exp(2 ) + 4*exp(2)*log(x))
Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {8 \, x^{3} e^{2} + 8 \, x^{2} e^{2} \log \left (x\right ) - {\left (x^{2} - x\right )} e^{\left (-x\right )} + 2 \, x}{4 \, {\left (x e^{2} + e^{2} \log \left (x\right )\right )}} \]
integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1 )*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2 +8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {8 \, x^{3} e^{2} + 8 \, x^{2} e^{2} \log \left (x\right ) - x^{2} e^{\left (-x\right )} + x e^{\left (-x\right )} + 2 \, x}{4 \, {\left (x e^{2} + e^{2} \log \left (x\right )\right )}} \]
integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1 )*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2 +8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm=\
Time = 11.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 3.81 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {\frac {x\,{\mathrm {e}}^{-x-2}\,\left (2\,{\mathrm {e}}^x-x+2\,x^2-x^3+1\right )}{4\,\left (x+1\right )}-\frac {x\,{\mathrm {e}}^{-x-2}\,\ln \left (x\right )\,\left (2\,{\mathrm {e}}^x-3\,x+x^2+1\right )}{4\,\left (x+1\right )}}{x+\ln \left (x\right )}-\frac {1}{2\,{\mathrm {e}}^2+2\,x\,{\mathrm {e}}^2}+2\,x^2+\frac {{\mathrm {e}}^{-x}\,\left (\frac {{\mathrm {e}}^{-2}\,x^3}{4}-\frac {3\,{\mathrm {e}}^{-2}\,x^2}{4}+\frac {{\mathrm {e}}^{-2}\,x}{4}\right )}{x+1} \]
int((x + exp(x)*(16*x^3*exp(2) - 2) + log(x)*(exp(x)*(32*x^2*exp(2) + 2) - 3*x + x^2 + 1) - 2*x^2 + x^3 + 16*x*exp(2)*exp(x)*log(x)^2 - 1)/(4*x^2*ex p(2)*exp(x) + 4*exp(2)*exp(x)*log(x)^2 + 8*x*exp(2)*exp(x)*log(x)),x)