Integrand size = 146, antiderivative size = 23 \[ \int \frac {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx=x+\log \left (e^{-\frac {e^{-2+x}}{-9 x+x^2}}+x\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(50\) vs. \(2(23)=46\).
Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.17 \[ \int \frac {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx=\frac {e^2 x+\frac {e^x}{9 x-x^2}+e^2 \log \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}{e^2} \]
Integrate[(E^(3 - x)*(81*x^2 + 63*x^3 - 17*x^4 + x^5) + (E*(-9 + 11*x - x^ 2) + E^(3 - x)*(81*x^2 - 18*x^3 + x^4))/E^(E^(-2 + x)/(-9*x + x^2)))/(E^(3 - x - E^(-2 + x)/(-9*x + x^2))*(81*x^2 - 18*x^3 + x^4) + E^(3 - x)*(81*x^ 3 - 18*x^4 + x^5)),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.
\(\displaystyle \int \frac {e^{-\frac {e^{x-2}}{x^2-9 x}} \left (e \left (-x^2+11 x-9\right )+e^{3-x} \left (x^4-18 x^3+81 x^2\right )\right )+e^{3-x} \left (x^5-17 x^4+63 x^3+81 x^2\right )}{e^{3-x} \left (x^5-18 x^4+81 x^3\right )+e^{-\frac {e^{x-2}}{x^2-9 x}-x+3} \left (x^4-18 x^3+81 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x+\frac {e^{x-2}}{(x-9) x}-3} \left (e^{-\frac {e^{x-2}}{x^2-9 x}} \left (e \left (-x^2+11 x-9\right )+e^{3-x} \left (x^4-18 x^3+81 x^2\right )\right )+e^{3-x} \left (x^5-17 x^4+63 x^3+81 x^2\right )\right )}{(9-x)^2 x^2 \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^2 (x-9)^2 x^2+e^{\frac {e^{x-2}}{(x-9) x}+2} (x-9)^2 (x+1) x^2-e^x \left (x^2-11 x+9\right )}{e^2 (9-x)^2 x^2 \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^2 (9-x)^2 x^2+e^{2-\frac {e^{x-2}}{(9-x) x}} (9-x)^2 (x+1) x^2-e^x \left (x^2-11 x+9\right )}{(9-x)^2 x^2 \left (e^{-\frac {e^{x-2}}{(9-x) x}} x+1\right )}dx}{e^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {e^2 \left (e^{\frac {e^{x-2}}{(x-9) x}} x+e^{\frac {e^{x-2}}{(x-9) x}}+1\right )}{e^{\frac {e^{x-2}}{(x-9) x}} x+1}-\frac {e^x \left (x^2-11 x+9\right )}{(x-9)^2 x^2 \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}\right )dx}{e^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{9} \int \frac {e^x}{x^2 \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}dx+\frac {1}{9} \int \frac {e^x}{(x-9)^2 \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}dx-\frac {1}{9} \int \frac {e^x}{(x-9) \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}dx-e^2 \int \frac {1}{x \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}dx+\frac {1}{9} \int \frac {e^x}{x \left (e^{\frac {e^{x-2}}{(x-9) x}} x+1\right )}dx+e^2 x+e^2 \log (x)}{e^2}\) |
Int[(E^(3 - x)*(81*x^2 + 63*x^3 - 17*x^4 + x^5) + (E*(-9 + 11*x - x^2) + E ^(3 - x)*(81*x^2 - 18*x^3 + x^4))/E^(E^(-2 + x)/(-9*x + x^2)))/(E^(3 - x - E^(-2 + x)/(-9*x + x^2))*(81*x^2 - 18*x^3 + x^4) + E^(3 - x)*(81*x^3 - 18 *x^4 + x^5)),x]
3.18.60.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 7.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22
method | result | size |
parallelrisch | \(36+\ln \left ({\mathrm e}^{-\frac {{\mathrm e} \,{\mathrm e}^{-3+x}}{x \left (x -9\right )}}+x \right )+x\) | \(28\) |
risch | \(x -\frac {{\mathrm e}^{-2+x}}{x \left (x -9\right )}+\frac {{\mathrm e}^{-2+x}}{x^{2}-9 x}+\ln \left ({\mathrm e}^{-\frac {{\mathrm e}^{-2+x}}{x \left (x -9\right )}}+x \right )\) | \(49\) |
