Integrand size = 86, antiderivative size = 28 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=e^2-\frac {2}{x-\frac {4}{2-\frac {e^5}{2}+4 x^4}} \]
Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=\frac {2 \left (-4+e^5-8 x^4\right )}{-8+4 x-e^5 x+8 x^5} \]
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]
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 |
| \(\displaystyle \frac {8}{5} \left (4-e^5\right )^2 \int \frac {1}{\left (-8 x^5-4 \left (1-\frac {e^5}{4}\right ) x+8\right )^2}dx+640 \int \frac {x^3}{\left (-8 x^5-4 \left (1-\frac {e^5}{4}\right ) x+8\right )^2}dx+16 \int \frac {x^3}{8 x^5+4 \left (1-\frac {e^5}{4}\right ) x-8}dx+\frac {2 \left (4-e^5\right )}{5 \left (-8 x^5-\left (4-e^5\right ) x+8\right )}\) |
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]
3.28.42.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Time = 0.58 (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 | \(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (64+64 \textit {\_Z}^{10}+\left (-16 \,{\mathrm e}^{5}+64\right ) \textit {\_Z}^{6}-128 \textit {\_Z}^{5}+\left (-8 \,{\mathrm e}^{5}+{\mathrm e}^{10}+16\right ) \textit {\_Z}^{2}+\left (16 \,{\mathrm e}^{5}-64\right ) \textit {\_Z} \right )}{\sum }\frac {\left (16+64 \textit {\_R}^{8}+16 \left (4-{\mathrm e}^{5}\right ) \textit {\_R}^{4}+256 \textit {\_R}^{3}-8 \,{\mathrm e}^{5}+{\mathrm e}^{10}\right ) \ln \left (x -\textit {\_R} \right )}{32-320 \textit {\_R}^{9}+48 \textit {\_R}^{5} {\mathrm e}^{5}-192 \textit {\_R}^{5}+320 \textit {\_R}^{4}-\textit {\_R} \,{\mathrm e}^{10}+8 \textit {\_R} \,{\mathrm e}^{5}-8 \,{\mathrm e}^{5}-16 \textit {\_R}}\right )\) | \(128\) |
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)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=-\frac {2 \, {\left (8 \, x^{4} - e^{5} + 4\right )}}{8 \, x^{5} - x e^{5} + 4 \, x - 8} \]
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=\
Time = 0.70 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=\frac {- 16 x^{4} - 8 + 2 e^{5}}{8 x^{5} + x \left (4 - e^{5}\right ) - 8} \]
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)
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=-\frac {2 \, {\left (8 \, x^{4} - e^{5} + 4\right )}}{8 \, x^{5} - x {\left (e^{5} - 4\right )} - 8} \]
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=\
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=-\frac {2 \, {\left (8 \, x^{4} - e^{5} + 4\right )}}{8 \, x^{5} - x e^{5} + 4 \, x - 8} \]
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=\
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {32+2 e^{10}+512 x^3+128 x^4+128 x^8+e^5 \left (-16-32 x^4\right )}{64-64 x+16 x^2+e^{10} x^2-128 x^5+64 x^6+64 x^{10}+e^5 \left (16 x-8 x^2-16 x^6\right )} \, dx=\frac {16\,x^4-2\,{\mathrm {e}}^5+8}{-8\,x^5+\left ({\mathrm {e}}^5-4\right )\,x+8} \]