Integrand size = 141, antiderivative size = 26 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)} \] Output:
exp(x)/(exp(x)-5)/exp(exp(4)/x-2*ln(2))/(1+x)
Time = 1.97 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)} \] Input:
Integrate[(E^x*(E^4*(-5 - 5*x) - 5*x^3) + E^(2*x)*(-x^2 + E^4*(1 + x)))/(E ^(x + (E^4 - x*Log[4])/x)*(-10*x^2 - 20*x^3 - 10*x^4) + E^(2*x + (E^4 - x* Log[4])/x)*(x^2 + 2*x^3 + x^4) + E^((E^4 - x*Log[4])/x)*(25*x^2 + 50*x^3 + 25*x^4)),x]
Output:
(4*E^(-(E^4/x) + x))/((-5 + E^x)*(1 + 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.
\(\displaystyle \int \frac {e^x \left (e^4 (-5 x-5)-5 x^3\right )+e^{2 x} \left (e^4 (x+1)-x^2\right )}{\left (-10 x^4-20 x^3-10 x^2\right ) e^{x+\frac {e^4-x \log (4)}{x}}+\left (x^4+2 x^3+x^2\right ) e^{2 x+\frac {e^4-x \log (4)}{x}}+\left (25 x^4+50 x^3+25 x^2\right ) e^{\frac {e^4-x \log (4)}{x}}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 e^{x-\frac {e^4}{x}} \left (-5 x^3-e^x x^2+e^{x+4} (x+1)-5 e^4 (x+1)\right )}{\left (5-e^x\right )^2 x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {e^{x-\frac {e^4}{x}} \left (5 x^3+e^x x^2-e^{x+4} (x+1)+5 e^4 (x+1)\right )}{\left (5-e^x\right )^2 x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {e^{x-\frac {e^4}{x}} \left (5 x^3+e^x x^2-e^{x+4} (x+1)+5 e^4 (x+1)\right )}{\left (5-e^x\right )^2 x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (\frac {e^{x-\frac {e^4}{x}} \left (x^2-e^4 x-e^4\right )}{\left (-5+e^x\right ) x^2 (x+1)^2}+\frac {5 e^{x-\frac {e^4}{x}}}{\left (-5+e^x\right )^2 (x+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (-\int \frac {e^{x+4-\frac {e^4}{x}}}{\left (-5+e^x\right ) x^2}dx+\int \frac {e^{x+4-\frac {e^4}{x}}}{\left (-5+e^x\right ) x}dx+\int \frac {e^{x-\frac {e^4}{x}}}{\left (-5+e^x\right ) (x+1)^2}dx+5 \int \frac {e^{x-\frac {e^4}{x}}}{\left (-5+e^x\right )^2 (x+1)}dx-\int \frac {e^{x+4-\frac {e^4}{x}}}{\left (-5+e^x\right ) (x+1)}dx\right )\) |
Input:
Int[(E^x*(E^4*(-5 - 5*x) - 5*x^3) + E^(2*x)*(-x^2 + E^4*(1 + x)))/(E^(x + (E^4 - x*Log[4])/x)*(-10*x^2 - 20*x^3 - 10*x^4) + E^(2*x + (E^4 - x*Log[4] )/x)*(x^2 + 2*x^3 + x^4) + E^((E^4 - x*Log[4])/x)*(25*x^2 + 50*x^3 + 25*x^ 4)),x]
Output:
$Aborted
Time = 2.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{-\frac {-x^{2}+{\mathrm e}^{4}}{x}}}{\left (1+x \right ) \left ({\mathrm e}^{x}-5\right )}\) | \(28\) |
norman | \(\frac {{\mathrm e}^{x} {\mathrm e}^{-\frac {-2 x \ln \left (2\right )+{\mathrm e}^{4}}{x}}}{\left ({\mathrm e}^{x}-5\right ) \left (1+x \right )}\) | \(30\) |
parallelrisch | \(\frac {{\mathrm e}^{x} {\mathrm e}^{-\frac {-2 x \ln \left (2\right )+{\mathrm e}^{4}}{x}}}{\left ({\mathrm e}^{x}-5\right ) \left (1+x \right )}\) | \(30\) |
Input:
int((((1+x)*exp(4)-x^2)*exp(x)^2+((-5*x-5)*exp(4)-5*x^3)*exp(x))/((x^4+2*x ^3+x^2)*exp((-2*x*ln(2)+exp(4))/x)*exp(x)^2+(-10*x^4-20*x^3-10*x^2)*exp((- 2*x*ln(2)+exp(4))/x)*exp(x)+(25*x^4+50*x^3+25*x^2)*exp((-2*x*ln(2)+exp(4)) /x)),x,method=_RETURNVERBOSE)
Output:
4/(1+x)/(exp(x)-5)*exp(-(-x^2+exp(4))/x)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {e^{\left (3 \, x\right )}}{{\left (x + 1\right )} e^{\left (x + \frac {2 \, x^{2} - 2 \, x \log \left (2\right ) + e^{4}}{x}\right )} - 5 \, {\left (x + 1\right )} e^{\left (\frac {2 \, x^{2} - 2 \, x \log \left (2\right ) + e^{4}}{x}\right )}} \] Input:
