Integrand size = 99, antiderivative size = 26 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx=6+\left (e^{\frac {\left (e^2+x\right ) \left (4+x^2\right )}{x}}-x\right )^2+\log (x) \] Output:
ln(x)+6+(exp((x^2+4)*(x+exp(2))/x)-x)^2
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx=\left (e^{\frac {\left (e^2+x\right ) \left (4+x^2\right )}{x}}-x\right )^2+\log (x) \] Input:
Integrate[(x + 2*x^3 + E^((2*(4*x + x^3 + E^2*(4 + x^2)))/x)*(4*x^3 + E^2* (-8 + 2*x^2)) + E^((4*x + x^3 + E^2*(4 + x^2))/x)*(-2*x^2 - 4*x^4 + E^2*(8 *x - 2*x^3)))/x^2,x]
Output:
(E^(((E^2 + x)*(4 + x^2))/x) - x)^2 + Log[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.
\(\displaystyle \int \frac {2 x^3+e^{\frac {2 \left (x^3+e^2 \left (x^2+4\right )+4 x\right )}{x}} \left (4 x^3+e^2 \left (2 x^2-8\right )\right )+e^{\frac {x^3+e^2 \left (x^2+4\right )+4 x}{x}} \left (-4 x^4+e^2 \left (8 x-2 x^3\right )-2 x^2\right )+x}{x^2} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {2 x^2+1}{x}+\frac {2 e^{\frac {2 \left (x+e^2\right ) \left (x^2+4\right )}{x}} \left (2 x^3+e^2 x^2-4 e^2\right )}{x^2}-\frac {2 e^{\frac {\left (x+e^2\right ) \left (x^2+4\right )}{x}} \left (2 x^3+e^2 x^2+x-4 e^2\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int e^{\frac {2 \left (x+e^2\right ) \left (x^2+4\right )}{x}+2}dx-8 \int \frac {e^{\frac {2 \left (x+e^2\right ) \left (x^2+4\right )}{x}+2}}{x^2}dx+4 \int e^{\frac {2 \left (x+e^2\right ) \left (x^2+4\right )}{x}} xdx+x^2+\frac {2 e^{\frac {\left (x+e^2\right ) \left (x^2+4\right )}{x}} \left (-2 x^3-e^2 x^2+4 e^2\right )}{\left (-\frac {\left (x^2+4\right ) \left (x+e^2\right )}{x^2}+\frac {x^2+4}{x}+2 \left (x+e^2\right )\right ) x}+\log (x)\) |
Input:
Int[(x + 2*x^3 + E^((2*(4*x + x^3 + E^2*(4 + x^2)))/x)*(4*x^3 + E^2*(-8 + 2*x^2)) + E^((4*x + x^3 + E^2*(4 + x^2))/x)*(-2*x^2 - 4*x^4 + E^2*(8*x - 2 *x^3)))/x^2,x]
Output:
$Aborted
Time = 16.76 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50
method | result | size |
risch | \(\ln \left (x \right )+x^{2}+{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) \left (x +{\mathrm e}^{2}\right )}{x}}-2 x \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) \left (x +{\mathrm e}^{2}\right )}{x}}\) | \(39\) |
parts | \(\ln \left (x \right )+x^{2}+{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{2}+2 x^{3}+8 x}{x}}-2 x \,{\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{2}+x^{3}+4 x}{x}}\) | \(52\) |
parallelrisch | \(x^{2}-2 x \,{\mathrm e}^{\frac {x^{2} {\mathrm e}^{2}+x^{3}+4 \,{\mathrm e}^{2}+4 x}{x}}+{\mathrm e}^{\frac {2 x^{2} {\mathrm e}^{2}+2 x^{3}+8 \,{\mathrm e}^{2}+8 x}{x}}+\ln \left (x \right )\) | \(56\) |
norman | \(\frac {x^{3}+x \,{\mathrm e}^{\frac {2 \left (x^{2}+4\right ) {\mathrm e}^{2}+2 x^{3}+8 x}{x}}-2 x^{2} {\mathrm e}^{\frac {\left (x^{2}+4\right ) {\mathrm e}^{2}+x^{3}+4 x}{x}}}{x}+\ln \left (x \right )\) | \(61\) |
Input:
int((((2*x^2-8)*exp(2)+4*x^3)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)^2+((-2*x^3+8 *x)*exp(2)-4*x^4-2*x^2)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)+2*x^3+x)/x^2,x,met hod=_RETURNVERBOSE)
Output:
ln(x)+x^2+exp(2*(x^2+4)*(x+exp(2))/x)-2*x*exp((x^2+4)*(x+exp(2))/x)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx=x^{2} - 2 \, x e^{\left (\frac {x^{3} + {\left (x^{2} + 4\right )} e^{2} + 4 \, x}{x}\right )} + e^{\left (\frac {2 \, {\left (x^{3} + {\left (x^{2} + 4\right )} e^{2} + 4 \, x\right )}}{x}\right )} + \log \left (x\right ) \] Input:
integrate((((2*x^2-8)*exp(2)+4*x^3)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)^2+((-2 *x^3+8*x)*exp(2)-4*x^4-2*x^2)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)+2*x^3+x)/x^2 ,x, algorithm="fricas")
Output:
x^2 - 2*x*e^((x^3 + (x^2 + 4)*e^2 + 4*x)/x) + e^(2*(x^3 + (x^2 + 4)*e^2 + 4*x)/x) + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).
Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx=x^{2} - 2 x e^{\frac {x^{3} + 4 x + \left (x^{2} + 4\right ) e^{2}}{x}} + e^{\frac {2 \left (x^{3} + 4 x + \left (x^{2} + 4\right ) e^{2}\right )}{x}} + \log {\left (x \right )} \] Input:
integrate((((2*x**2-8)*exp(2)+4*x**3)*exp(((x**2+4)*exp(2)+x**3+4*x)/x)**2 +((-2*x**3+8*x)*exp(2)-4*x**4-2*x**2)*exp(((x**2+4)*exp(2)+x**3+4*x)/x)+2* x**3+x)/x**2,x)
Output:
x**2 - 2*x*exp((x**3 + 4*x + (x**2 + 4)*exp(2))/x) + exp(2*(x**3 + 4*x + ( x**2 + 4)*exp(2))/x) + log(x)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx=x^{2} - 2 \, x e^{\left (x^{2} + x e^{2} + \frac {4 \, e^{2}}{x} + 4\right )} + e^{\left (2 \, x^{2} + 2 \, x e^{2} + \frac {8 \, e^{2}}{x} + 8\right )} + \log \left (x\right ) \] Input:
integrate((((2*x^2-8)*exp(2)+4*x^3)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)^2+((-2 *x^3+8*x)*exp(2)-4*x^4-2*x^2)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)+2*x^3+x)/x^2 ,x, algorithm="maxima")
Output:
x^2 - 2*x*e^(x^2 + x*e^2 + 4*e^2/x + 4) + e^(2*x^2 + 2*x*e^2 + 8*e^2/x + 8 ) + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (24) = 48\).
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx=x^{2} - 2 \, x e^{\left (\frac {x^{3} + x^{2} e^{2} + 4 \, x + 4 \, e^{2}}{x}\right )} + e^{\left (\frac {2 \, {\left (x^{3} + x^{2} e^{2} + 4 \, x + 4 \, e^{2}\right )}}{x}\right )} + \log \left (x\right ) \] Input:
integrate((((2*x^2-8)*exp(2)+4*x^3)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)^2+((-2 *x^3+8*x)*exp(2)-4*x^4-2*x^2)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)+2*x^3+x)/x^2 ,x, algorithm="giac")
Output:
x^2 - 2*x*e^((x^3 + x^2*e^2 + 4*x + 4*e^2)/x) + e^(2*(x^3 + x^2*e^2 + 4*x + 4*e^2)/x) + log(x)
Time = 3.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx={\mathrm {e}}^{2\,x\,{\mathrm {e}}^2+\frac {8\,{\mathrm {e}}^2}{x}+2\,x^2+8}+\ln \left (x\right )-2\,x\,{\mathrm {e}}^{x\,{\mathrm {e}}^2+\frac {4\,{\mathrm {e}}^2}{x}+x^2+4}+x^2 \] Input:
int((x - exp((4*x + x^3 + exp(2)*(x^2 + 4))/x)*(2*x^2 - exp(2)*(8*x - 2*x^ 3) + 4*x^4) + 2*x^3 + exp((2*(4*x + x^3 + exp(2)*(x^2 + 4)))/x)*(exp(2)*(2 *x^2 - 8) + 4*x^3))/x^2,x)
Output:
exp(2*x*exp(2) + (8*exp(2))/x + 2*x^2 + 8) + log(x) - 2*x*exp(x*exp(2) + ( 4*exp(2))/x + x^2 + 4) + x^2
Time = 13.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42 \[ \int \frac {x+2 x^3+e^{\frac {2 \left (4 x+x^3+e^2 \left (4+x^2\right )\right )}{x}} \left (4 x^3+e^2 \left (-8+2 x^2\right )\right )+e^{\frac {4 x+x^3+e^2 \left (4+x^2\right )}{x}} \left (-2 x^2-4 x^4+e^2 \left (8 x-2 x^3\right )\right )}{x^2} \, dx=e^{\frac {2 e^{2} x^{2}+2 x^{3}+8 e^{2}}{x}} e^{8}-2 e^{\frac {e^{2} x^{2}+x^{3}+4 e^{2}}{x}} e^{4} x +\mathrm {log}\left (x \right )+x^{2} \] Input:
int((((2*x^2-8)*exp(2)+4*x^3)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)^2+((-2*x^3+8 *x)*exp(2)-4*x^4-2*x^2)*exp(((x^2+4)*exp(2)+x^3+4*x)/x)+2*x^3+x)/x^2,x)
Output:
e**((2*e**2*x**2 + 8*e**2 + 2*x**3)/x)*e**8 - 2*e**((e**2*x**2 + 4*e**2 + x**3)/x)*e**4*x + log(x) + x**2