Integrand size = 122, antiderivative size = 27 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=\left (x+x^2-e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2\right )^2 \] Output:
(x^2-exp(5*(exp(x)-ln(x))/x)^4*x^2+x)^2
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(27)=54\).
Time = 0.48 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=-2 e^{\frac {20 e^x}{x}} x^{3-\frac {20}{x}} (1+x)+x^2 \left (1+2 x+x^2+e^{\frac {40 e^x}{x}} x^{2-\frac {40}{x}}\right ) \] Input:
Integrate[2*x + 6*x^2 + 4*x^3 + E^((8*(5*E^x - 5*Log[x]))/x)*(-40*x^2 + 4* x^3 + E^x*(-40*x^2 + 40*x^3) + 40*x^2*Log[x]) + E^((4*(5*E^x - 5*Log[x]))/ x)*(40*x + 34*x^2 - 8*x^3 + E^x*(40*x - 40*x^3) + (-40*x - 40*x^2)*Log[x]) ,x]
Output:
-2*E^((20*E^x)/x)*x^(3 - 20/x)*(1 + x) + x^2*(1 + 2*x + x^2 + E^((40*E^x)/ x)*x^(2 - 40/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.
\(\displaystyle \int \left (4 x^3+6 x^2+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (4 x^3-40 x^2+40 x^2 \log (x)+e^x \left (40 x^3-40 x^2\right )\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (-8 x^3+e^x \left (40 x-40 x^3\right )+34 x^2+\left (-40 x^2-40 x\right ) \log (x)+40 x\right )+2 x\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x^3dx-8 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^3dx+34 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2dx-40 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2 \log (x)dx+40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} xdx+40 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} xdx-40 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \log (x)dx+x^4+2 x^3+x^2-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+x^2 (-\log (x))+e^x \left (x^2-x^3\right )\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}\) |
Input:
Int[2*x + 6*x^2 + 4*x^3 + E^((8*(5*E^x - 5*Log[x]))/x)*(-40*x^2 + 4*x^3 + E^x*(-40*x^2 + 40*x^3) + 40*x^2*Log[x]) + E^((4*(5*E^x - 5*Log[x]))/x)*(40 *x + 34*x^2 - 8*x^3 + E^x*(40*x - 40*x^3) + (-40*x - 40*x^2)*Log[x]),x]
Output:
$Aborted
Time = 3.43 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89
method | result | size |
risch | \(x^{4} {\mathrm e}^{\frac {-40 \ln \left (x \right )+40 \,{\mathrm e}^{x}}{x}}-2 x^{3} \left (1+x \right ) {\mathrm e}^{\frac {-20 \ln \left (x \right )+20 \,{\mathrm e}^{x}}{x}}+x^{4}+2 x^{3}+x^{2}\) | \(51\) |
parallelrisch | \(x^{4} {\mathrm e}^{-\frac {40 \left (\ln \left (x \right )-{\mathrm e}^{x}\right )}{x}}-2 \,{\mathrm e}^{-\frac {20 \left (\ln \left (x \right )-{\mathrm e}^{x}\right )}{x}} x^{4}-2 \,{\mathrm e}^{-\frac {20 \left (\ln \left (x \right )-{\mathrm e}^{x}\right )}{x}} x^{3}+x^{4}+2 x^{3}+x^{2}\) | \(72\) |
Input:
int((40*x^2*ln(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*ln(x)+5*exp (x))/x)^8+((-40*x^2-40*x)*ln(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^2+40*x)*e xp((-5*ln(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x,method=_RETURNVERBOSE)
Output:
x^4*exp(40*(exp(x)-ln(x))/x)-2*x^3*(1+x)*exp(20*(exp(x)-ln(x))/x)+x^4+2*x^ 3+x^2
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=x^{4} e^{\left (\frac {40 \, {\left (e^{x} - \log \left (x\right )\right )}}{x}\right )} + x^{4} + 2 \, x^{3} + x^{2} - 2 \, {\left (x^{4} + x^{3}\right )} e^{\left (\frac {20 \, {\left (e^{x} - \log \left (x\right )\right )}}{x}\right )} \] Input:
integrate((40*x^2*log(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*log( x)+5*exp(x))/x)^8+((-40*x^2-40*x)*log(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^ 2+40*x)*exp((-5*log(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x, algorithm="fricas ")
Output:
x^4*e^(40*(e^x - log(x))/x) + x^4 + 2*x^3 + x^2 - 2*(x^4 + x^3)*e^(20*(e^x - log(x))/x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=x^{4} e^{\frac {8 \cdot \left (5 e^{x} - 5 \log {\left (x \right )}\right )}{x}} + x^{4} + 2 x^{3} + x^{2} + \left (- 2 x^{4} - 2 x^{3}\right ) e^{\frac {4 \cdot \left (5 e^{x} - 5 \log {\left (x \right )}\right )}{x}} \] Input:
integrate((40*x**2*ln(x)+(40*x**3-40*x**2)*exp(x)+4*x**3-40*x**2)*exp((-5* ln(x)+5*exp(x))/x)**8+((-40*x**2-40*x)*ln(x)+(-40*x**3+40*x)*exp(x)-8*x**3 +34*x**2+40*x)*exp((-5*ln(x)+5*exp(x))/x)**4+4*x**3+6*x**2+2*x,x)
Output:
x**4*exp(8*(5*exp(x) - 5*log(x))/x) + x**4 + 2*x**3 + x**2 + (-2*x**4 - 2* x**3)*exp(4*(5*exp(x) - 5*log(x))/x)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=x^{4} e^{\left (\frac {40 \, e^{x}}{x} - \frac {40 \, \log \left (x\right )}{x}\right )} + x^{4} + 2 \, x^{3} + x^{2} - 2 \, {\left (x^{4} + x^{3}\right )} e^{\left (\frac {20 \, e^{x}}{x} - \frac {20 \, \log \left (x\right )}{x}\right )} \] Input:
integrate((40*x^2*log(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*log( x)+5*exp(x))/x)^8+((-40*x^2-40*x)*log(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^ 2+40*x)*exp((-5*log(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x, algorithm="maxima ")
Output:
x^4*e^(40*e^x/x - 40*log(x)/x) + x^4 + 2*x^3 + x^2 - 2*(x^4 + x^3)*e^(20*e ^x/x - 20*log(x)/x)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (28) = 56\).
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.41 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=x^{4} e^{\left (\frac {40 \, {\left (e^{x} - \log \left (x\right )\right )}}{x}\right )} - 2 \, x^{4} e^{\left (\frac {20 \, {\left (e^{x} - \log \left (x\right )\right )}}{x}\right )} + x^{4} - 2 \, x^{3} e^{\left (\frac {20 \, {\left (e^{x} - \log \left (x\right )\right )}}{x}\right )} + 2 \, x^{3} + x^{2} \] Input:
integrate((40*x^2*log(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*log( x)+5*exp(x))/x)^8+((-40*x^2-40*x)*log(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^ 2+40*x)*exp((-5*log(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x, algorithm="giac")
Output:
x^4*e^(40*(e^x - log(x))/x) - 2*x^4*e^(20*(e^x - log(x))/x) + x^4 - 2*x^3* e^(20*(e^x - log(x))/x) + 2*x^3 + x^2
Time = 4.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.33 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=x^2+2\,x^3+x^4-\frac {{\mathrm {e}}^{\frac {20\,{\mathrm {e}}^x}{x}}\,\left (2\,x^4+2\,x^3\right )}{x^{20/x}}+\frac {x^4\,{\mathrm {e}}^{\frac {40\,{\mathrm {e}}^x}{x}}}{x^{40/x}} \] Input:
int(2*x - exp((8*(5*exp(x) - 5*log(x)))/x)*(exp(x)*(40*x^2 - 40*x^3) - 40* x^2*log(x) + 40*x^2 - 4*x^3) + 6*x^2 + 4*x^3 + exp((4*(5*exp(x) - 5*log(x) ))/x)*(40*x + exp(x)*(40*x - 40*x^3) - log(x)*(40*x + 40*x^2) + 34*x^2 - 8 *x^3),x)
Output:
x^2 + 2*x^3 + x^4 - (exp((20*exp(x))/x)*(2*x^3 + 2*x^4))/x^(20/x) + (x^4*e xp((40*exp(x))/x))/x^(40/x)
Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.85 \[ \int \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx=\frac {x^{2} \left (e^{\frac {40 e^{x}}{x}} x^{2}-2 e^{\frac {20 e^{x}+20 \,\mathrm {log}\left (x \right )}{x}} x^{2}-2 e^{\frac {20 e^{x}+20 \,\mathrm {log}\left (x \right )}{x}} x +e^{\frac {40 \,\mathrm {log}\left (x \right )}{x}} x^{2}+2 e^{\frac {40 \,\mathrm {log}\left (x \right )}{x}} x +e^{\frac {40 \,\mathrm {log}\left (x \right )}{x}}\right )}{e^{\frac {40 \,\mathrm {log}\left (x \right )}{x}}} \] Input:
int((40*x^2*log(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*log(x)+5*e xp(x))/x)^8+((-40*x^2-40*x)*log(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^2+40*x )*exp((-5*log(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x)
Output:
(x**2*(e**((40*e**x)/x)*x**2 - 2*e**((20*e**x + 20*log(x))/x)*x**2 - 2*e** ((20*e**x + 20*log(x))/x)*x + e**((40*log(x))/x)*x**2 + 2*e**((40*log(x))/ x)*x + e**((40*log(x))/x)))/e**((40*log(x))/x)