Integrand size = 85, antiderivative size = 24 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{e^{-e^x+2 x} (-2+x) x^2 \log ^4(2)} \] Output:
exp(ln(2)^4*exp(ln(x*(-2+x))-exp(x))*x*exp(x)^2)
Time = 2.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{e^{-e^x+2 x} (-2+x) x^2 \log ^4(2)} \] Input:
Integrate[(E^(-E^x + E^(-E^x + 2*x)*x*(-2*x + x^2)*Log[2]^4)*(-2*x + x^2)* (E^(3*x)*(2*x - x^2)*Log[2]^4 + E^(2*x)*(-4 - x + 2*x^2)*Log[2]^4))/(-2 + x),x]
Output:
E^(E^(-E^x + 2*x)*(-2 + x)*x^2*Log[2]^4)
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.
\(\displaystyle \int \frac {\left (x^2-2 x\right ) e^{e^{2 x-e^x} x \left (x^2-2 x\right ) \log ^4(2)-e^x} \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (2 x^2-x-4\right ) \log ^4(2)\right )}{x-2} \, dx\) |
\(\Big \downarrow \) 2014 |
\(\displaystyle \int x \left (e^{2 x} \left (-2 x^2+x+4\right ) \log ^4(2)-e^{3 x} \left (2 x-x^2\right ) \log ^4(2)\right ) \left (-\exp \left (-e^{2 x-e^x} x \left (2 x-x^2\right ) \log ^4(2)-e^x\right )\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \exp \left (-e^{2 x-e^x} x \log ^4(2) \left (2 x-x^2\right )-e^x\right ) x \left (e^{2 x} \left (-2 x^2+x+4\right ) \log ^4(2)-e^{3 x} \left (2 x-x^2\right ) \log ^4(2)\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\int \exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+2 x-e^x\right ) x \left (e^x x^2-2 x^2-2 e^x x+x+4\right ) \log ^4(2)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\log ^4(2) \int \exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+2 x-e^x\right ) x \left (e^x x^2-2 x^2-2 e^x x+x+4\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\log ^4(2) \int \left (\exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+3 x-e^x\right ) (x-2) x^2-\exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+2 x-e^x\right ) x \left (2 x^2-x-4\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log ^4(2) \left (4 \int \exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+2 x-e^x\right ) xdx+\int \exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+2 x-e^x\right ) x^2dx-2 \int \exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+3 x-e^x\right ) x^2dx-2 \int \exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+2 x-e^x\right ) x^3dx+\int \exp \left (-e^{2 x-e^x} \left (2 x-x^2\right ) \log ^4(2) x+3 x-e^x\right ) x^3dx\right )\) |
Input:
Int[(E^(-E^x + E^(-E^x + 2*x)*x*(-2*x + x^2)*Log[2]^4)*(-2*x + x^2)*(E^(3* x)*(2*x - x^2)*Log[2]^4 + E^(2*x)*(-4 - x + 2*x^2)*Log[2]^4))/(-2 + x),x]
Output:
$Aborted
Time = 199.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \({\mathrm e}^{x \ln \left (2\right )^{4} {\mathrm e}^{2 x} {\mathrm e}^{\ln \left (x^{2}-2 x \right )-{\mathrm e}^{x}}}\) | \(26\) |
risch | \({\mathrm e}^{\ln \left (2\right )^{4} \left (-2+x \right ) x^{2} {\mathrm e}^{2 x -\frac {i \pi \operatorname {csgn}\left (i x \left (-2+x \right )\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i x \left (-2+x \right )\right )^{2} \operatorname {csgn}\left (i x \right )}{2}+\frac {i \pi \operatorname {csgn}\left (i x \left (-2+x \right )\right )^{2} \operatorname {csgn}\left (i \left (-2+x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x \left (-2+x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-2+x \right )\right )}{2}-{\mathrm e}^{x}}}\) | \(100\) |
