Integrand size = 87, antiderivative size = 23 \[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=\frac {e^5}{-7-e^{2 x}+\frac {1}{-1+x}-x} \] Output:
exp(5)/(1/(-1+x)-7-exp(x)^2-x)
Time = 1.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=\frac {e^5 (1-x)}{-8+e^{2 x} (-1+x)+6 x+x^2} \] Input:
Integrate[(E^5*(2 - 2*x + x^2) + E^(5 + 2*x)*(2 - 4*x + 2*x^2))/(64 - 96*x + 20*x^2 + 12*x^3 + x^4 + E^(4*x)*(1 - 2*x + x^2) + E^(2*x)*(16 - 28*x + 10*x^2 + 2*x^3)),x]
Output:
(E^5*(1 - x))/(-8 + E^(2*x)*(-1 + x) + 6*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.
\(\displaystyle \int \frac {e^5 \left (x^2-2 x+2\right )+e^{2 x+5} \left (2 x^2-4 x+2\right )}{x^4+12 x^3+20 x^2+e^{4 x} \left (x^2-2 x+1\right )+e^{2 x} \left (2 x^3+10 x^2-28 x+16\right )-96 x+64} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^5 \left (x^2+2 e^{2 x} (x-1)^2-2 x+2\right )}{\left (-x^2-6 x-e^{2 x} (x-1)+8\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^5 \int \frac {2 e^{2 x} (1-x)^2+x^2-2 x+2}{\left (-x^2-6 x+e^{2 x} (1-x)+8\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle e^5 \int \left (\frac {2 (x-1)}{x^2+e^{2 x} x+6 x-e^{2 x}-8}-\frac {2 x^3+9 x^2-26 x+14}{\left (x^2+e^{2 x} x+6 x-e^{2 x}-8\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^5 \left (-14 \int \frac {1}{\left (x^2+e^{2 x} x+6 x-e^{2 x}-8\right )^2}dx+26 \int \frac {x}{\left (x^2+e^{2 x} x+6 x-e^{2 x}-8\right )^2}dx-9 \int \frac {x^2}{\left (x^2+e^{2 x} x+6 x-e^{2 x}-8\right )^2}dx-2 \int \frac {1}{x^2+e^{2 x} x+6 x-e^{2 x}-8}dx+2 \int \frac {x}{x^2+e^{2 x} x+6 x-e^{2 x}-8}dx-2 \int \frac {x^3}{\left (x^2+e^{2 x} x+6 x-e^{2 x}-8\right )^2}dx\right )\) |
Input:
Int[(E^5*(2 - 2*x + x^2) + E^(5 + 2*x)*(2 - 4*x + 2*x^2))/(64 - 96*x + 20* x^2 + 12*x^3 + x^4 + E^(4*x)*(1 - 2*x + x^2) + E^(2*x)*(16 - 28*x + 10*x^2 + 2*x^3)),x]
Output:
$Aborted
Time = 0.48 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30
method | result | size |
risch | \(-\frac {{\mathrm e}^{5} \left (-1+x \right )}{x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x}+x^{2}+6 x -8}\) | \(30\) |
norman | \(\frac {-x \,{\mathrm e}^{5}+{\mathrm e}^{5}}{x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x}+x^{2}+6 x -8}\) | \(32\) |
parallelrisch | \(-\frac {x \,{\mathrm e}^{5}-{\mathrm e}^{5}}{x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x}+x^{2}+6 x -8}\) | \(34\) |
Input:
int(((2*x^2-4*x+2)*exp(5)*exp(x)^2+(x^2-2*x+2)*exp(5))/((x^2-2*x+1)*exp(x) ^4+(2*x^3+10*x^2-28*x+16)*exp(x)^2+x^4+12*x^3+20*x^2-96*x+64),x,method=_RE TURNVERBOSE)
