Integrand size = 93, antiderivative size = 26 \[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=\frac {16 \left (e^{\frac {2 e^{-2+x} (-4+x)}{x}}+x^2\right )^2}{x^2} \] Output:
16*(exp(2/exp(-ln(x)+2-x)*(-4+x)/x^2)+x^2)^2/x^2
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=\frac {16 \left (e^{\frac {2 e^{-2+x} (-4+x)}{x}}+x^2\right )^2}{x^2} \] Input:
Integrate[(E^(-2 + x)*(32*E^(2 - x)*x^5 + E^((2*E^(-2 + x)*(-8 + 2*x))/x)* (256 - 256*x - 32*E^(2 - x)*x + 64*x^2) + E^((E^(-2 + x)*(-8 + 2*x))/x)*(2 56*x^2 - 256*x^3 + 64*x^4)))/x^4,x]
Output:
(16*(E^((2*E^(-2 + x)*(-4 + x))/x) + x^2)^2)/x^2
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.
\(\displaystyle \int \frac {e^{x-2} \left (32 e^{2-x} x^5+e^{\frac {2 e^{x-2} (2 x-8)}{x}} \left (64 x^2-32 e^{2-x} x-256 x+256\right )+e^{\frac {e^{x-2} (2 x-8)}{x}} \left (64 x^4-256 x^3+256 x^2\right )\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {64 e^{\frac {2 e^{x-2} (x-4)}{x}+x-2} (x-2)^2}{x^2}+\frac {32 e^{\frac {4 e^{x-2} (x-4)}{x}-2} \left (2 e^x x^2-8 e^x x-e^2 x+8 e^x\right )}{x^4}+32 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 256 \int \frac {e^{\frac {2 e^{x-2} (x-4)}{x}+x-2}}{x^2}dx+64 \int e^{\frac {2 e^{x-2} (x-4)}{x}+x-2}dx-256 \int \frac {e^{\frac {2 e^{x-2} (x-4)}{x}+x-2}}{x}dx+16 x^2+\frac {16 e^{-\frac {4 e^{x-2} (4-x)}{x}-2} \left (e^x x^2-4 e^x x+4 e^x\right )}{\left (\frac {e^{x-2} (4-x)}{x^2}-\frac {e^{x-2} (4-x)}{x}+\frac {e^{x-2}}{x}\right ) x^4}\) |
Input:
Int[(E^(-2 + x)*(32*E^(2 - x)*x^5 + E^((2*E^(-2 + x)*(-8 + 2*x))/x)*(256 - 256*x - 32*E^(2 - x)*x + 64*x^2) + E^((E^(-2 + x)*(-8 + 2*x))/x)*(256*x^2 - 256*x^3 + 64*x^4)))/x^4,x]
Output:
$Aborted
Time = 1.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54
method | result | size |
risch | \(16 x^{2}+\frac {16 \,{\mathrm e}^{\frac {4 \left (x -4\right ) {\mathrm e}^{-2+x}}{x}}}{x^{2}}+32 \,{\mathrm e}^{\frac {2 \left (x -4\right ) {\mathrm e}^{-2+x}}{x}}\) | \(40\) |
parallelrisch | \(\frac {16 x^{4}+32 \,{\mathrm e}^{\frac {2 \left (x -4\right ) {\mathrm e}^{-2+x}}{x}} x^{2}+16 \,{\mathrm e}^{\frac {4 \left (x -4\right ) {\mathrm e}^{-2+x}}{x}}}{x^{2}}\) | \(62\) |
Input:
int(((-32*x^2*exp(-ln(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp(-ln(x) +2-x))^2+(64*x^4-256*x^3+256*x^2)*exp((2*x-8)/x^2/exp(-ln(x)+2-x))+32*x^6* exp(-ln(x)+2-x))/x^5/exp(-ln(x)+2-x),x,method=_RETURNVERBOSE)
Output:
16*x^2+16/x^2*exp(4*(x-4)*exp(-2+x)/x)+32*exp(2*(x-4)*exp(-2+x)/x)
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=\frac {16 \, {\left (x^{4} + 2 \, x^{2} e^{\left (\frac {2 \, {\left (x - 4\right )} e^{\left (x + \log \left (x\right ) - 2\right )}}{x^{2}}\right )} + e^{\left (\frac {4 \, {\left (x - 4\right )} e^{\left (x + \log \left (x\right ) - 2\right )}}{x^{2}}\right )}\right )}}{x^{2}} \] Input:
integrate(((-32*x^2*exp(-log(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp (-log(x)+2-x))^2+(64*x^4-256*x^3+256*x^2)*exp((2*x-8)/x^2/exp(-log(x)+2-x) )+32*x^6*exp(-log(x)+2-x))/x^5/exp(-log(x)+2-x),x, algorithm="fricas")
Output:
16*(x^4 + 2*x^2*e^(2*(x - 4)*e^(x + log(x) - 2)/x^2) + e^(4*(x - 4)*e^(x + log(x) - 2)/x^2))/x^2
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=16 x^{2} + \frac {32 x^{2} e^{\frac {\left (2 x - 8\right ) e^{x - 2}}{x}} + 16 e^{\frac {2 \cdot \left (2 x - 8\right ) e^{x - 2}}{x}}}{x^{2}} \] Input:
