Integrand size = 108, antiderivative size = 28 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{-3+x-e^{2 x} \left (5+2 x-\frac {\log ((-3+x) x)}{x}\right )} \]
Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{-3+x-e^{2 x} (5+2 x)} ((-3+x) x)^{\frac {e^{2 x}}{x}} \]
Integrate[(E^((-3*x + x^2 + E^(2*x)*(-5*x - 2*x^2) + E^(2*x)*Log[-3*x + x^ 2])/x)*(-3*x^2 + x^3 + E^(2*x)*(-3 + 2*x + 36*x^2 - 4*x^4) + E^(2*x)*(3 - 7*x + 2*x^2)*Log[-3*x + x^2]))/(-3*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.
\(\displaystyle \int \frac {\left (x^3-3 x^2+e^{2 x} \left (2 x^2-7 x+3\right ) \log \left (x^2-3 x\right )+e^{2 x} \left (-4 x^4+36 x^2+2 x-3\right )\right ) \exp \left (\frac {x^2+e^{2 x} \left (-2 x^2-5 x\right )+e^{2 x} \log \left (x^2-3 x\right )-3 x}{x}\right )}{x^3-3 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^3-3 x^2+e^{2 x} \left (2 x^2-7 x+3\right ) \log \left (x^2-3 x\right )+e^{2 x} \left (-4 x^4+36 x^2+2 x-3\right )\right ) \exp \left (\frac {x^2+e^{2 x} \left (-2 x^2-5 x\right )+e^{2 x} \log \left (x^2-3 x\right )-3 x}{x}\right )}{(x-3) x^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-x^3+3 x^2-e^{2 x} \left (2 x^2-7 x+3\right ) \log \left (x^2-3 x\right )-e^{2 x} \left (-4 x^4+36 x^2+2 x-3\right )\right ) \exp \left (\frac {e^{2 x} \left (-2 x^2-5 x\right )}{x}+\frac {e^{2 x} \log \left (x^2-3 x\right )}{x}+x-3\right )}{(3-x) x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\exp \left (\frac {e^{2 x} \left (-2 x^2-5 x\right )}{x}+\frac {e^{2 x} \log \left (x^2-3 x\right )}{x}+x-3\right )-\frac {\left (4 x^4-36 x^2-2 x^2 \log ((x-3) x)-2 x+7 x \log ((x-3) x)-3 \log ((x-3) x)+3\right ) \exp \left (\frac {e^{2 x} \left (-2 x^2-5 x\right )}{x}+\frac {e^{2 x} \log \left (x^2-3 x\right )}{x}+3 x-3\right )}{(x-3) x^2}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}-\frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}} \left (4 x^4-36 x^2-2 x^2 \log ((x-3) x)-2 x+7 x \log ((x-3) x)-3 \log ((x-3) x)+3\right )}{(x-3) x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx+\int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx}{x-3}dx+\int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx}{x}dx-\log (-((3-x) x)) \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx+\int e^{x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}dx-12 \int e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}dx+\frac {1}{3} \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x-3}dx-\frac {1}{3} \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx-4 \int e^{3 x-e^{2 x} (2 x+5)-3} x ((x-3) x)^{\frac {e^{2 x}}{x}}dx-2 \int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx}{x-3}dx-2 \int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx}{x}dx+2 \log (-((3-x) x)) \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx\) |
Int[(E^((-3*x + x^2 + E^(2*x)*(-5*x - 2*x^2) + E^(2*x)*Log[-3*x + x^2])/x) *(-3*x^2 + x^3 + E^(2*x)*(-3 + 2*x + 36*x^2 - 4*x^4) + E^(2*x)*(3 - 7*x + 2*x^2)*Log[-3*x + x^2]))/(-3*x^2 + x^3),x]
