Integrand size = 58, antiderivative size = 27 \[ \int \frac {e^{4-x} \left (-6+6 x-2 x^2-x^3+x^4\right )}{180+60 x-175 x^2+30 x^3+55 x^4-30 x^5+5 x^6} \, dx=-\frac {e^{4-x} x}{5 \left (6+x-(3-x) x^2\right )} \]
Time = 1.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{4-x} \left (-6+6 x-2 x^2-x^3+x^4\right )}{180+60 x-175 x^2+30 x^3+55 x^4-30 x^5+5 x^6} \, dx=-\frac {e^{4-x} x}{5 \left (6+x-3 x^2+x^3\right )} \]
Integrate[(E^(4 - x)*(-6 + 6*x - 2*x^2 - x^3 + x^4))/(180 + 60*x - 175*x^2 + 30*x^3 + 55*x^4 - 30*x^5 + 5*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.
\(\displaystyle \int \frac {e^{4-x} \left (x^4-x^3-2 x^2+6 x-6\right )}{5 x^6-30 x^5+55 x^4+30 x^3-175 x^2+60 x+180} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \frac {e^{4-x} \left (x^4-x^3-2 x^2+6 x-6\right )}{5 \left (x^3-3 x^2+x+6\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {e^{4-x} \left (-x^4+x^3+2 x^2-6 x+6\right )}{\left (x^3-3 x^2+x+6\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {e^{4-x} \left (-x^4+x^3+2 x^2-6 x+6\right )}{\left (x^3-3 x^2+x+6\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {e^{4-x} (-x-2)}{x^3-3 x^2+x+6}+\frac {e^{4-x} \left (-3 x^2+2 x+18\right )}{\left (x^3-3 x^2+x+6\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-18 \int \frac {e^{4-x}}{\left (x^3-3 x^2+x+6\right )^2}dx-2 \int \frac {e^{4-x} x}{\left (x^3-3 x^2+x+6\right )^2}dx+3 \int \frac {e^{4-x} x^2}{\left (x^3-3 x^2+x+6\right )^2}dx+2 \int \frac {e^{4-x}}{x^3-3 x^2+x+6}dx+\int \frac {e^{4-x} x}{x^3-3 x^2+x+6}dx\right )\) |
Int[(E^(4 - x)*(-6 + 6*x - 2*x^2 - x^3 + x^4))/(180 + 60*x - 175*x^2 + 30* x^3 + 55*x^4 - 30*x^5 + 5*x^6),x]
3.6.24.3.1 Defintions of rubi rules used
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[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-\frac {x \,{\mathrm e}^{-x +4}}{5 \left (x^{3}-3 x^{2}+x +6\right )}\) | \(23\) |
gosper | \(-\frac {{\mathrm e}^{4} x \,{\mathrm e}^{-x}}{5 \left (x^{3}-3 x^{2}+x +6\right )}\) | \(25\) |
norman | \(-\frac {{\mathrm e}^{4} x \,{\mathrm e}^{-x}}{5 \left (x^{3}-3 x^{2}+x +6\right )}\) | \(25\) |
parallelrisch | \(-\frac {{\mathrm e}^{4} x \,{\mathrm e}^{-x}}{5 \left (x^{3}-3 x^{2}+x +6\right )}\) | \(25\) |
default | \(\frac {{\mathrm e}^{4} \left (\frac {{\mathrm e}^{-x} \left (151 x^{2}-861 x -1206\right )}{643 x^{3}-1929 x^{2}+643 x +3858}-\frac {\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}-3 \textit {\_Z}^{2}+\textit {\_Z} +6\right )}{\sum }\frac {\left (151 \textit {\_R1}^{2}-67 \textit {\_R1} -546\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{3 \textit {\_R1}^{2}-6 \textit {\_R1} +1}\right )}{643}-\frac {6 \,{\mathrm e}^{-x} \left (12 x^{2}+21 x -49\right )}{643 \left (x^{3}-3 x^{2}+x +6\right )}+\frac {6 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}-3 \textit {\_Z}^{2}+\textit {\_Z} +6\right )}{\sum }\frac {\left (12 \textit {\_R1}^{2}+33 \textit {\_R1} +29\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{3 \textit {\_R1}^{2}-6 \textit {\_R1} +1}\right )}{643}+\frac {6 \,{\mathrm e}^{-x} \left (57 x^{2}-61 x -72\right )}{643 \left (x^{3}-3 x^{2}+x +6\right )}-\frac {6 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}-3 \textit {\_Z}^{2}+\textit {\_Z} +6\right )}{\sum }\frac {\left (57 \textit {\_R1}^{2}-4 \textit {\_R1} -23\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{3 \textit {\_R1}^{2}-6 \textit {\_R1} +1}\right )}{643}-\frac {2 \,{\mathrm e}^{-x} \left (110 x^{2}-129 x -342\right )}{643 \left (x^{3}-3 x^{2}+x +6\right )}+\frac {2 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}-3 \textit {\_Z}^{2}+\textit {\_Z} +6\right )}{\sum }\frac {\left (110 \textit {\_R1}^{2}-19 \textit {\_R1} -270\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{3 \textit {\_R1}^{2}-6 \textit {\_R1} +1}\right )}{643}-\frac {{\mathrm e}^{-x} \left (201 x^{2}-452 x -660\right )}{643 \left (x^{3}-3 x^{2}+x +6\right )}+\frac {\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}-3 \textit {\_Z}^{2}+\textit {\_Z} +6\right )}{\sum }\frac {\left (201 \textit {\_R1}^{2}-251 \textit {\_R1} -318\right ) {\mathrm e}^{-\textit {\_R1}} \operatorname {Ei}_{1}\left (x -\textit {\_R1} \right )}{3 \textit {\_R1}^{2}-6 \textit {\_R1} +1}\right )}{643}\right )}{5}\) | \(408\) |
int((x^4-x^3-2*x^2+6*x-6)*exp(2)^2/(5*x^6-30*x^5+55*x^4+30*x^3-175*x^2+60* x+180)/exp(x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4-x} \left (-6+6 x-2 x^2-x^3+x^4\right )}{180+60 x-175 x^2+30 x^3+55 x^4-30 x^5+5 x^6} \, dx=-\frac {x e^{\left (-x + 4\right )}}{5 \, {\left (x^{3} - 3 \, x^{2} + x + 6\right )}} \]
integrate((x^4-x^3-2*x^2+6*x-6)*exp(2)^2/(5*x^6-30*x^5+55*x^4+30*x^3-175*x ^2+60*x+180)/exp(x),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4-x} \left (-6+6 x-2 x^2-x^3+x^4\right )}{180+60 x-175 x^2+30 x^3+55 x^4-30 x^5+5 x^6} \, dx=- \frac {x e^{4} e^{- x}}{5 x^{3} - 15 x^{2} + 5 x + 30} \]
integrate((x**4-x**3-2*x**2+6*x-6)*exp(2)**2/(5*x**6-30*x**5+55*x**4+30*x* *3-175*x**2+60*x+180)/exp(x),x)
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4-x} \left (-6+6 x-2 x^2-x^3+x^4\right )}{180+60 x-175 x^2+30 x^3+55 x^4-30 x^5+5 x^6} \, dx=-\frac {x e^{\left (-x + 4\right )}}{5 \, {\left (x^{3} - 3 \, x^{2} + x + 6\right )}} \]
integrate((x^4-x^3-2*x^2+6*x-6)*exp(2)^2/(5*x^6-30*x^5+55*x^4+30*x^3-175*x ^2+60*x+180)/exp(x),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {e^{4-x} \left (-6+6 x-2 x^2-x^3+x^4\right )}{180+60 x-175 x^2+30 x^3+55 x^4-30 x^5+5 x^6} \, dx=-\frac {{\left (x - 4\right )} e^{\left (-x + 4\right )} + 4 \, e^{\left (-x + 4\right )}}{5 \, {\left ({\left (x - 4\right )}^{3} + 9 \, {\left (x - 4\right )}^{2} + 25 \, x - 74\right )}} \]
integrate((x^4-x^3-2*x^2+6*x-6)*exp(2)^2/(5*x^6-30*x^5+55*x^4+30*x^3-175*x ^2+60*x+180)/exp(x),x, algorithm=\
Time = 8.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4-x} \left (-6+6 x-2 x^2-x^3+x^4\right )}{180+60 x-175 x^2+30 x^3+55 x^4-30 x^5+5 x^6} \, dx=-\frac {x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^4}{5\,\left (x^3-3\,x^2+x+6\right )} \]