Integrand size = 86, antiderivative size = 28 \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\frac {-2+x}{16-x+\frac {1}{4} e^{-5-x} (5-x) x} \] Output:
(-2+x)/(16-x+1/4*x/exp(x)/exp(5)*(5-x))
Time = 3.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\frac {-56 e^{5+x}+(-5+x) x}{4 e^{5+x} (-16+x)+(-5+x) x} \] Input:
Integrate[(224*E^(10 + 2*x) + E^(5 + x)*(40 - 56*x + 32*x^2 - 4*x^3))/(25* x^2 - 10*x^3 + x^4 + E^(10 + 2*x)*(4096 - 512*x + 16*x^2) + E^(5 + x)*(640 *x - 168*x^2 + 8*x^3)),x]
Output:
(-56*E^(5 + x) + (-5 + x)*x)/(4*E^(5 + x)*(-16 + x) + (-5 + x)*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.
\(\displaystyle \int \frac {e^{x+5} \left (-4 x^3+32 x^2-56 x+40\right )+224 e^{2 x+10}}{x^4-10 x^3+25 x^2+e^{2 x+10} \left (16 x^2-512 x+4096\right )+e^{x+5} \left (8 x^3-168 x^2+640 x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 e^{x+5} \left (-x^3+8 x^2-14 x+56 e^{x+5}+10\right )}{\left (4 e^{x+5} (x-16)+(x-5) x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {e^{x+5} \left (-x^3+8 x^2-14 x+56 e^{x+5}+10\right )}{\left (4 e^{x+5} (16-x)+(5-x) x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 4 \int \left (\frac {14 e^{x+5}}{(x-16) \left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )}-\frac {e^{x+5} \left (x^4-24 x^3+156 x^2-304 x+160\right )}{(x-16) \left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (-144 \int \frac {e^{x+5}}{\left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )^2}dx-2464 \int \frac {e^{x+5}}{(x-16) \left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )^2}dx-28 \int \frac {e^{x+5} x}{\left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )^2}dx+8 \int \frac {e^{x+5} x^2}{\left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )^2}dx+14 \int \frac {e^{x+5}}{(x-16) \left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )}dx-\int \frac {e^{x+5} x^3}{\left (x^2+4 e^{x+5} x-5 x-64 e^{x+5}\right )^2}dx\right )\) |
Input:
Int[(224*E^(10 + 2*x) + E^(5 + x)*(40 - 56*x + 32*x^2 - 4*x^3))/(25*x^2 - 10*x^3 + x^4 + E^(10 + 2*x)*(4096 - 512*x + 16*x^2) + E^(5 + x)*(640*x - 1 68*x^2 + 8*x^3)),x]
Output:
$Aborted
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36
method | result | size |
norman | \(\frac {8 \,{\mathrm e}^{5} {\mathrm e}^{x}-4 x \,{\mathrm e}^{5} {\mathrm e}^{x}}{4 x \,{\mathrm e}^{5} {\mathrm e}^{x}-64 \,{\mathrm e}^{5} {\mathrm e}^{x}+x^{2}-5 x}\) | \(38\) |
parallelrisch | \(\frac {8 \,{\mathrm e}^{5} {\mathrm e}^{x}-4 x \,{\mathrm e}^{5} {\mathrm e}^{x}}{4 x \,{\mathrm e}^{5} {\mathrm e}^{x}-64 \,{\mathrm e}^{5} {\mathrm e}^{x}+x^{2}-5 x}\) | \(38\) |
risch | \(-\frac {14}{x -16}+\frac {\left (x^{2}-7 x +10\right ) x}{\left (x -16\right ) \left (4 x \,{\mathrm e}^{5+x}-64 \,{\mathrm e}^{5+x}+x^{2}-5 x \right )}\) | \(46\) |
Input:
int((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/((16*x^2 -512*x+4096)*exp(5)^2*exp(x)^2+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x^4-10* x^3+25*x^2),x,method=_RETURNVERBOSE)
Output:
(8*exp(5)*exp(x)-4*x*exp(5)*exp(x))/(4*x*exp(5)*exp(x)-64*exp(5)*exp(x)+x^ 2-5*x)
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\frac {x^{2} - 5 \, x - 56 \, e^{\left (x + 5\right )}}{x^{2} + 4 \, {\left (x - 16\right )} e^{\left (x + 5\right )} - 5 \, x} \] Input:
integrate((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/(( 16*x^2-512*x+4096)*exp(5)^2*exp(x)^2+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x ^4-10*x^3+25*x^2),x, algorithm="fricas")
Output:
(x^2 - 5*x - 56*e^(x + 5))/(x^2 + 4*(x - 16)*e^(x + 5) - 5*x)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (19) = 38\).
Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\frac {x^{3} - 7 x^{2} + 10 x}{x^{3} - 21 x^{2} + 80 x + \left (4 x^{2} e^{5} - 128 x e^{5} + 1024 e^{5}\right ) e^{x}} - \frac {14}{x - 16} \] Input:
integrate((224*exp(5)**2*exp(x)**2+(-4*x**3+32*x**2-56*x+40)*exp(5)*exp(x) )/((16*x**2-512*x+4096)*exp(5)**2*exp(x)**2+(8*x**3-168*x**2+640*x)*exp(5) *exp(x)+x**4-10*x**3+25*x**2),x)
Output:
(x**3 - 7*x**2 + 10*x)/(x**3 - 21*x**2 + 80*x + (4*x**2*exp(5) - 128*x*exp (5) + 1024*exp(5))*exp(x)) - 14/(x - 16)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\frac {x^{2} - 5 \, x - 56 \, e^{\left (x + 5\right )}}{x^{2} + 4 \, {\left (x e^{5} - 16 \, e^{5}\right )} e^{x} - 5 \, x} \] Input:
integrate((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/(( 16*x^2-512*x+4096)*exp(5)^2*exp(x)^2+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x ^4-10*x^3+25*x^2),x, algorithm="maxima")
Output:
(x^2 - 5*x - 56*e^(x + 5))/(x^2 + 4*(x*e^5 - 16*e^5)*e^x - 5*x)
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\frac {x^{2} - 5 \, x - 56 \, e^{\left (x + 5\right )}}{x^{2} + 4 \, x e^{\left (x + 5\right )} - 5 \, x - 64 \, e^{\left (x + 5\right )}} \] Input:
integrate((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/(( 16*x^2-512*x+4096)*exp(5)^2*exp(x)^2+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x ^4-10*x^3+25*x^2),x, algorithm="giac")
Output:
(x^2 - 5*x - 56*e^(x + 5))/(x^2 + 4*x*e^(x + 5) - 5*x - 64*e^(x + 5))
Timed out. \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\int \frac {224\,{\mathrm {e}}^{2\,x+10}-{\mathrm {e}}^{x+5}\,\left (4\,x^3-32\,x^2+56\,x-40\right )}{{\mathrm {e}}^{x+5}\,\left (8\,x^3-168\,x^2+640\,x\right )+{\mathrm {e}}^{2\,x+10}\,\left (16\,x^2-512\,x+4096\right )+25\,x^2-10\,x^3+x^4} \,d x \] Input:
int((224*exp(2*x)*exp(10) - exp(5)*exp(x)*(56*x - 32*x^2 + 4*x^3 - 40))/(2 5*x^2 - 10*x^3 + x^4 + exp(5)*exp(x)*(640*x - 168*x^2 + 8*x^3) + exp(2*x)* exp(10)*(16*x^2 - 512*x + 4096)),x)
Output:
int((224*exp(2*x + 10) - exp(x + 5)*(56*x - 32*x^2 + 4*x^3 - 40))/(exp(x + 5)*(640*x - 168*x^2 + 8*x^3) + exp(2*x + 10)*(16*x^2 - 512*x + 4096) + 25 *x^2 - 10*x^3 + x^4), x)
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx=\frac {4 e^{x} e^{5} \left (-x +2\right )}{4 e^{x} e^{5} x -64 e^{x} e^{5}+x^{2}-5 x} \] Input:
int((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/((16*x^2 -512*x+4096)*exp(5)^2*exp(x)^2+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x^4-10* x^3+25*x^2),x)
Output:
(4*e**x*e**5*( - x + 2))/(4*e**x*e**5*x - 64*e**x*e**5 + x**2 - 5*x)