integrate((((1+x)*exp(4)-x^2)*exp(x)^2+((-5*x-5)*exp(4)-5*x^3)*exp(x))/((x ^4+2*x^3+x^2)*exp((-2*x*log(2)+exp(4))/x)*exp(x)^2+(-10*x^4-20*x^3-10*x^2) *exp((-2*x*log(2)+exp(4))/x)*exp(x)+(25*x^4+50*x^3+25*x^2)*exp((-2*x*log(2 )+exp(4))/x)),x, algorithm="fricas")
Output:
e^(3*x)/((x + 1)*e^(x + (2*x^2 - 2*x*log(2) + e^4)/x) - 5*(x + 1)*e^((2*x^ 2 - 2*x*log(2) + e^4)/x))
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {e^{x} e^{- \frac {- 2 x \log {\left (2 \right )} + e^{4}}{x}}}{x e^{x} - 5 x + e^{x} - 5} \] Input:
integrate((((1+x)*exp(4)-x**2)*exp(x)**2+((-5*x-5)*exp(4)-5*x**3)*exp(x))/ ((x**4+2*x**3+x**2)*exp((-2*x*ln(2)+exp(4))/x)*exp(x)**2+(-10*x**4-20*x**3 -10*x**2)*exp((-2*x*ln(2)+exp(4))/x)*exp(x)+(25*x**4+50*x**3+25*x**2)*exp( (-2*x*ln(2)+exp(4))/x)),x)
Output:
exp(x)*exp(-(-2*x*log(2) + exp(4))/x)/(x*exp(x) - 5*x + exp(x) - 5)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 \, e^{\left (x - \frac {e^{4}}{x}\right )}}{{\left (x + 1\right )} e^{x} - 5 \, x - 5} \] Input:
integrate((((1+x)*exp(4)-x^2)*exp(x)^2+((-5*x-5)*exp(4)-5*x^3)*exp(x))/((x ^4+2*x^3+x^2)*exp((-2*x*log(2)+exp(4))/x)*exp(x)^2+(-10*x^4-20*x^3-10*x^2) *exp((-2*x*log(2)+exp(4))/x)*exp(x)+(25*x^4+50*x^3+25*x^2)*exp((-2*x*log(2 )+exp(4))/x)),x, algorithm="maxima")
Output:
4*e^(x - e^4/x)/((x + 1)*e^x - 5*x - 5)
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (26) = 52\).
Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.85 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 \, e^{\left (\frac {2 \, {\left (2 \, x^{2} + e^{4}\right )}}{x}\right )}}{x e^{\left (\frac {2 \, x^{2} + e^{4}}{x} + \frac {2 \, {\left (x^{2} + e^{4}\right )}}{x}\right )} - 5 \, x e^{\left (\frac {3 \, {\left (x^{2} + e^{4}\right )}}{x}\right )} + e^{\left (\frac {2 \, x^{2} + e^{4}}{x} + \frac {2 \, {\left (x^{2} + e^{4}\right )}}{x}\right )} - 5 \, e^{\left (\frac {3 \, {\left (x^{2} + e^{4}\right )}}{x}\right )}} \] Input:
integrate((((1+x)*exp(4)-x^2)*exp(x)^2+((-5*x-5)*exp(4)-5*x^3)*exp(x))/((x ^4+2*x^3+x^2)*exp((-2*x*log(2)+exp(4))/x)*exp(x)^2+(-10*x^4-20*x^3-10*x^2) *exp((-2*x*log(2)+exp(4))/x)*exp(x)+(25*x^4+50*x^3+25*x^2)*exp((-2*x*log(2 )+exp(4))/x)),x, algorithm="giac")
Output:
4*e^(2*(2*x^2 + e^4)/x)/(x*e^((2*x^2 + e^4)/x + 2*(x^2 + e^4)/x) - 5*x*e^( 3*(x^2 + e^4)/x) + e^((2*x^2 + e^4)/x + 2*(x^2 + e^4)/x) - 5*e^(3*(x^2 + e ^4)/x))
Time = 3.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=-\frac {4\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^4}{x}}\,{\mathrm {e}}^x}{5\,x-{\mathrm {e}}^x-x\,{\mathrm {e}}^x+5} \] Input:
int((exp(2*x)*(exp(4)*(x + 1) - x^2) - exp(x)*(5*x^3 + exp(4)*(5*x + 5)))/ (exp((exp(4) - 2*x*log(2))/x)*(25*x^2 + 50*x^3 + 25*x^4) + exp(2*x)*exp((e xp(4) - 2*x*log(2))/x)*(x^2 + 2*x^3 + x^4) - exp((exp(4) - 2*x*log(2))/x)* exp(x)*(10*x^2 + 20*x^3 + 10*x^4)),x)
Output:
-(4*exp(-exp(4)/x)*exp(x))/(5*x - exp(x) - x*exp(x) + 5)
Time = 6.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 e^{x}}{e^{\frac {e^{4}}{x}} \left (e^{x} x +e^{x}-5 x -5\right )} \] Input:
int((((1+x)*exp(4)-x^2)*exp(x)^2+((-5*x-5)*exp(4)-5*x^3)*exp(x))/((x^4+2*x ^3+x^2)*exp((-2*x*log(2)+exp(4))/x)*exp(x)^2+(-10*x^4-20*x^3-10*x^2)*exp(( -2*x*log(2)+exp(4))/x)*exp(x)+(25*x^4+50*x^3+25*x^2)*exp((-2*x*log(2)+exp( 4))/x)),x)
Output:
(4*e**x)/(e**(e**4/x)*(e**x*x + e**x - 5*x - 5))