Integrand size = 57, antiderivative size = 29 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=\frac {1}{32 e^5 x (4+x)}-\left (x^2-x^3\right )^2 \] Output:
1/exp(5)/x/(128+32*x)-(-x^3+x^2)^2
Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=-\frac {16 e^5 (-1+x)^2 x^4-\frac {1}{2 x (4+x)}}{16 e^5} \] Input:
Integrate[(-2 - x + E^5*(-1024*x^5 + 2048*x^6 - 320*x^7 - 608*x^8 - 96*x^9 ))/(E^5*(256*x^2 + 128*x^3 + 16*x^4)),x]
Output:
-1/16*(16*E^5*(-1 + x)^2*x^4 - 1/(2*x*(4 + x)))/E^5
Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {27, 27, 2026, 2007, 2123, 2009}
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.
\(\displaystyle \int \frac {e^5 \left (-96 x^9-608 x^8-320 x^7+2048 x^6-1024 x^5\right )-x-2}{e^5 \left (16 x^4+128 x^3+256 x^2\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {x+32 e^5 \left (3 x^9+19 x^8+10 x^7-64 x^6+32 x^5\right )+2}{16 \left (x^4+8 x^3+16 x^2\right )}dx}{e^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {x+32 e^5 \left (3 x^9+19 x^8+10 x^7-64 x^6+32 x^5\right )+2}{x^4+8 x^3+16 x^2}dx}{16 e^5}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle -\frac {\int \frac {x+32 e^5 \left (3 x^9+19 x^8+10 x^7-64 x^6+32 x^5\right )+2}{x^2 \left (x^2+8 x+16\right )}dx}{16 e^5}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -\frac {\int \frac {x+32 e^5 \left (3 x^9+19 x^8+10 x^7-64 x^6+32 x^5\right )+2}{x^2 (x+4)^2}dx}{16 e^5}\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle -\frac {\int \left (96 e^5 x^5-160 e^5 x^4+64 e^5 x^3-\frac {1}{8 (x+4)^2}+\frac {1}{8 x^2}\right )dx}{16 e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {16 e^5 x^6-32 e^5 x^5+16 e^5 x^4+\frac {1}{8 (x+4)}-\frac {1}{8 x}}{16 e^5}\) |
Input:
Int[(-2 - x + E^5*(-1024*x^5 + 2048*x^6 - 320*x^7 - 608*x^8 - 96*x^9))/(E^ 5*(256*x^2 + 128*x^3 + 16*x^4)),x]
Output:
-1/16*(-1/8*1/x + 16*E^5*x^4 - 32*E^5*x^5 + 16*E^5*x^6 + 1/(8*(4 + x)))/E^ 5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-x^{6}+2 x^{5}-x^{4}+\frac {{\mathrm e}^{-5}}{32 x \left (4+x \right )}\) | \(29\) |
norman | \(\frac {-4 x^{5}+7 x^{6}-2 x^{7}-x^{8}+\frac {{\mathrm e}^{-5}}{32}}{\left (4+x \right ) x}\) | \(37\) |
default | \(\frac {{\mathrm e}^{-5} \left (-16 x^{6} {\mathrm e}^{5}+32 x^{5} {\mathrm e}^{5}-16 x^{4} {\mathrm e}^{5}-\frac {1}{8 \left (4+x \right )}+\frac {1}{8 x}\right )}{16}\) | \(41\) |
gosper | \(-\frac {\left (32 \,{\mathrm e}^{5} x^{8}+64 \,{\mathrm e}^{5} x^{7}-224 x^{6} {\mathrm e}^{5}+128 x^{5} {\mathrm e}^{5}-1\right ) {\mathrm e}^{-5}}{32 x \left (4+x \right )}\) | \(45\) |
parallelrisch | \(-\frac {\left (32 \,{\mathrm e}^{5} x^{8}+64 \,{\mathrm e}^{5} x^{7}-224 x^{6} {\mathrm e}^{5}+128 x^{5} {\mathrm e}^{5}-1\right ) {\mathrm e}^{-5}}{32 x \left (4+x \right )}\) | \(45\) |
