Integrand size = 107, antiderivative size = 33 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^3-e^x \left (-4+\log \left (e^x+\frac {2-x}{2 x}+2 x\right )\right ) \] Output:
-(ln(2*x+exp(x)+1/2*(2-x)/x)-4)*exp(x)-exp(3)
Time = 5.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^x \left (-4+\log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\right ) \] Input:
Integrate[(6*E^(2*x)*x^2 + E^x*(2 + 8*x - 8*x^2 + 16*x^3) + (-2*E^(2*x)*x^ 2 + E^x*(-2*x + x^2 - 4*x^3))*Log[(2 - x + 2*E^x*x + 4*x^2)/(2*x)])/(2*x - x^2 + 2*E^x*x^2 + 4*x^3),x]
Output:
-(E^x*(-4 + Log[-1/2 + E^x + x^(-1) + 2*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.
\(\displaystyle \int \frac {6 e^{2 x} x^2+e^x \left (16 x^3-8 x^2+8 x+2\right )+\left (e^x \left (-4 x^3+x^2-2 x\right )-2 e^{2 x} x^2\right ) \log \left (\frac {4 x^2+2 e^x x-x+2}{2 x}\right )}{4 x^3+2 e^x x^2-x^2+2 x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 x^3-5 x^2+2 x+2}{2 x^2}-\frac {16 x^5-24 x^4+21 x^3-4 x^2+2 x+4}{2 x^2 \left (4 x^2+2 e^x x-x+2\right )}-e^x \left (\log \left (2 x+e^x+\frac {1}{x}-\frac {1}{2}\right )-3\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {1}{4 x^2+2 e^x x-x+2}dx-2 \int \frac {e^x}{4 x^2+2 e^x x-x+2}dx-2 \int \frac {1}{x^2 \left (4 x^2+2 e^x x-x+2\right )}dx-\int \frac {1}{x \left (4 x^2+2 e^x x-x+2\right )}dx-2 \int \frac {e^x}{x \left (4 x^2+2 e^x x-x+2\right )}dx-\frac {21}{2} \int \frac {x}{4 x^2+2 e^x x-x+2}dx+5 \int \frac {e^x x}{4 x^2+2 e^x x-x+2}dx+12 \int \frac {x^2}{4 x^2+2 e^x x-x+2}dx-4 \int \frac {e^x x^2}{4 x^2+2 e^x x-x+2}dx-8 \int \frac {x^3}{4 x^2+2 e^x x-x+2}dx+x^2-\frac {5 x}{2}+4 e^x-\frac {1}{x}+\log (x)-e^x \log \left (2 x+e^x+\frac {1}{x}-\frac {1}{2}\right )\) |
Input:
Int[(6*E^(2*x)*x^2 + E^x*(2 + 8*x - 8*x^2 + 16*x^3) + (-2*E^(2*x)*x^2 + E^ x*(-2*x + x^2 - 4*x^3))*Log[(2 - x + 2*E^x*x + 4*x^2)/(2*x)])/(2*x - x^2 + 2*E^x*x^2 + 4*x^3),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(74\) vs. \(2(28)=56\).
