Integrand size = 86, antiderivative size = 28 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=e^2-\frac {2}{x-\frac {4}{2-\frac {e^5}{2}+4 x^4}} \] Output:
exp(2)-2/(x-4/(4*x^4+2-1/2*exp(5)))
Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=\frac {2 \left (-4+e^5-8 x^4\right )}{-8+4 x-e^5 x+8 x^5} \] Input:
Integrate[(32 + 2*E^10 + 512*x^3 + 128*x^4 + 128*x^8 + E^5*(-16 - 32*x^4)) /(64 - 64*x + 16*x^2 + E^10*x^2 - 128*x^5 + 64*x^6 + 64*x^10 + E^5*(16*x - 8*x^2 - 16*x^6)),x]
Output:
(2*(-4 + E^5 - 8*x^4))/(-8 + 4*x - E^5*x + 8*x^5)
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.
\(\displaystyle \int \frac {128 x^8+128 x^4+e^5 \left (-32 x^4-16\right )+512 x^3+2 e^{10}+32}{64 x^{10}+64 x^6-128 x^5+e^{10} x^2+16 x^2+e^5 \left (-16 x^6-8 x^2+16 x\right )-64 x+64} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {128 x^8+128 x^4+e^5 \left (-32 x^4-16\right )+512 x^3+2 e^{10}+32}{64 x^{10}+64 x^6-128 x^5+\left (16+e^{10}\right ) x^2+e^5 \left (-16 x^6-8 x^2+16 x\right )-64 x+64}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (\frac {16 x^3}{8 x^5+4 \left (1-\frac {e^5}{4}\right ) x-8}+\frac {2 \left (8 \left (4-e^5\right ) x^4+320 x^3+\left (e^5-4\right )^2\right )}{\left (-8 x^5-4 \left (1-\frac {e^5}{4}\right ) x+8\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8}{5} \left (4-e^5\right )^2 \int \frac {1}{\left (-8 x^5-4 \left (1-\frac {e^5}{4}\right ) x+8\right )^2}dx+640 \int \frac {x^3}{\left (-8 x^5-4 \left (1-\frac {e^5}{4}\right ) x+8\right )^2}dx+16 \int \frac {x^3}{8 x^5+4 \left (1-\frac {e^5}{4}\right ) x-8}dx+\frac {2 \left (4-e^5\right )}{5 \left (-8 x^5-\left (4-e^5\right ) x+8\right )}\) |
Input:
Int[(32 + 2*E^10 + 512*x^3 + 128*x^4 + 128*x^8 + E^5*(-16 - 32*x^4))/(64 - 64*x + 16*x^2 + E^10*x^2 - 128*x^5 + 64*x^6 + 64*x^10 + E^5*(16*x - 8*x^2 - 16*x^6)),x]
Output:
$Aborted
Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00
method | result | size |
gosper | \(-\frac {2 \left (-8 x^{4}+{\mathrm e}^{5}-4\right )}{-8 x^{5}+x \,{\mathrm e}^{5}-4 x +8}\) | \(28\) |
norman | \(\frac {8+16 x^{4}-2 \,{\mathrm e}^{5}}{-8 x^{5}+x \,{\mathrm e}^{5}-4 x +8}\) | \(29\) |
risch | \(\frac {8+16 x^{4}-2 \,{\mathrm e}^{5}}{-8 x^{5}+x \,{\mathrm e}^{5}-4 x +8}\) | \(29\) |
parallelrisch | \(-\frac {-64-128 x^{4}+16 \,{\mathrm e}^{5}}{8 \left (-8 x^{5}+x \,{\mathrm e}^{5}-4 x +8\right )}\) | \(30\) |
default | \(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (64+64 \textit {\_Z}^{10}+\left (-16 \,{\mathrm e}^{5}+64\right ) \textit {\_Z}^{6}-128 \textit {\_Z}^{5}+\left (-8 \,{\mathrm e}^{5}+{\mathrm e}^{10}+16\right ) \textit {\_Z}^{2}+\left (16 \,{\mathrm e}^{5}-64\right ) \textit {\_Z} \right )}{\sum }\frac {\left (16+64 \textit {\_R}^{8}+16 \left (4-{\mathrm e}^{5}\right ) \textit {\_R}^{4}+256 \textit {\_R}^{3}-8 \,{\mathrm e}^{5}+{\mathrm e}^{10}\right ) \ln \left (x -\textit {\_R} \right )}{32-320 \textit {\_R}^{9}+48 \textit {\_R}^{5} {\mathrm e}^{5}-192 \textit {\_R}^{5}+320 \textit {\_R}^{4}-\textit {\_R} \,{\mathrm e}^{10}+8 \textit {\_R} \,{\mathrm e}^{5}-8 \,{\mathrm e}^{5}-16 \textit {\_R}}\right )\) | \(128\) |
