Integrand size = 85, antiderivative size = 21 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=x-\frac {5}{e^{-8 x}+\frac {x^2}{5+x}} \] Output:
x-5/(x^2/(5+x)+exp(-8*x))
Time = 3.35 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x (5+x)+e^{8 x} \left (-25-5 x+x^3\right )}{5+x+e^{8 x} x^2} \] Input:
Integrate[(50*x + 5*x^2 + x^4 + (25 + 10*x + x^2)/E^(16*x) + (-1000 - 400* x - 30*x^2 + 2*x^3)/E^(8*x))/(x^4 + (25 + 10*x + x^2)/E^(16*x) + (10*x^2 + 2*x^3)/E^(8*x)),x]
Output:
(x*(5 + x) + E^(8*x)*(-25 - 5*x + x^3))/(5 + x + E^(8*x)*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.
\(\displaystyle \int \frac {x^4+5 x^2+e^{-16 x} \left (x^2+10 x+25\right )+e^{-8 x} \left (2 x^3-30 x^2-400 x-1000\right )+50 x}{x^4+e^{-16 x} \left (x^2+10 x+25\right )+e^{-8 x} \left (2 x^3+10 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{16 x} \left (x^4+5 x^2+e^{-16 x} \left (x^2+10 x+25\right )+e^{-8 x} \left (2 x^3-30 x^2-400 x-1000\right )+50 x\right )}{\left (e^{8 x} x^2+x+5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {40 e^{16 x} x^2}{e^{8 x} x^2+x+5}+\frac {5 e^{16 x} \left (8 x^2+41 x+10\right ) x}{\left (e^{8 x} x^2+x+5\right )^2}-40 e^{8 x}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 50 \int \frac {e^{16 x} x}{\left (e^{8 x} x^2+x+5\right )^2}dx+205 \int \frac {e^{16 x} x^2}{\left (e^{8 x} x^2+x+5\right )^2}dx+40 \int \frac {e^{16 x} x^2}{e^{8 x} x^2+x+5}dx+40 \int \frac {e^{16 x} x^3}{\left (e^{8 x} x^2+x+5\right )^2}dx+x-5 e^{8 x}\) |
Input:
Int[(50*x + 5*x^2 + x^4 + (25 + 10*x + x^2)/E^(16*x) + (-1000 - 400*x - 30 *x^2 + 2*x^3)/E^(8*x))/(x^4 + (25 + 10*x + x^2)/E^(16*x) + (10*x^2 + 2*x^3 )/E^(8*x)),x]
Output:
$Aborted
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
risch | \(x -\frac {5 \left (5+x \right )}{{\mathrm e}^{-8 x} x +x^{2}+5 \,{\mathrm e}^{-8 x}}\) | \(26\) |
norman | \(\frac {-25+x^{3}+{\mathrm e}^{-8 x} x^{2}+5 \,{\mathrm e}^{-8 x} x -5 x}{{\mathrm e}^{-8 x} x +x^{2}+5 \,{\mathrm e}^{-8 x}}\) | \(43\) |
parallelrisch | \(\frac {x^{3}+{\mathrm e}^{-8 x} x^{2}-25-5 x^{2}-5 x -25 \,{\mathrm e}^{-8 x}}{{\mathrm e}^{-8 x} x +x^{2}+5 \,{\mathrm e}^{-8 x}}\) | \(47\) |
Input:
int(((x^2+10*x+25)*exp(-8*x)^2+(2*x^3-30*x^2-400*x-1000)*exp(-8*x)+x^4+5*x ^2+50*x)/((x^2+10*x+25)*exp(-8*x)^2+(2*x^3+10*x^2)*exp(-8*x)+x^4),x,method =_RETURNVERBOSE)
Output:
x-5*(5+x)/(exp(-8*x)*x+x^2+5*exp(-8*x))
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x^{3} + {\left (x^{2} + 5 \, x\right )} e^{\left (-8 \, x\right )} - 5 \, x - 25}{x^{2} + {\left (x + 5\right )} e^{\left (-8 \, x\right )}} \] Input:
integrate(((x^2+10*x+25)*exp(-8*x)^2+(2*x^3-30*x^2-400*x-1000)*exp(-8*x)+x ^4+5*x^2+50*x)/((x^2+10*x+25)*exp(-8*x)^2+(2*x^3+10*x^2)*exp(-8*x)+x^4),x, algorithm="fricas")
Output:
(x^3 + (x^2 + 5*x)*e^(-8*x) - 5*x - 25)/(x^2 + (x + 5)*e^(-8*x))
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=x + \frac {- 5 x - 25}{x^{2} + \left (x + 5\right ) e^{- 8 x}} \] Input:
integrate(((x**2+10*x+25)*exp(-8*x)**2+(2*x**3-30*x**2-400*x-1000)*exp(-8* x)+x**4+5*x**2+50*x)/((x**2+10*x+25)*exp(-8*x)**2+(2*x**3+10*x**2)*exp(-8* x)+x**4),x)
Output:
x + (-5*x - 25)/(x**2 + (x + 5)*exp(-8*x))
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x^{2} + {\left (x^{3} - 5 \, x - 25\right )} e^{\left (8 \, x\right )} + 5 \, x}{x^{2} e^{\left (8 \, x\right )} + x + 5} \] Input:
integrate(((x^2+10*x+25)*exp(-8*x)^2+(2*x^3-30*x^2-400*x-1000)*exp(-8*x)+x ^4+5*x^2+50*x)/((x^2+10*x+25)*exp(-8*x)^2+(2*x^3+10*x^2)*exp(-8*x)+x^4),x, algorithm="maxima")
Output:
(x^2 + (x^3 - 5*x - 25)*e^(8*x) + 5*x)/(x^2*e^(8*x) + x + 5)
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x^{3} + x^{2} e^{\left (-8 \, x\right )} + 5 \, x e^{\left (-8 \, x\right )} - 5 \, x - 25}{x^{2} + x e^{\left (-8 \, x\right )} + 5 \, e^{\left (-8 \, x\right )}} \] Input:
integrate(((x^2+10*x+25)*exp(-8*x)^2+(2*x^3-30*x^2-400*x-1000)*exp(-8*x)+x ^4+5*x^2+50*x)/((x^2+10*x+25)*exp(-8*x)^2+(2*x^3+10*x^2)*exp(-8*x)+x^4),x, algorithm="giac")
Output:
(x^3 + x^2*e^(-8*x) + 5*x*e^(-8*x) - 5*x - 25)/(x^2 + x*e^(-8*x) + 5*e^(-8 *x))
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=x-\frac {25\,{\mathrm {e}}^{8\,x}+5\,x\,{\mathrm {e}}^{8\,x}}{x+x^2\,{\mathrm {e}}^{8\,x}+5} \] Input:
int((50*x - exp(-8*x)*(400*x + 30*x^2 - 2*x^3 + 1000) + exp(-16*x)*(10*x + x^2 + 25) + 5*x^2 + x^4)/(exp(-8*x)*(10*x^2 + 2*x^3) + exp(-16*x)*(10*x + x^2 + 25) + x^4),x)
Output:
x - (25*exp(8*x) + 5*x*exp(8*x))/(x + x^2*exp(8*x) + 5)
\[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=-1000 \left (\int \frac {e^{8 x}}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right )+\int \frac {x^{2}}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x +\int \frac {e^{16 x} x^{4}}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x +5 \left (\int \frac {e^{16 x} x^{2}}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right )+50 \left (\int \frac {e^{16 x} x}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right )+2 \left (\int \frac {e^{8 x} x^{3}}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right )-30 \left (\int \frac {e^{8 x} x^{2}}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right )-400 \left (\int \frac {e^{8 x} x}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right )+10 \left (\int \frac {x}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right )+25 \left (\int \frac {1}{e^{16 x} x^{4}+2 e^{8 x} x^{3}+10 e^{8 x} x^{2}+x^{2}+10 x +25}d x \right ) \] Input:
int(((x^2+10*x+25)*exp(-8*x)^2+(2*x^3-30*x^2-400*x-1000)*exp(-8*x)+x^4+5*x ^2+50*x)/((x^2+10*x+25)*exp(-8*x)^2+(2*x^3+10*x^2)*exp(-8*x)+x^4),x)
Output:
- 1000*int(e**(8*x)/(e**(16*x)*x**4 + 2*e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x**2 + 10*x + 25),x) + int(x**2/(e**(16*x)*x**4 + 2*e**(8*x)*x**3 + 10*e **(8*x)*x**2 + x**2 + 10*x + 25),x) + int((e**(16*x)*x**4)/(e**(16*x)*x**4 + 2*e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x**2 + 10*x + 25),x) + 5*int((e**( 16*x)*x**2)/(e**(16*x)*x**4 + 2*e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x**2 + 10*x + 25),x) + 50*int((e**(16*x)*x)/(e**(16*x)*x**4 + 2*e**(8*x)*x**3 + 1 0*e**(8*x)*x**2 + x**2 + 10*x + 25),x) + 2*int((e**(8*x)*x**3)/(e**(16*x)* x**4 + 2*e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x**2 + 10*x + 25),x) - 30*int( (e**(8*x)*x**2)/(e**(16*x)*x**4 + 2*e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x** 2 + 10*x + 25),x) - 400*int((e**(8*x)*x)/(e**(16*x)*x**4 + 2*e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x**2 + 10*x + 25),x) + 10*int(x/(e**(16*x)*x**4 + 2* e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x**2 + 10*x + 25),x) + 25*int(1/(e**(16 *x)*x**4 + 2*e**(8*x)*x**3 + 10*e**(8*x)*x**2 + x**2 + 10*x + 25),x)