Time = 0.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.27
method | result | size |
parallelrisch | \(2-{\mathrm e}^{x} \ln \left (\frac {2 \,{\mathrm e}^{x} x +4 x^{2}-x +2}{2 x}\right )-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (\frac {1}{2}+x^{2}+\frac {{\mathrm e}^{x} x}{2}-\frac {x}{4}\right )}{2}+4 \,{\mathrm e}^{x}-\frac {\ln \left (\frac {2 \,{\mathrm e}^{x} x +4 x^{2}-x +2}{2 x}\right )}{2}\) | \(75\) |
risch | \(-{\mathrm e}^{x} \ln \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )+{\mathrm e}^{x} \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )\right ) {\operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )}^{2} {\mathrm e}^{x}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}-\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{x}}{2}+\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )}^{3} {\mathrm e}^{x}}{2}-{\mathrm e}^{x} \ln \left (2\right )+4 \,{\mathrm e}^{x}\) | \(193\) |
Input:
int(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*ln(1/2*(2*exp(x)*x+4*x^2-x+ 2)/x)+6*exp(x)^2*x^2+(16*x^3-8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x^3-x^2+ 2*x),x,method=_RETURNVERBOSE)
Output:
2-exp(x)*ln(1/2*(2*exp(x)*x+4*x^2-x+2)/x)-1/2*ln(x)+1/2*ln(1/2+x^2+1/2*exp (x)*x-1/4*x)+4*exp(x)-1/2*ln(1/2*(2*exp(x)*x+4*x^2-x+2)/x)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^{x} \log \left (\frac {4 \, x^{2} + 2 \, x e^{x} - x + 2}{2 \, x}\right ) + 4 \, e^{x} \] Input:
integrate(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*log(1/2*(2*exp(x)*x+4 *x^2-x+2)/x)+6*exp(x)^2*x^2+(16*x^3-8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x ^3-x^2+2*x),x, algorithm="fricas")
Output:
-e^x*log(1/2*(4*x^2 + 2*x*e^x - x + 2)/x) + 4*e^x
Timed out. \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=\text {Timed out} \] Input:
integrate(((-2*exp(x)**2*x**2+(-4*x**3+x**2-2*x)*exp(x))*ln(1/2*(2*exp(x)* x+4*x**2-x+2)/x)+6*exp(x)**2*x**2+(16*x**3-8*x**2+8*x+2)*exp(x))/(2*exp(x) *x**2+4*x**3-x**2+2*x),x)
Output:
Timed out
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx={\left (\log \left (2\right ) + \log \left (x\right ) + 4\right )} e^{x} - e^{x} \log \left (4 \, x^{2} + 2 \, x e^{x} - x + 2\right ) \] Input:
integrate(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*log(1/2*(2*exp(x)*x+4 *x^2-x+2)/x)+6*exp(x)^2*x^2+(16*x^3-8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x ^3-x^2+2*x),x, algorithm="maxima")
Output:
(log(2) + log(x) + 4)*e^x - e^x*log(4*x^2 + 2*x*e^x - x + 2)
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-e^{x} \log \left (\frac {4 \, x^{2} + 2 \, x e^{x} - x + 2}{2 \, x}\right ) + 4 \, e^{x} \] Input:
integrate(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*log(1/2*(2*exp(x)*x+4 *x^2-x+2)/x)+6*exp(x)^2*x^2+(16*x^3-8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x ^3-x^2+2*x),x, algorithm="giac")
Output:
-e^x*log(1/2*(4*x^2 + 2*x*e^x - x + 2)/x) + 4*e^x
Time = 2.84 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=-{\mathrm {e}}^x\,\left (\ln \left (\frac {x\,{\mathrm {e}}^x-\frac {x}{2}+2\,x^2+1}{x}\right )-4\right ) \] Input:
int((6*x^2*exp(2*x) - log((x*exp(x) - x/2 + 2*x^2 + 1)/x)*(2*x^2*exp(2*x) + exp(x)*(2*x - x^2 + 4*x^3)) + exp(x)*(8*x - 8*x^2 + 16*x^3 + 2))/(2*x + 2*x^2*exp(x) - x^2 + 4*x^3),x)
Output:
-exp(x)*(log((x*exp(x) - x/2 + 2*x^2 + 1)/x) - 4)
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {6 e^{2 x} x^2+e^x \left (2+8 x-8 x^2+16 x^3\right )+\left (-2 e^{2 x} x^2+e^x \left (-2 x+x^2-4 x^3\right )\right ) \log \left (\frac {2-x+2 e^x x+4 x^2}{2 x}\right )}{2 x-x^2+2 e^x x^2+4 x^3} \, dx=e^{x} \left (-\mathrm {log}\left (\frac {2 e^{x} x +4 x^{2}-x +2}{2 x}\right )+4\right ) \] Input:
int(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*log(1/2*(2*exp(x)*x+4*x^2-x +2)/x)+6*exp(x)^2*x^2+(16*x^3-8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x^3-x^2 +2*x),x)
Output:
e**x*( - log((2*e**x*x + 4*x**2 - x + 2)/(2*x)) + 4)