orering | \(\frac {\left (-8 x^{4}+{\mathrm e}^{5}-4\right ) \left (-8 x^{5}+x \,{\mathrm e}^{5}-4 x +8\right ) \left (2 \,{\mathrm e}^{10}+\left (-32 x^{4}-16\right ) {\mathrm e}^{5}+128 x^{8}+128 x^{4}+512 x^{3}+32\right )}{\left (-64 x^{8}+16 x^{4} {\mathrm e}^{5}-64 x^{4}-256 x^{3}-{\mathrm e}^{10}+8 \,{\mathrm e}^{5}-16\right ) \left ({\mathrm e}^{10} x^{2}+\left (-16 x^{6}-8 x^{2}+16 x \right ) {\mathrm e}^{5}+64 x^{10}+64 x^{6}-128 x^{5}+16 x^{2}-64 x +64\right )}\) | \(144\) |
Input:
int((2*exp(5)^2+(-32*x^4-16)*exp(5)+128*x^8+128*x^4+512*x^3+32)/(x^2*exp(5 )^2+(-16*x^6-8*x^2+16*x)*exp(5)+64*x^10+64*x^6-128*x^5+16*x^2-64*x+64),x,m ethod=_RETURNVERBOSE)
Output:
-2*(-8*x^4+exp(5)-4)/(-8*x^5+x*exp(5)-4*x+8)
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=-\frac {2 \, {\left (8 \, x^{4} - e^{5} + 4\right )}}{8 \, x^{5} - x e^{5} + 4 \, x - 8} \] Input:
integrate((2*exp(5)^2+(-32*x^4-16)*exp(5)+128*x^8+128*x^4+512*x^3+32)/(x^2 *exp(5)^2+(-16*x^6-8*x^2+16*x)*exp(5)+64*x^10+64*x^6-128*x^5+16*x^2-64*x+6 4),x, algorithm="fricas")
Output:
-2*(8*x^4 - e^5 + 4)/(8*x^5 - x*e^5 + 4*x - 8)
Time = 0.76 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=\frac {- 16 x^{4} - 8 + 2 e^{5}}{8 x^{5} + x \left (4 - e^{5}\right ) - 8} \] Input:
integrate((2*exp(5)**2+(-32*x**4-16)*exp(5)+128*x**8+128*x**4+512*x**3+32) /(x**2*exp(5)**2+(-16*x**6-8*x**2+16*x)*exp(5)+64*x**10+64*x**6-128*x**5+1 6*x**2-64*x+64),x)
Output:
(-16*x**4 - 8 + 2*exp(5))/(8*x**5 + x*(4 - exp(5)) - 8)
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=-\frac {2 \, {\left (8 \, x^{4} - e^{5} + 4\right )}}{8 \, x^{5} - x {\left (e^{5} - 4\right )} - 8} \] Input:
integrate((2*exp(5)^2+(-32*x^4-16)*exp(5)+128*x^8+128*x^4+512*x^3+32)/(x^2 *exp(5)^2+(-16*x^6-8*x^2+16*x)*exp(5)+64*x^10+64*x^6-128*x^5+16*x^2-64*x+6 4),x, algorithm="maxima")
Output:
-2*(8*x^4 - e^5 + 4)/(8*x^5 - x*(e^5 - 4) - 8)
Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=-\frac {2 \, {\left (8 \, x^{4} - e^{5} + 4\right )}}{8 \, x^{5} - x e^{5} + 4 \, x - 8} \] Input:
integrate((2*exp(5)^2+(-32*x^4-16)*exp(5)+128*x^8+128*x^4+512*x^3+32)/(x^2 *exp(5)^2+(-16*x^6-8*x^2+16*x)*exp(5)+64*x^10+64*x^6-128*x^5+16*x^2-64*x+6 4),x, algorithm="giac")
Output:
-2*(8*x^4 - e^5 + 4)/(8*x^5 - x*e^5 + 4*x - 8)
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=\frac {16\,x^4-2\,{\mathrm {e}}^5+8}{-8\,x^5+\left ({\mathrm {e}}^5-4\right )\,x+8} \] Input:
int((2*exp(10) - exp(5)*(32*x^4 + 16) + 512*x^3 + 128*x^4 + 128*x^8 + 32)/ (x^2*exp(10) - exp(5)*(8*x^2 - 16*x + 16*x^6) - 64*x + 16*x^2 - 128*x^5 + 64*x^6 + 64*x^10 + 64),x)
Output:
(16*x^4 - 2*exp(5) + 8)/(x*(exp(5) - 4) - 8*x^5 + 8)
Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=\frac {x \left (e^{10}-8 e^{5} x^{4}-8 e^{5}+32 x^{4}+64 x^{3}+16\right )}{4 e^{5} x -32 x^{5}-16 x +32} \] Input:
int((2*exp(5)^2+(-32*x^4-16)*exp(5)+128*x^8+128*x^4+512*x^3+32)/(x^2*exp(5 )^2+(-16*x^6-8*x^2+16*x)*exp(5)+64*x^10+64*x^6-128*x^5+16*x^2-64*x+64),x)
Output:
(x*(e**10 - 8*e**5*x**4 - 8*e**5 + 32*x**4 + 64*x**3 + 16))/(4*(e**5*x - 8 *x**5 - 4*x + 8))