Input:
int(((-x^2+2*x)*ln(2)^4*exp(x)^3+(2*x^2-x-4)*ln(2)^4*exp(x)^2)*exp(ln(x^2- 2*x)-exp(x))*exp(x*ln(2)^4*exp(x)^2*exp(ln(x^2-2*x)-exp(x)))/(-2+x),x,meth od=_RETURNVERBOSE)
Output:
exp(x*ln(2)^4*exp(x)^2*exp(ln(x^2-2*x)-exp(x)))
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{\left (x e^{\left (2 \, x - e^{x} + \log \left (x^{2} - 2 \, x\right )\right )} \log \left (2\right )^{4}\right )} \] Input:
integrate(((-x^2+2*x)*log(2)^4*exp(x)^3+(2*x^2-x-4)*log(2)^4*exp(x)^2)*exp (log(x^2-2*x)-exp(x))*exp(x*log(2)^4*exp(x)^2*exp(log(x^2-2*x)-exp(x)))/(- 2+x),x, algorithm="fricas")
Output:
e^(x*e^(2*x - e^x + log(x^2 - 2*x))*log(2)^4)
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{x \left (x^{2} - 2 x\right ) e^{2 x} e^{- e^{x}} \log {\left (2 \right )}^{4}} \] Input:
integrate(((-x**2+2*x)*ln(2)**4*exp(x)**3+(2*x**2-x-4)*ln(2)**4*exp(x)**2) *exp(ln(x**2-2*x)-exp(x))*exp(x*ln(2)**4*exp(x)**2*exp(ln(x**2-2*x)-exp(x) ))/(-2+x),x)
Output:
exp(x*(x**2 - 2*x)*exp(2*x)*exp(-exp(x))*log(2)**4)
Time = 0.41 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{\left (x^{3} e^{\left (2 \, x - e^{x}\right )} \log \left (2\right )^{4} - 2 \, x^{2} e^{\left (2 \, x - e^{x}\right )} \log \left (2\right )^{4}\right )} \] Input:
integrate(((-x^2+2*x)*log(2)^4*exp(x)^3+(2*x^2-x-4)*log(2)^4*exp(x)^2)*exp (log(x^2-2*x)-exp(x))*exp(x*log(2)^4*exp(x)^2*exp(log(x^2-2*x)-exp(x)))/(- 2+x),x, algorithm="maxima")
Output:
e^(x^3*e^(2*x - e^x)*log(2)^4 - 2*x^2*e^(2*x - e^x)*log(2)^4)
\[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=\int { -\frac {{\left ({\left (x^{2} - 2 \, x\right )} e^{\left (3 \, x\right )} \log \left (2\right )^{4} - {\left (2 \, x^{2} - x - 4\right )} e^{\left (2 \, x\right )} \log \left (2\right )^{4}\right )} e^{\left (x e^{\left (2 \, x - e^{x} + \log \left (x^{2} - 2 \, x\right )\right )} \log \left (2\right )^{4} - e^{x} + \log \left (x^{2} - 2 \, x\right )\right )}}{x - 2} \,d x } \] Input:
integrate(((-x^2+2*x)*log(2)^4*exp(x)^3+(2*x^2-x-4)*log(2)^4*exp(x)^2)*exp (log(x^2-2*x)-exp(x))*exp(x*log(2)^4*exp(x)^2*exp(log(x^2-2*x)-exp(x)))/(- 2+x),x, algorithm="giac")
Output:
integrate(-((x^2 - 2*x)*e^(3*x)*log(2)^4 - (2*x^2 - x - 4)*e^(2*x)*log(2)^ 4)*e^(x*e^(2*x - e^x + log(x^2 - 2*x))*log(2)^4 - e^x + log(x^2 - 2*x))/(x - 2), x)
Time = 0.47 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx={\mathrm {e}}^{x^3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\ln \left (2\right )}^4}\,{\mathrm {e}}^{-2\,x^2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\ln \left (2\right )}^4} \] Input:
int(-(exp(log(x^2 - 2*x) - exp(x))*exp(x*exp(2*x)*exp(log(x^2 - 2*x) - exp (x))*log(2)^4)*(exp(2*x)*log(2)^4*(x - 2*x^2 + 4) - exp(3*x)*log(2)^4*(2*x - x^2)))/(x - 2),x)
Output:
exp(x^3*exp(2*x)*exp(-exp(x))*log(2)^4)*exp(-2*x^2*exp(2*x)*exp(-exp(x))*l og(2)^4)
Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=\frac {e^{\frac {e^{2 x} \mathrm {log}\left (2\right )^{4} x^{3}}{e^{e^{x}}}}}{e^{\frac {2 e^{2 x} \mathrm {log}\left (2\right )^{4} x^{2}}{e^{e^{x}}}}} \] Input:
int(((-x^2+2*x)*log(2)^4*exp(x)^3+(2*x^2-x-4)*log(2)^4*exp(x)^2)*exp(log(x ^2-2*x)-exp(x))*exp(x*log(2)^4*exp(x)^2*exp(log(x^2-2*x)-exp(x)))/(-2+x),x )
Output:
e**((e**(2*x)*log(2)**4*x**3)/e**(e**x))/e**((2*e**(2*x)*log(2)**4*x**2)/e **(e**x))