orering | \(\frac {\left (32 \,{\mathrm e}^{5} x^{8}+64 \,{\mathrm e}^{5} x^{7}-224 x^{6} {\mathrm e}^{5}+128 x^{5} {\mathrm e}^{5}-1\right ) x \left (4+x \right ) \left (\left (-96 x^{9}-608 x^{8}-320 x^{7}+2048 x^{6}-1024 x^{5}\right ) {\mathrm e}^{5}-x -2\right ) {\mathrm e}^{-5}}{2 \left (96 \,{\mathrm e}^{5} x^{9}+608 \,{\mathrm e}^{5} x^{8}+320 \,{\mathrm e}^{5} x^{7}-2048 x^{6} {\mathrm e}^{5}+1024 x^{5} {\mathrm e}^{5}+x +2\right ) \left (16 x^{4}+128 x^{3}+256 x^{2}\right )}\) | \(133\) |
Input:
int(((-96*x^9-608*x^8-320*x^7+2048*x^6-1024*x^5)*exp(5)-x-2)/(16*x^4+128*x ^3+256*x^2)/exp(5),x,method=_RETURNVERBOSE)
Output:
-x^6+2*x^5-x^4+1/32*exp(-5)/x/(4+x)
Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=-\frac {{\left (32 \, {\left (x^{8} + 2 \, x^{7} - 7 \, x^{6} + 4 \, x^{5}\right )} e^{5} - 1\right )} e^{\left (-5\right )}}{32 \, {\left (x^{2} + 4 \, x\right )}} \] Input:
integrate(((-96*x^9-608*x^8-320*x^7+2048*x^6-1024*x^5)*exp(5)-x-2)/(16*x^4 +128*x^3+256*x^2)/exp(5),x, algorithm="fricas")
Output:
-1/32*(32*(x^8 + 2*x^7 - 7*x^6 + 4*x^5)*e^5 - 1)*e^(-5)/(x^2 + 4*x)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=- x^{6} + 2 x^{5} - x^{4} + \frac {1}{32 x^{2} e^{5} + 128 x e^{5}} \] Input:
integrate(((-96*x**9-608*x**8-320*x**7+2048*x**6-1024*x**5)*exp(5)-x-2)/(1 6*x**4+128*x**3+256*x**2)/exp(5),x)
Output:
-x**6 + 2*x**5 - x**4 + 1/(32*x**2*exp(5) + 128*x*exp(5))
Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=-\frac {1}{32} \, {\left (32 \, x^{6} e^{5} - 64 \, x^{5} e^{5} + 32 \, x^{4} e^{5} - \frac {1}{x^{2} + 4 \, x}\right )} e^{\left (-5\right )} \] Input:
integrate(((-96*x^9-608*x^8-320*x^7+2048*x^6-1024*x^5)*exp(5)-x-2)/(16*x^4 +128*x^3+256*x^2)/exp(5),x, algorithm="maxima")
Output:
-1/32*(32*x^6*e^5 - 64*x^5*e^5 + 32*x^4*e^5 - 1/(x^2 + 4*x))*e^(-5)
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=-\frac {1}{32} \, {\left (32 \, x^{6} e^{5} - 64 \, x^{5} e^{5} + 32 \, x^{4} e^{5} - \frac {1}{x^{2} + 4 \, x}\right )} e^{\left (-5\right )} \] Input:
integrate(((-96*x^9-608*x^8-320*x^7+2048*x^6-1024*x^5)*exp(5)-x-2)/(16*x^4 +128*x^3+256*x^2)/exp(5),x, algorithm="giac")
Output:
-1/32*(32*x^6*e^5 - 64*x^5*e^5 + 32*x^4*e^5 - 1/(x^2 + 4*x))*e^(-5)
Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=\frac {1}{2\,\left (16\,{\mathrm {e}}^5\,x^2+64\,{\mathrm {e}}^5\,x\right )}-x^4+2\,x^5-x^6 \] Input:
int(-(exp(-5)*(x + exp(5)*(1024*x^5 - 2048*x^6 + 320*x^7 + 608*x^8 + 96*x^ 9) + 2))/(256*x^2 + 128*x^3 + 16*x^4),x)
Output:
1/(2*(64*x*exp(5) + 16*x^2*exp(5))) - x^4 + 2*x^5 - x^6
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {-2-x+e^5 \left (-1024 x^5+2048 x^6-320 x^7-608 x^8-96 x^9\right )}{e^5 \left (256 x^2+128 x^3+16 x^4\right )} \, dx=\frac {-32 e^{5} x^{8}-64 e^{5} x^{7}+224 e^{5} x^{6}-128 e^{5} x^{5}+1}{32 e^{5} x \left (x +4\right )} \] Input:
int(((-96*x^9-608*x^8-320*x^7+2048*x^6-1024*x^5)*exp(5)-x-2)/(16*x^4+128*x ^3+256*x^2)/exp(5),x)
Output:
( - 32*e**5*x**8 - 64*e**5*x**7 + 224*e**5*x**6 - 128*e**5*x**5 + 1)/(32*e **5*x*(x + 4))