Integrand size = 108, antiderivative size = 35 \[ \int \frac {-10 e^8-e^4 x^2+x^6 \left (-250-100 x-10 x^2\right )+x^3 \left (-75 x-30 x^2-3 x^3+e^4 (100+20 x)\right )}{5 e^{12} x^2+e^8 x^3 \left (-50 x^2-10 x^3\right )+e^4 x^6 \left (125 x^2+50 x^3+5 x^4\right )} \, dx=\frac {2+\frac {1}{5} \left (x-\frac {x}{-x^3+\frac {e^4}{5+x}}\right )}{e^4 x} \]
Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {-10 e^8-e^4 x^2+x^6 \left (-250-100 x-10 x^2\right )+x^3 \left (-75 x-30 x^2-3 x^3+e^4 (100+20 x)\right )}{5 e^{12} x^2+e^8 x^3 \left (-50 x^2-10 x^3\right )+e^4 x^6 \left (125 x^2+50 x^3+5 x^4\right )} \, dx=\frac {\frac {10}{x}+\frac {5+x}{-e^4+5 x^3+x^4}}{5 e^4} \]
Integrate[(-10*E^8 - E^4*x^2 + x^6*(-250 - 100*x - 10*x^2) + x^3*(-75*x - 30*x^2 - 3*x^3 + E^4*(100 + 20*x)))/(5*E^12*x^2 + E^8*x^3*(-50*x^2 - 10*x^ 3) + E^4*x^6*(125*x^2 + 50*x^3 + 5*x^4)),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.
\(\displaystyle \int \frac {-e^4 x^2+\left (-10 x^2-100 x-250\right ) x^6+\left (-3 x^3-30 x^2-75 x+e^4 (20 x+100)\right ) x^3-10 e^8}{5 e^{12} x^2+e^8 \left (-10 x^3-50 x^2\right ) x^3+e^4 \left (5 x^4+50 x^3+125 x^2\right ) x^6} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-e^4 x^2+\left (-10 x^2-100 x-250\right ) x^6+\left (-3 x^3-30 x^2-75 x+e^4 (20 x+100)\right ) x^3-10 e^8}{x^2 \left (5 e^4 x^8+50 e^4 x^7+125 e^4 x^6-10 e^8 x^4-50 e^8 x^3+5 e^{12}\right )}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (-\frac {2}{e^4 x^2}+\frac {3}{5 e^4 \left (-x^4-5 x^3+e^4\right )}+\frac {-15 x^3-75 x^2-4 e^4}{5 e^4 \left (-x^4-5 x^3+e^4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4}{5} \int \frac {1}{\left (-x^4-5 x^3+e^4\right )^2}dx+\frac {3 \int \frac {1}{-x^4-5 x^3+e^4}dx}{5 e^4}-\frac {15 \int \frac {x^2}{\left (-x^4-5 x^3+e^4\right )^2}dx}{4 e^4}-\frac {3}{4 e^4 \left (-x^4-5 x^3+e^4\right )}+\frac {2}{e^4 x}\) |
Int[(-10*E^8 - E^4*x^2 + x^6*(-250 - 100*x - 10*x^2) + x^3*(-75*x - 30*x^2 - 3*x^3 + E^4*(100 + 20*x)))/(5*E^12*x^2 + E^8*x^3*(-50*x^2 - 10*x^3) + E ^4*x^6*(125*x^2 + 50*x^3 + 5*x^4)),x]
3.22.18.3.1 Defintions of rubi rules used
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[(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 = 0.40 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29
method | result | size |
risch | \(\frac {\left (-2 x^{4}-10 x^{3}-\frac {x^{2}}{5}+2 \,{\mathrm e}^{4}-x \right ) {\mathrm e}^{-4}}{x \left (-x^{4}-5 x^{3}+{\mathrm e}^{4}\right )}\) | \(45\) |
parallelrisch | \(-\frac {\left (10 x^{4}+50 x^{3}+x^{2}+5 x -10 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-4}}{5 x \left (-x^{4}-5 x^{3}+{\mathrm e}^{4}\right )}\) | \(46\) |
gosper | \(\frac {\left (-10 x^{4}-50 x^{3}-x^{2}+10 \,{\mathrm e}^{4}-5 x \right ) {\mathrm e}^{-4}}{5 x \left (-x^{4}-5 x^{3}+{\mathrm e}^{4}\right )}\) | \(48\) |
norman | \(\frac {2-2 \,{\mathrm e}^{-4} x^{4}-{\mathrm e}^{-4} x -\frac {{\mathrm e}^{-4} x^{2}}{5}-10 \,{\mathrm e}^{-4} x^{3}}{x \left (-x^{4}-5 x^{3}+{\mathrm e}^{4}\right )}\) | \(56\) |