3.14.96.3.1 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 = 1.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {{\mathrm e}^{2 x} \ln \left (x^{2}-3 x \right )+\left (-2 x^{2}-5 x \right ) {\mathrm e}^{2 x}+x^{2}-3 x}{x}}\) | \(40\) |
risch | \(\left (-3+x \right )^{\frac {{\mathrm e}^{2 x}}{x}} x^{\frac {{\mathrm e}^{2 x}}{x}} {\mathrm e}^{-\frac {i {\mathrm e}^{2 x} \operatorname {csgn}\left (i x \left (-3+x \right )\right )^{3} \pi -i {\mathrm e}^{2 x} \operatorname {csgn}\left (i x \left (-3+x \right )\right )^{2} \operatorname {csgn}\left (i x \right ) \pi -i {\mathrm e}^{2 x} \operatorname {csgn}\left (i x \left (-3+x \right )\right )^{2} \operatorname {csgn}\left (i \left (-3+x \right )\right ) \pi +i {\mathrm e}^{2 x} \operatorname {csgn}\left (i x \left (-3+x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-3+x \right )\right ) \pi +4 \,{\mathrm e}^{2 x} x^{2}+10 x \,{\mathrm e}^{2 x}-2 x^{2}+6 x}{2 x}}\) | \(149\) |
int(((2*x^2-7*x+3)*exp(x)^2*ln(x^2-3*x)+(-4*x^4+36*x^2+2*x-3)*exp(x)^2+x^3 -3*x^2)*exp((exp(x)^2*ln(x^2-3*x)+(-2*x^2-5*x)*exp(x)^2+x^2-3*x)/x)/(x^3-3 *x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\left (\frac {x^{2} - {\left (2 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (x^{2} - 3 \, x\right ) - 3 \, x}{x}\right )} \]
integrate(((2*x^2-7*x+3)*exp(x)^2*log(x^2-3*x)+(-4*x^4+36*x^2+2*x-3)*exp(x )^2+x^3-3*x^2)*exp((exp(x)^2*log(x^2-3*x)+(-2*x^2-5*x)*exp(x)^2+x^2-3*x)/x )/(x^3-3*x^2),x, algorithm=\
Time = 0.49 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\frac {x^{2} - 3 x + \left (- 2 x^{2} - 5 x\right ) e^{2 x} + e^{2 x} \log {\left (x^{2} - 3 x \right )}}{x}} \]
integrate(((2*x**2-7*x+3)*exp(x)**2*ln(x**2-3*x)+(-4*x**4+36*x**2+2*x-3)*e xp(x)**2+x**3-3*x**2)*exp((exp(x)**2*ln(x**2-3*x)+(-2*x**2-5*x)*exp(x)**2+ x**2-3*x)/x)/(x**3-3*x**2),x)
Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\left (-2 \, x e^{\left (2 \, x\right )} + x + \frac {e^{\left (2 \, x\right )} \log \left (x - 3\right )}{x} + \frac {e^{\left (2 \, x\right )} \log \left (x\right )}{x} - 5 \, e^{\left (2 \, x\right )} - 3\right )} \]
integrate(((2*x^2-7*x+3)*exp(x)^2*log(x^2-3*x)+(-4*x^4+36*x^2+2*x-3)*exp(x )^2+x^3-3*x^2)*exp((exp(x)^2*log(x^2-3*x)+(-2*x^2-5*x)*exp(x)^2+x^2-3*x)/x )/(x^3-3*x^2),x, algorithm=\
Time = 0.40 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\left (-2 \, x e^{\left (2 \, x\right )} + x + \frac {e^{\left (2 \, x\right )} \log \left (x^{2} - 3 \, x\right )}{x} - 5 \, e^{\left (2 \, x\right )} - 3\right )} \]
integrate(((2*x^2-7*x+3)*exp(x)^2*log(x^2-3*x)+(-4*x^4+36*x^2+2*x-3)*exp(x )^2+x^3-3*x^2)*exp((exp(x)^2*log(x^2-3*x)+(-2*x^2-5*x)*exp(x)^2+x^2-3*x)/x )/(x^3-3*x^2),x, algorithm=\
Time = 10.37 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx={\mathrm {e}}^{-5\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^x\,{\left (x^2-3\,x\right )}^{\frac {{\mathrm {e}}^{2\,x}}{x}} \]