integrate(((-32*x**2*exp(-ln(x)+2-x)+64*x**2-256*x+256)*exp((2*x-8)/x**2/e xp(-ln(x)+2-x))**2+(64*x**4-256*x**3+256*x**2)*exp((2*x-8)/x**2/exp(-ln(x) +2-x))+32*x**6*exp(-ln(x)+2-x))/x**5/exp(-ln(x)+2-x),x)
Output:
16*x**2 + (32*x**2*exp((2*x - 8)*exp(x - 2)/x) + 16*exp(2*(2*x - 8)*exp(x - 2)/x))/x**2
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=16 \, x^{2} + \frac {16 \, {\left (2 \, x^{2} e^{\left (-\frac {8 \, e^{\left (x - 2\right )}}{x} + 2 \, e^{\left (x - 2\right )}\right )} + e^{\left (-\frac {16 \, e^{\left (x - 2\right )}}{x} + 4 \, e^{\left (x - 2\right )}\right )}\right )}}{x^{2}} \] Input:
integrate(((-32*x^2*exp(-log(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp (-log(x)+2-x))^2+(64*x^4-256*x^3+256*x^2)*exp((2*x-8)/x^2/exp(-log(x)+2-x) )+32*x^6*exp(-log(x)+2-x))/x^5/exp(-log(x)+2-x),x, algorithm="maxima")
Output:
16*x^2 + 16*(2*x^2*e^(-8*e^(x - 2)/x + 2*e^(x - 2)) + e^(-16*e^(x - 2)/x + 4*e^(x - 2)))/x^2
\[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=\int { \frac {32 \, {\left (x^{6} e^{\left (-x - \log \left (x\right ) + 2\right )} - {\left (x^{2} e^{\left (-x - \log \left (x\right ) + 2\right )} - 2 \, x^{2} + 8 \, x - 8\right )} e^{\left (\frac {4 \, {\left (x - 4\right )} e^{\left (x + \log \left (x\right ) - 2\right )}}{x^{2}}\right )} + 2 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )} e^{\left (\frac {2 \, {\left (x - 4\right )} e^{\left (x + \log \left (x\right ) - 2\right )}}{x^{2}}\right )}\right )} e^{\left (x + \log \left (x\right ) - 2\right )}}{x^{5}} \,d x } \] Input:
integrate(((-32*x^2*exp(-log(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp (-log(x)+2-x))^2+(64*x^4-256*x^3+256*x^2)*exp((2*x-8)/x^2/exp(-log(x)+2-x) )+32*x^6*exp(-log(x)+2-x))/x^5/exp(-log(x)+2-x),x, algorithm="giac")
Output:
integrate(32*(x^6*e^(-x - log(x) + 2) - (x^2*e^(-x - log(x) + 2) - 2*x^2 + 8*x - 8)*e^(4*(x - 4)*e^(x + log(x) - 2)/x^2) + 2*(x^4 - 4*x^3 + 4*x^2)*e ^(2*(x - 4)*e^(x + log(x) - 2)/x^2))*e^(x + log(x) - 2)/x^5, x)
Time = 1.95 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=32\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {8\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x}}+16\,x^2+\frac {16\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {16\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x}}}{x^2} \] Input:
int((exp(x + log(x) - 2)*(32*x^6*exp(2 - log(x) - x) + exp((exp(x + log(x) - 2)*(2*x - 8))/x^2)*(256*x^2 - 256*x^3 + 64*x^4) - exp((2*exp(x + log(x) - 2)*(2*x - 8))/x^2)*(256*x + 32*x^2*exp(2 - log(x) - x) - 64*x^2 - 256)) )/x^5,x)
Output:
32*exp(2*exp(-2)*exp(x))*exp(-(8*exp(-2)*exp(x))/x) + 16*x^2 + (16*exp(4*e xp(-2)*exp(x))*exp(-(16*exp(-2)*exp(x))/x))/x^2
Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.85 \[ \int \frac {e^{-2+x} \left (32 e^{2-x} x^5+e^{\frac {2 e^{-2+x} (-8+2 x)}{x}} \left (256-256 x-32 e^{2-x} x+64 x^2\right )+e^{\frac {e^{-2+x} (-8+2 x)}{x}} \left (256 x^2-256 x^3+64 x^4\right )\right )}{x^4} \, dx=\frac {16 e^{\frac {4 e^{x}}{e^{2}}}+32 e^{\frac {2 e^{x} x +8 e^{x}}{e^{2} x}} x^{2}+16 e^{\frac {16 e^{x}}{e^{2} x}} x^{4}}{e^{\frac {16 e^{x}}{e^{2} x}} x^{2}} \] Input:
int(((-32*x^2*exp(-log(x)+2-x)+64*x^2-256*x+256)*exp((2*x-8)/x^2/exp(-log( x)+2-x))^2+(64*x^4-256*x^3+256*x^2)*exp((2*x-8)/x^2/exp(-log(x)+2-x))+32*x ^6*exp(-log(x)+2-x))/x^5/exp(-log(x)+2-x),x)
Output:
(16*(e**((4*e**x)/e**2) + 2*e**((2*e**x*x + 8*e**x)/(e**2*x))*x**2 + e**(( 16*e**x)/(e**2*x))*x**4))/(e**((16*e**x)/(e**2*x))*x**2)