int(((-10*x^2-100*x-250)*x^6+((20*x+100)*exp(4)-3*x^3-30*x^2-75*x)*x^3-10* exp(4)^2-x^2*exp(4))/((5*x^4+50*x^3+125*x^2)*exp(4)*x^6+(-10*x^3-50*x^2)*e xp(4)^2*x^3+5*x^2*exp(4)^3),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {-10 e^8-e^4 x^2+x^6 \left (-250-100 x-10 x^2\right )+x^3 \left (-75 x-30 x^2-3 x^3+e^4 (100+20 x)\right )}{5 e^{12} x^2+e^8 x^3 \left (-50 x^2-10 x^3\right )+e^4 x^6 \left (125 x^2+50 x^3+5 x^4\right )} \, dx=-\frac {10 \, x^{4} + 50 \, x^{3} + x^{2} + 5 \, x - 10 \, e^{4}}{5 \, {\left (x e^{8} - {\left (x^{5} + 5 \, x^{4}\right )} e^{4}\right )}} \]
integrate(((-10*x^2-100*x-250)*x^6+((20*x+100)*exp(4)-3*x^3-30*x^2-75*x)*x ^3-10*exp(4)^2-x^2*exp(4))/((5*x^4+50*x^3+125*x^2)*exp(4)*x^6+(-10*x^3-50* x^2)*exp(4)^2*x^3+5*x^2*exp(4)^3),x, algorithm=\
Time = 0.99 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.31 \[ \int \frac {-10 e^8-e^4 x^2+x^6 \left (-250-100 x-10 x^2\right )+x^3 \left (-75 x-30 x^2-3 x^3+e^4 (100+20 x)\right )}{5 e^{12} x^2+e^8 x^3 \left (-50 x^2-10 x^3\right )+e^4 x^6 \left (125 x^2+50 x^3+5 x^4\right )} \, dx=- \frac {- 10 x^{4} - 50 x^{3} - x^{2} - 5 x + 10 e^{4}}{5 x^{5} e^{4} + 25 x^{4} e^{4} - 5 x e^{8}} \]
integrate(((-10*x**2-100*x-250)*x**6+((20*x+100)*exp(4)-3*x**3-30*x**2-75* x)*x**3-10*exp(4)**2-x**2*exp(4))/((5*x**4+50*x**3+125*x**2)*exp(4)*x**6+( -10*x**3-50*x**2)*exp(4)**2*x**3+5*x**2*exp(4)**3),x)
-(-10*x**4 - 50*x**3 - x**2 - 5*x + 10*exp(4))/(5*x**5*exp(4) + 25*x**4*ex p(4) - 5*x*exp(8))
Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {-10 e^8-e^4 x^2+x^6 \left (-250-100 x-10 x^2\right )+x^3 \left (-75 x-30 x^2-3 x^3+e^4 (100+20 x)\right )}{5 e^{12} x^2+e^8 x^3 \left (-50 x^2-10 x^3\right )+e^4 x^6 \left (125 x^2+50 x^3+5 x^4\right )} \, dx=\frac {10 \, x^{4} + 50 \, x^{3} + x^{2} + 5 \, x - 10 \, e^{4}}{5 \, {\left (x^{5} e^{4} + 5 \, x^{4} e^{4} - x e^{8}\right )}} \]
integrate(((-10*x^2-100*x-250)*x^6+((20*x+100)*exp(4)-3*x^3-30*x^2-75*x)*x ^3-10*exp(4)^2-x^2*exp(4))/((5*x^4+50*x^3+125*x^2)*exp(4)*x^6+(-10*x^3-50* x^2)*exp(4)^2*x^3+5*x^2*exp(4)^3),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \frac {-10 e^8-e^4 x^2+x^6 \left (-250-100 x-10 x^2\right )+x^3 \left (-75 x-30 x^2-3 x^3+e^4 (100+20 x)\right )}{5 e^{12} x^2+e^8 x^3 \left (-50 x^2-10 x^3\right )+e^4 x^6 \left (125 x^2+50 x^3+5 x^4\right )} \, dx=\frac {{\left (10 \, x^{4} + 50 \, x^{3} + x^{2} + 5 \, x - 10 \, e^{4}\right )} e^{\left (-4\right )}}{5 \, {\left (x^{5} + 5 \, x^{4} - x e^{4}\right )}} \]
integrate(((-10*x^2-100*x-250)*x^6+((20*x+100)*exp(4)-3*x^3-30*x^2-75*x)*x ^3-10*exp(4)^2-x^2*exp(4))/((5*x^4+50*x^3+125*x^2)*exp(4)*x^6+(-10*x^3-50* x^2)*exp(4)^2*x^3+5*x^2*exp(4)^3),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {-10 e^8-e^4 x^2+x^6 \left (-250-100 x-10 x^2\right )+x^3 \left (-75 x-30 x^2-3 x^3+e^4 (100+20 x)\right )}{5 e^{12} x^2+e^8 x^3 \left (-50 x^2-10 x^3\right )+e^4 x^6 \left (125 x^2+50 x^3+5 x^4\right )} \, dx=\frac {\frac {x}{5}+1}{{\mathrm {e}}^4\,x^4+5\,{\mathrm {e}}^4\,x^3-{\mathrm {e}}^8}+\frac {2\,{\mathrm {e}}^{-4}}{x} \]