Integrand size = 97, antiderivative size = 18 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{e^{x \left (8+\frac {1}{x^2}+2 x+\log (-2+x)\right )}} \]
Time = 5.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x} \]
Integrate[(E^(E^((1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x) + (1 + 8*x^2 + 2 *x^3 + x^2*Log[-2 + x])/x)*(2 - x - 16*x^2 + x^3 + 4*x^4 + (-2*x^2 + x^3)* Log[-2 + x]))/(-2*x^2 + x^3),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.
\(\displaystyle \int \frac {\left (4 x^4+x^3-16 x^2+\left (x^3-2 x^2\right ) \log (x-2)-x+2\right ) \exp \left (\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}+e^{\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}}\right )}{x^3-2 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (4 x^4+x^3-16 x^2+\left (x^3-2 x^2\right ) \log (x-2)-x+2\right ) \exp \left (\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}+e^{\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}}\right )}{(x-2) x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\log (x-2) \exp \left (\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}+e^{\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}}\right )+\frac {\left (4 x^4+x^3-16 x^2-x+2\right ) \exp \left (\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}+e^{\frac {2 x^3+8 x^2+x^2 \log (x-2)+1}{x}}\right )}{(x-2) x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 9 \int \exp \left (\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}+e^{\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}}\right )dx+2 \int \frac {\exp \left (\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}+e^{\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}}\right )}{x-2}dx-\int \frac {\exp \left (\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}+e^{\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}}\right )}{x^2}dx+4 \int \exp \left (\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}+e^{\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}}\right ) xdx+\int \exp \left (\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}+e^{\frac {2 x^3+\log (x-2) x^2+8 x^2+1}{x}}\right ) \log (x-2)dx\) |
Int[(E^(E^((1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x) + (1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x)*(2 - x - 16*x^2 + x^3 + 4*x^4 + (-2*x^2 + x^3)*Log[-2 + x]))/(-2*x^2 + x^3),x]
3.13.33.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 1.60 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39
method | result | size |
risch | \({\mathrm e}^{\left (-2+x \right )^{x} {\mathrm e}^{\frac {2 x^{3}+8 x^{2}+1}{x}}}\) | \(25\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {x^{2} \ln \left (-2+x \right )+2 x^{3}+8 x^{2}+1}{x}}}\) | \(27\) |
int(((x^3-2*x^2)*ln(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*ln(-2+x)+2*x^3+8* x^2+1)/x)*exp(exp((x^2*ln(-2+x)+2*x^3+8*x^2+1)/x))/(x^3-2*x^2),x,method=_R ETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (16) = 32\).
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 4.33 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + x e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )} + 1}{x} - \frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )} \]
integrate(((x^3-2*x^2)*log(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*log(-2+x)+ 2*x^3+8*x^2+1)/x)*exp(exp((x^2*log(-2+x)+2*x^3+8*x^2+1)/x))/(x^3-2*x^2),x, algorithm=\
e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + x*e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x) + 1)/x - (2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x)
Time = 0.72 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{e^{\frac {2 x^{3} + x^{2} \log {\left (x - 2 \right )} + 8 x^{2} + 1}{x}}} \]
integrate(((x**3-2*x**2)*ln(-2+x)+4*x**4+x**3-16*x**2-x+2)*exp((x**2*ln(-2 +x)+2*x**3+8*x**2+1)/x)*exp(exp((x**2*ln(-2+x)+2*x**3+8*x**2+1)/x))/(x**3- 2*x**2),x)
Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{\left (e^{\left (2 \, x^{2} + x \log \left (x - 2\right ) + 8 \, x + \frac {1}{x}\right )}\right )} \]
integrate(((x^3-2*x^2)*log(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*log(-2+x)+ 2*x^3+8*x^2+1)/x)*exp(exp((x^2*log(-2+x)+2*x^3+8*x^2+1)/x))/(x^3-2*x^2),x, algorithm=\
\[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=\int { \frac {{\left (4 \, x^{4} + x^{3} - 16 \, x^{2} + {\left (x^{3} - 2 \, x^{2}\right )} \log \left (x - 2\right ) - x + 2\right )} e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x} + e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )}\right )}}{x^{3} - 2 \, x^{2}} \,d x } \]
integrate(((x^3-2*x^2)*log(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*log(-2+x)+ 2*x^3+8*x^2+1)/x)*exp(exp((x^2*log(-2+x)+2*x^3+8*x^2+1)/x))/(x^3-2*x^2),x, algorithm=\
integrate((4*x^4 + x^3 - 16*x^2 + (x^3 - 2*x^2)*log(x - 2) - x + 2)*e^((2* x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x + e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x))/(x^3 - 2*x^2), x)
Time = 13.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx={\mathrm {e}}^{{\mathrm {e}}^{8\,x}\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{2\,x^2}\,{\left (x-2\right )}^x} \]