norman | \(\frac {\left (x^{3} {\mathrm e}^{-x +3}-81 x \,{\mathrm e}^{-x +3}\right ) {\mathrm e}^{-3+x}}{x \left (x -9\right )}+\ln \left ({\mathrm e}^{-\frac {{\mathrm e} \,{\mathrm e}^{-3+x}}{x^{2}-9 x}}+x \right )\) | \(64\) |
int((((x^4-18*x^3+81*x^2)*exp(-x+3)+(-x^2+11*x-9)*exp(1))*exp(-exp(1)/(x^2 -9*x)/exp(-x+3))+(x^5-17*x^4+63*x^3+81*x^2)*exp(-x+3))/((x^4-18*x^3+81*x^2 )*exp(-x+3)*exp(-exp(1)/(x^2-9*x)/exp(-x+3))+(x^5-18*x^4+81*x^3)*exp(-x+3) ),x,method=_RETURNVERBOSE)
Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx=x + \log \left (x + e^{\left (-\frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x}\right )}\right ) \]
integrate((((x^4-18*x^3+81*x^2)*exp(-x+3)+(-x^2+11*x-9)*exp(1))*exp(-exp(1 )/(x^2-9*x)/exp(-x+3))+(x^5-17*x^4+63*x^3+81*x^2)*exp(-x+3))/((x^4-18*x^3+ 81*x^2)*exp(-x+3)*exp(-exp(1)/(x^2-9*x)/exp(-x+3))+(x^5-18*x^4+81*x^3)*exp (-x+3)),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx=x + \log {\left (x + e^{- \frac {e e^{x - 3}}{x^{2} - 9 x}} \right )} \]
integrate((((x**4-18*x**3+81*x**2)*exp(-x+3)+(-x**2+11*x-9)*exp(1))*exp(-e xp(1)/(x**2-9*x)/exp(-x+3))+(x**5-17*x**4+63*x**3+81*x**2)*exp(-x+3))/((x* *4-18*x**3+81*x**2)*exp(-x+3)*exp(-exp(1)/(x**2-9*x)/exp(-x+3))+(x**5-18*x **4+81*x**3)*exp(-x+3)),x)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (21) = 42\).
Time = 0.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.91 \[ \int \frac {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx=\frac {9 \, x^{2} e^{2} - 81 \, x e^{2} - e^{x}}{9 \, {\left (x e^{2} - 9 \, e^{2}\right )}} + \log \left (x\right ) + \log \left (\frac {x e^{\left (\frac {e^{x}}{9 \, {\left (x e^{2} - 9 \, e^{2}\right )}}\right )} + e^{\left (\frac {e^{\left (x - 2\right )}}{9 \, x}\right )}}{x}\right ) \]
integrate((((x^4-18*x^3+81*x^2)*exp(-x+3)+(-x^2+11*x-9)*exp(1))*exp(-exp(1 )/(x^2-9*x)/exp(-x+3))+(x^5-17*x^4+63*x^3+81*x^2)*exp(-x+3))/((x^4-18*x^3+ 81*x^2)*exp(-x+3)*exp(-exp(1)/(x^2-9*x)/exp(-x+3))+(x^5-18*x^4+81*x^3)*exp (-x+3)),x, algorithm=\
1/9*(9*x^2*e^2 - 81*x*e^2 - e^x)/(x*e^2 - 9*e^2) + log(x) + log((x*e^(1/9* e^x/(x*e^2 - 9*e^2)) + e^(1/9*e^(x - 2)/x))/x)
\[ \int \frac {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx=\int { \frac {{\left (x^{5} - 17 \, x^{4} + 63 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x + 3\right )} - {\left ({\left (x^{2} - 11 \, x + 9\right )} e - {\left (x^{4} - 18 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x + 3\right )}\right )} e^{\left (-\frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x}\right )}}{{\left (x^{4} - 18 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x - \frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x} + 3\right )} + {\left (x^{5} - 18 \, x^{4} + 81 \, x^{3}\right )} e^{\left (-x + 3\right )}} \,d x } \]
integrate((((x^4-18*x^3+81*x^2)*exp(-x+3)+(-x^2+11*x-9)*exp(1))*exp(-exp(1 )/(x^2-9*x)/exp(-x+3))+(x^5-17*x^4+63*x^3+81*x^2)*exp(-x+3))/((x^4-18*x^3+ 81*x^2)*exp(-x+3)*exp(-exp(1)/(x^2-9*x)/exp(-x+3))+(x^5-18*x^4+81*x^3)*exp (-x+3)),x, algorithm=\
integrate(((x^5 - 17*x^4 + 63*x^3 + 81*x^2)*e^(-x + 3) - ((x^2 - 11*x + 9) *e - (x^4 - 18*x^3 + 81*x^2)*e^(-x + 3))*e^(-e^(x - 2)/(x^2 - 9*x)))/((x^4 - 18*x^3 + 81*x^2)*e^(-x - e^(x - 2)/(x^2 - 9*x) + 3) + (x^5 - 18*x^4 + 8 1*x^3)*e^(-x + 3)), x)
Time = 13.87 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, dx=x+\ln \left (x+{\mathrm {e}}^{-\frac {{\mathrm {e}}^{x-2}}{x\,\left (x-9\right )}}\right ) \]