Integrand size = 108, antiderivative size = 28 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{-3+x-e^{2 x} \left (5+2 x-\frac {\log ((-3+x) x)}{x}\right )} \] Output:
exp(x-(2*x-ln(x*(-3+x))/x+5)*exp(x)^2-3)
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{-3+x-e^{2 x} (5+2 x)} ((-3+x) x)^{\frac {e^{2 x}}{x}} \] Input:
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]
Output:
E^(-3 + x - E^(2*x)*(5 + 2*x))*((-3 + x)*x)^(E^(2*x)/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 |
\(\displaystyle \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx+\int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx}{x-3}dx+\int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx}{x}dx-\log (-((3-x) x)) \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x^2}dx+\int e^{x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}dx-12 \int e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}dx+\frac {1}{3} \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x-3}dx-\frac {1}{3} \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx-4 \int e^{3 x-e^{2 x} (2 x+5)-3} x ((x-3) x)^{\frac {e^{2 x}}{x}}dx-2 \int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx}{x-3}dx-2 \int \frac {\int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx}{x}dx+2 \log (-((3-x) x)) \int \frac {e^{3 x-e^{2 x} (2 x+5)-3} ((x-3) x)^{\frac {e^{2 x}}{x}}}{x}dx\) |
Input:
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]
Output:
$Aborted
Time = 1.76 (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 | \(x^{\frac {{\mathrm e}^{2 x}}{x}} \left (-3+x \right )^{\frac {{\mathrm e}^{2 x}}{x}} {\mathrm e}^{-\frac {i {\mathrm e}^{2 x} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (i x \left (-3+x \right )\right )-i {\mathrm e}^{2 x} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (-3+x \right )\right )^{2}-i {\mathrm e}^{2 x} \pi \,\operatorname {csgn}\left (i \left (-3+x \right )\right ) \operatorname {csgn}\left (i x \left (-3+x \right )\right )^{2}+i {\mathrm e}^{2 x} \pi \operatorname {csgn}\left (i x \left (-3+x \right )\right )^{3}+4 \,{\mathrm e}^{2 x} x^{2}+10 x \,{\mathrm e}^{2 x}-2 x^{2}+6 x}{2 x}}\) | \(149\) |
Input:
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)
Output:
exp((exp(x)^2*ln(x^2-3*x)+(-2*x^2-5*x)*exp(x)^2+x^2-3*x)/x)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\left (\frac {x^{2} - {\left (2 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (x^{2} - 3 \, x\right ) - 3 \, x}{x}\right )} \] Input:
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="fricas")
Output:
e^((x^2 - (2*x^2 + 5*x)*e^(2*x) + e^(2*x)*log(x^2 - 3*x) - 3*x)/x)
Time = 0.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\frac {x^{2} - 3 x + \left (- 2 x^{2} - 5 x\right ) e^{2 x} + e^{2 x} \log {\left (x^{2} - 3 x \right )}}{x}} \] Input:
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)
Output:
exp((x**2 - 3*x + (-2*x**2 - 5*x)*exp(2*x) + exp(2*x)*log(x**2 - 3*x))/x)
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\left (-2 \, x e^{\left (2 \, x\right )} + x + \frac {e^{\left (2 \, x\right )} \log \left (x - 3\right )}{x} + \frac {e^{\left (2 \, x\right )} \log \left (x\right )}{x} - 5 \, e^{\left (2 \, x\right )} - 3\right )} \] Input:
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="maxima")
Output:
e^(-2*x*e^(2*x) + x + e^(2*x)*log(x - 3)/x + e^(2*x)*log(x)/x - 5*e^(2*x) - 3)
Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=e^{\left (-2 \, x e^{\left (2 \, x\right )} + x + \frac {e^{\left (2 \, x\right )} \log \left (x^{2} - 3 \, x\right )}{x} - 5 \, e^{\left (2 \, x\right )} - 3\right )} \] Input:
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="giac")
Output:
e^(-2*x*e^(2*x) + x + e^(2*x)*log(x^2 - 3*x)/x - 5*e^(2*x) - 3)
Time = 3.70 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx={\mathrm {e}}^{-5\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^x\,{\left (x^2-3\,x\right )}^{\frac {{\mathrm {e}}^{2\,x}}{x}} \] Input:
int(-(exp(-(3*x + exp(2*x)*(5*x + 2*x^2) - exp(2*x)*log(x^2 - 3*x) - x^2)/ x)*(exp(2*x)*(2*x + 36*x^2 - 4*x^4 - 3) - 3*x^2 + x^3 + exp(2*x)*log(x^2 - 3*x)*(2*x^2 - 7*x + 3)))/(3*x^2 - x^3),x)
Output:
exp(-5*exp(2*x))*exp(-3)*exp(-2*x*exp(2*x))*exp(x)*(x^2 - 3*x)^(exp(2*x)/x )
Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {e^{\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}} \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx=\frac {e^{\frac {e^{2 x} \mathrm {log}\left (x^{2}-3 x \right )+x^{2}}{x}}}{e^{2 e^{2 x} x +5 e^{2 x}} e^{3}} \] Input:
int(((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)
Output:
e**((e**(2*x)*log(x**2 - 3*x) + x**2)/x)/(e**(2*e**(2*x)*x + 5*e**(2*x))*e **3)