Output:
-exp(5)*(-1+x)/(x*exp(2*x)-exp(2*x)+x^2+6*x-8)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=-\frac {{\left (x - 1\right )} e^{10}}{{\left (x^{2} + 6 \, x - 8\right )} e^{5} + {\left (x - 1\right )} e^{\left (2 \, x + 5\right )}} \] Input:
integrate(((2*x^2-4*x+2)*exp(5)*exp(x)^2+(x^2-2*x+2)*exp(5))/((x^2-2*x+1)* exp(x)^4+(2*x^3+10*x^2-28*x+16)*exp(x)^2+x^4+12*x^3+20*x^2-96*x+64),x, alg orithm="fricas")
Output:
-(x - 1)*e^10/((x^2 + 6*x - 8)*e^5 + (x - 1)*e^(2*x + 5))
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=\frac {- x e^{5} + e^{5}}{x^{2} + 6 x + \left (x - 1\right ) e^{2 x} - 8} \] Input:
integrate(((2*x**2-4*x+2)*exp(5)*exp(x)**2+(x**2-2*x+2)*exp(5))/((x**2-2*x +1)*exp(x)**4+(2*x**3+10*x**2-28*x+16)*exp(x)**2+x**4+12*x**3+20*x**2-96*x +64),x)
Output:
(-x*exp(5) + exp(5))/(x**2 + 6*x + (x - 1)*exp(2*x) - 8)
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=-\frac {x e^{5} - e^{5}}{x^{2} + {\left (x - 1\right )} e^{\left (2 \, x\right )} + 6 \, x - 8} \] Input:
integrate(((2*x^2-4*x+2)*exp(5)*exp(x)^2+(x^2-2*x+2)*exp(5))/((x^2-2*x+1)* exp(x)^4+(2*x^3+10*x^2-28*x+16)*exp(x)^2+x^4+12*x^3+20*x^2-96*x+64),x, alg orithm="maxima")
Output:
-(x*e^5 - e^5)/(x^2 + (x - 1)*e^(2*x) + 6*x - 8)
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.70 \[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=-\frac {2 \, {\left ({\left (2 \, x + 5\right )} e^{10} - 7 \, e^{10}\right )}}{{\left (2 \, x + 5\right )}^{2} e^{5} + 2 \, {\left (2 \, x + 5\right )} e^{5} + 2 \, {\left (2 \, x + 5\right )} e^{\left (2 \, x + 5\right )} - 67 \, e^{5} - 14 \, e^{\left (2 \, x + 5\right )}} \] Input:
integrate(((2*x^2-4*x+2)*exp(5)*exp(x)^2+(x^2-2*x+2)*exp(5))/((x^2-2*x+1)* exp(x)^4+(2*x^3+10*x^2-28*x+16)*exp(x)^2+x^4+12*x^3+20*x^2-96*x+64),x, alg orithm="giac")
Output:
-2*((2*x + 5)*e^10 - 7*e^10)/((2*x + 5)^2*e^5 + 2*(2*x + 5)*e^5 + 2*(2*x + 5)*e^(2*x + 5) - 67*e^5 - 14*e^(2*x + 5))
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35 \[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=-\frac {\frac {{\mathrm {e}}^{2\,x+5}}{8}+\frac {x\,{\mathrm {e}}^5}{4}-\frac {x\,{\mathrm {e}}^{2\,x+5}}{8}-\frac {x^2\,{\mathrm {e}}^5}{8}}{6\,x-{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^{2\,x}+x^2-8} \] Input:
int((exp(5)*(x^2 - 2*x + 2) + exp(2*x)*exp(5)*(2*x^2 - 4*x + 2))/(exp(2*x) *(10*x^2 - 28*x + 2*x^3 + 16) - 96*x + exp(4*x)*(x^2 - 2*x + 1) + 20*x^2 + 12*x^3 + x^4 + 64),x)
Output:
-(exp(2*x + 5)/8 + (x*exp(5))/4 - (x*exp(2*x + 5))/8 - (x^2*exp(5))/8)/(6* x - exp(2*x) + x*exp(2*x) + x^2 - 8)
\[ \int \frac {e^5 \left (2-2 x+x^2\right )+e^{5+2 x} \left (2-4 x+2 x^2\right )}{64-96 x+20 x^2+12 x^3+x^4+e^{4 x} \left (1-2 x+x^2\right )+e^{2 x} \left (16-28 x+10 x^2+2 x^3\right )} \, dx=e^{5} \left (2 \left (\int \frac {e^{2 x}}{e^{4 x} x^{2}-2 e^{4 x} x +e^{4 x}+2 e^{2 x} x^{3}+10 e^{2 x} x^{2}-28 e^{2 x} x +16 e^{2 x}+x^{4}+12 x^{3}+20 x^{2}-96 x +64}d x \right )+\int \frac {x^{2}}{e^{4 x} x^{2}-2 e^{4 x} x +e^{4 x}+2 e^{2 x} x^{3}+10 e^{2 x} x^{2}-28 e^{2 x} x +16 e^{2 x}+x^{4}+12 x^{3}+20 x^{2}-96 x +64}d x +2 \left (\int \frac {e^{2 x} x^{2}}{e^{4 x} x^{2}-2 e^{4 x} x +e^{4 x}+2 e^{2 x} x^{3}+10 e^{2 x} x^{2}-28 e^{2 x} x +16 e^{2 x}+x^{4}+12 x^{3}+20 x^{2}-96 x +64}d x \right )-4 \left (\int \frac {e^{2 x} x}{e^{4 x} x^{2}-2 e^{4 x} x +e^{4 x}+2 e^{2 x} x^{3}+10 e^{2 x} x^{2}-28 e^{2 x} x +16 e^{2 x}+x^{4}+12 x^{3}+20 x^{2}-96 x +64}d x \right )-2 \left (\int \frac {x}{e^{4 x} x^{2}-2 e^{4 x} x +e^{4 x}+2 e^{2 x} x^{3}+10 e^{2 x} x^{2}-28 e^{2 x} x +16 e^{2 x}+x^{4}+12 x^{3}+20 x^{2}-96 x +64}d x \right )+2 \left (\int \frac {1}{e^{4 x} x^{2}-2 e^{4 x} x +e^{4 x}+2 e^{2 x} x^{3}+10 e^{2 x} x^{2}-28 e^{2 x} x +16 e^{2 x}+x^{4}+12 x^{3}+20 x^{2}-96 x +64}d x \right )\right ) \] Input:
int(((2*x^2-4*x+2)*exp(5)*exp(x)^2+(x^2-2*x+2)*exp(5))/((x^2-2*x+1)*exp(x) ^4+(2*x^3+10*x^2-28*x+16)*exp(x)^2+x^4+12*x^3+20*x^2-96*x+64),x)
Output:
e**5*(2*int(e**(2*x)/(e**(4*x)*x**2 - 2*e**(4*x)*x + e**(4*x) + 2*e**(2*x) *x**3 + 10*e**(2*x)*x**2 - 28*e**(2*x)*x + 16*e**(2*x) + x**4 + 12*x**3 + 20*x**2 - 96*x + 64),x) + int(x**2/(e**(4*x)*x**2 - 2*e**(4*x)*x + e**(4*x ) + 2*e**(2*x)*x**3 + 10*e**(2*x)*x**2 - 28*e**(2*x)*x + 16*e**(2*x) + x** 4 + 12*x**3 + 20*x**2 - 96*x + 64),x) + 2*int((e**(2*x)*x**2)/(e**(4*x)*x* *2 - 2*e**(4*x)*x + e**(4*x) + 2*e**(2*x)*x**3 + 10*e**(2*x)*x**2 - 28*e** (2*x)*x + 16*e**(2*x) + x**4 + 12*x**3 + 20*x**2 - 96*x + 64),x) - 4*int(( e**(2*x)*x)/(e**(4*x)*x**2 - 2*e**(4*x)*x + e**(4*x) + 2*e**(2*x)*x**3 + 1 0*e**(2*x)*x**2 - 28*e**(2*x)*x + 16*e**(2*x) + x**4 + 12*x**3 + 20*x**2 - 96*x + 64),x) - 2*int(x/(e**(4*x)*x**2 - 2*e**(4*x)*x + e**(4*x) + 2*e**( 2*x)*x**3 + 10*e**(2*x)*x**2 - 28*e**(2*x)*x + 16*e**(2*x) + x**4 + 12*x** 3 + 20*x**2 - 96*x + 64),x) + 2*int(1/(e**(4*x)*x**2 - 2*e**(4*x)*x + e**( 4*x) + 2*e**(2*x)*x**3 + 10*e**(2*x)*x**2 - 28*e**(2*x)*x + 16*e**(2*x) + x**4 + 12*x**3 + 20*x**2 - 96*x + 64),x))