Integrand size = 108, antiderivative size = 19 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=\frac {x}{\left (-1+e^{-e^{5+x}} x^2\right )^4} \] Output:
x/(-1+x^2/exp(exp(5)*exp(x)))^4
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=\frac {e^{4 e^{5+x}} x}{\left (e^{e^{5+x}}-x^2\right )^4} \] Input:
Integrate[(E^(5*E^(5 + x)) + E^(4*E^(5 + x))*(7*x^2 - 4*E^(5 + x)*x^3))/(E ^(5*E^(5 + x)) - 5*E^(4*E^(5 + x))*x^2 + 10*E^(3*E^(5 + x))*x^4 - 10*E^(2* E^(5 + x))*x^6 + 5*E^E^(5 + x)*x^8 - x^10),x]
Output:
(E^(4*E^(5 + x))*x)/(E^E^(5 + x) - x^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.
\(\displaystyle \int \frac {e^{4 e^{x+5}} \left (7 x^2-4 e^{x+5} x^3\right )+e^{5 e^{x+5}}}{-x^{10}+5 e^{e^{x+5}} x^8-10 e^{2 e^{x+5}} x^6+10 e^{3 e^{x+5}} x^4-5 e^{4 e^{x+5}} x^2+e^{5 e^{x+5}}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{4 e^{x+5}} \left (-4 e^{x+5} x^3+7 x^2+e^{e^{x+5}}\right )}{\left (e^{e^{x+5}}-x^2\right )^5}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{4 e^{x+5}} \left (7 x^2+e^{e^{x+5}}\right )}{\left (e^{e^{x+5}}-x^2\right )^5}+\frac {4 e^{x+4 e^{x+5}+5} x^3}{\left (x^2-e^{e^{x+5}}\right )^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{4 e^{x+5}}}{\left (e^{e^{x+5}}-x^2\right )^4}dx-8 \int \frac {e^{4 e^{x+5}} x^2}{\left (x^2-e^{e^{x+5}}\right )^5}dx+4 \int \frac {e^{x+4 e^{x+5}+5} x^3}{\left (x^2-e^{e^{x+5}}\right )^5}dx\) |
Input:
Int[(E^(5*E^(5 + x)) + E^(4*E^(5 + x))*(7*x^2 - 4*E^(5 + x)*x^3))/(E^(5*E^ (5 + x)) - 5*E^(4*E^(5 + x))*x^2 + 10*E^(3*E^(5 + x))*x^4 - 10*E^(2*E^(5 + x))*x^6 + 5*E^E^(5 + x)*x^8 - x^10),x]
Output:
$Aborted
Time = 4.42 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{4 \,{\mathrm e}^{5+x}}}{\left (-x^{2}+{\mathrm e}^{{\mathrm e}^{5+x}}\right )^{4}}\) | \(23\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{4 \,{\mathrm e}^{5+x}}}{x^{8}-4 x^{6} {\mathrm e}^{{\mathrm e}^{5} {\mathrm e}^{x}}+6 \,{\mathrm e}^{2 \,{\mathrm e}^{5+x}} x^{4}-4 \,{\mathrm e}^{3 \,{\mathrm e}^{5+x}} x^{2}+{\mathrm e}^{4 \,{\mathrm e}^{5+x}}}\) | \(62\) |
Input:
int((exp(exp(5)*exp(x))^5+(-4*x^3*exp(5)*exp(x)+7*x^2)*exp(exp(5)*exp(x))^ 4)/(exp(exp(5)*exp(x))^5-5*x^2*exp(exp(5)*exp(x))^4+10*x^4*exp(exp(5)*exp( x))^3-10*x^6*exp(exp(5)*exp(x))^2+5*x^8*exp(exp(5)*exp(x))-x^10),x,method= _RETURNVERBOSE)
Output:
x*exp(4*exp(5+x))/(-x^2+exp(exp(5+x)))^4
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (17) = 34\).
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.95 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=\frac {x e^{\left (4 \, e^{\left (x + 5\right )}\right )}}{x^{8} - 4 \, x^{6} e^{\left (e^{\left (x + 5\right )}\right )} + 6 \, x^{4} e^{\left (2 \, e^{\left (x + 5\right )}\right )} - 4 \, x^{2} e^{\left (3 \, e^{\left (x + 5\right )}\right )} + e^{\left (4 \, e^{\left (x + 5\right )}\right )}} \] Input:
integrate((exp(exp(5)*exp(x))^5+(-4*x^3*exp(5)*exp(x)+7*x^2)*exp(exp(5)*ex p(x))^4)/(exp(exp(5)*exp(x))^5-5*x^2*exp(exp(5)*exp(x))^4+10*x^4*exp(exp(5 )*exp(x))^3-10*x^6*exp(exp(5)*exp(x))^2+5*x^8*exp(exp(5)*exp(x))-x^10),x, algorithm="fricas")
Output:
x*e^(4*e^(x + 5))/(x^8 - 4*x^6*e^(e^(x + 5)) + 6*x^4*e^(2*e^(x + 5)) - 4*x ^2*e^(3*e^(x + 5)) + e^(4*e^(x + 5)))
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (15) = 30\).
Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 5.53 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=x + \frac {- x^{9} + 4 x^{7} e^{e^{5} e^{x}} - 6 x^{5} e^{2 e^{5} e^{x}} + 4 x^{3} e^{3 e^{5} e^{x}}}{x^{8} - 4 x^{6} e^{e^{5} e^{x}} + 6 x^{4} e^{2 e^{5} e^{x}} - 4 x^{2} e^{3 e^{5} e^{x}} + e^{4 e^{5} e^{x}}} \] Input:
integrate((exp(exp(5)*exp(x))**5+(-4*x**3*exp(5)*exp(x)+7*x**2)*exp(exp(5) *exp(x))**4)/(exp(exp(5)*exp(x))**5-5*x**2*exp(exp(5)*exp(x))**4+10*x**4*e xp(exp(5)*exp(x))**3-10*x**6*exp(exp(5)*exp(x))**2+5*x**8*exp(exp(5)*exp(x ))-x**10),x)
Output:
x + (-x**9 + 4*x**7*exp(exp(5)*exp(x)) - 6*x**5*exp(2*exp(5)*exp(x)) + 4*x **3*exp(3*exp(5)*exp(x)))/(x**8 - 4*x**6*exp(exp(5)*exp(x)) + 6*x**4*exp(2 *exp(5)*exp(x)) - 4*x**2*exp(3*exp(5)*exp(x)) + exp(4*exp(5)*exp(x)))
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (17) = 34\).
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.95 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=\frac {x e^{\left (4 \, e^{\left (x + 5\right )}\right )}}{x^{8} - 4 \, x^{6} e^{\left (e^{\left (x + 5\right )}\right )} + 6 \, x^{4} e^{\left (2 \, e^{\left (x + 5\right )}\right )} - 4 \, x^{2} e^{\left (3 \, e^{\left (x + 5\right )}\right )} + e^{\left (4 \, e^{\left (x + 5\right )}\right )}} \] Input:
integrate((exp(exp(5)*exp(x))^5+(-4*x^3*exp(5)*exp(x)+7*x^2)*exp(exp(5)*ex p(x))^4)/(exp(exp(5)*exp(x))^5-5*x^2*exp(exp(5)*exp(x))^4+10*x^4*exp(exp(5 )*exp(x))^3-10*x^6*exp(exp(5)*exp(x))^2+5*x^8*exp(exp(5)*exp(x))-x^10),x, algorithm="maxima")
Output:
x*e^(4*e^(x + 5))/(x^8 - 4*x^6*e^(e^(x + 5)) + 6*x^4*e^(2*e^(x + 5)) - 4*x ^2*e^(3*e^(x + 5)) + e^(4*e^(x + 5)))
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (17) = 34\).
Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.95 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=\frac {x e^{\left (4 \, e^{\left (x + 5\right )}\right )}}{x^{8} - 4 \, x^{6} e^{\left (e^{\left (x + 5\right )}\right )} + 6 \, x^{4} e^{\left (2 \, e^{\left (x + 5\right )}\right )} - 4 \, x^{2} e^{\left (3 \, e^{\left (x + 5\right )}\right )} + e^{\left (4 \, e^{\left (x + 5\right )}\right )}} \] Input:
integrate((exp(exp(5)*exp(x))^5+(-4*x^3*exp(5)*exp(x)+7*x^2)*exp(exp(5)*ex p(x))^4)/(exp(exp(5)*exp(x))^5-5*x^2*exp(exp(5)*exp(x))^4+10*x^4*exp(exp(5 )*exp(x))^3-10*x^6*exp(exp(5)*exp(x))^2+5*x^8*exp(exp(5)*exp(x))-x^10),x, algorithm="giac")
Output:
x*e^(4*e^(x + 5))/(x^8 - 4*x^6*e^(e^(x + 5)) + 6*x^4*e^(2*e^(x + 5)) - 4*x ^2*e^(3*e^(x + 5)) + e^(4*e^(x + 5)))
Time = 3.17 (sec) , antiderivative size = 250, normalized size of antiderivative = 13.16 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=x-\frac {6\,\left (x^7\,{\mathrm {e}}^{x+5}-2\,x^6\right )}{\left (2\,x-x^2\,{\mathrm {e}}^{x+5}\right )\,\left ({\mathrm {e}}^{2\,{\mathrm {e}}^5\,{\mathrm {e}}^x}-2\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^5\,{\mathrm {e}}^x}+x^4\right )}-\frac {4\,\left (x^5\,{\mathrm {e}}^{x+5}-2\,x^4\right )}{\left ({\mathrm {e}}^{{\mathrm {e}}^5\,{\mathrm {e}}^x}-x^2\right )\,\left (2\,x-x^2\,{\mathrm {e}}^{x+5}\right )}-\frac {x^{11}\,{\mathrm {e}}^{x+5}-2\,x^{10}}{\left (2\,x-x^2\,{\mathrm {e}}^{x+5}\right )\,\left ({\mathrm {e}}^{4\,{\mathrm {e}}^5\,{\mathrm {e}}^x}-4\,x^2\,{\mathrm {e}}^{3\,{\mathrm {e}}^5\,{\mathrm {e}}^x}+6\,x^4\,{\mathrm {e}}^{2\,{\mathrm {e}}^5\,{\mathrm {e}}^x}-4\,x^6\,{\mathrm {e}}^{{\mathrm {e}}^5\,{\mathrm {e}}^x}+x^8\right )}-\frac {4\,\left (x^9\,{\mathrm {e}}^{x+5}-2\,x^8\right )}{\left (2\,x-x^2\,{\mathrm {e}}^{x+5}\right )\,\left ({\mathrm {e}}^{3\,{\mathrm {e}}^5\,{\mathrm {e}}^x}-3\,x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^5\,{\mathrm {e}}^x}+3\,x^4\,{\mathrm {e}}^{{\mathrm {e}}^5\,{\mathrm {e}}^x}-x^6\right )} \] Input:
int((exp(5*exp(5)*exp(x)) + exp(4*exp(5)*exp(x))*(7*x^2 - 4*x^3*exp(5)*exp (x)))/(exp(5*exp(5)*exp(x)) - 5*x^2*exp(4*exp(5)*exp(x)) + 10*x^4*exp(3*ex p(5)*exp(x)) - 10*x^6*exp(2*exp(5)*exp(x)) + 5*x^8*exp(exp(5)*exp(x)) - x^ 10),x)
Output:
x - (6*(x^7*exp(x + 5) - 2*x^6))/((2*x - x^2*exp(x + 5))*(exp(2*exp(5)*exp (x)) - 2*x^2*exp(exp(5)*exp(x)) + x^4)) - (4*(x^5*exp(x + 5) - 2*x^4))/((e xp(exp(5)*exp(x)) - x^2)*(2*x - x^2*exp(x + 5))) - (x^11*exp(x + 5) - 2*x^ 10)/((2*x - x^2*exp(x + 5))*(exp(4*exp(5)*exp(x)) - 4*x^2*exp(3*exp(5)*exp (x)) + 6*x^4*exp(2*exp(5)*exp(x)) - 4*x^6*exp(exp(5)*exp(x)) + x^8)) - (4* (x^9*exp(x + 5) - 2*x^8))/((2*x - x^2*exp(x + 5))*(exp(3*exp(5)*exp(x)) - 3*x^2*exp(2*exp(5)*exp(x)) + 3*x^4*exp(exp(5)*exp(x)) - x^6))
Time = 0.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.79 \[ \int \frac {e^{5 e^{5+x}}+e^{4 e^{5+x}} \left (7 x^2-4 e^{5+x} x^3\right )}{e^{5 e^{5+x}}-5 e^{4 e^{5+x}} x^2+10 e^{3 e^{5+x}} x^4-10 e^{2 e^{5+x}} x^6+5 e^{e^{5+x}} x^8-x^{10}} \, dx=\frac {e^{4 e^{x} e^{5}} x}{e^{4 e^{x} e^{5}}-4 e^{3 e^{x} e^{5}} x^{2}+6 e^{2 e^{x} e^{5}} x^{4}-4 e^{e^{x} e^{5}} x^{6}+x^{8}} \] Input:
int((exp(exp(5)*exp(x))^5+(-4*x^3*exp(5)*exp(x)+7*x^2)*exp(exp(5)*exp(x))^ 4)/(exp(exp(5)*exp(x))^5-5*x^2*exp(exp(5)*exp(x))^4+10*x^4*exp(exp(5)*exp( x))^3-10*x^6*exp(exp(5)*exp(x))^2+5*x^8*exp(exp(5)*exp(x))-x^10),x)
Output:
(e**(4*e**x*e**5)*x)/(e**(4*e**x*e**5) - 4*e**(3*e**x*e**5)*x**2 + 6*e**(2 *e**x*e**5)*x**4 - 4*e**(e**x*e**5)*x**6 + x**8)