Integrand size = 72, antiderivative size = 29 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=3 e^{-4-x} x \left (1-\log \left (\left (-5+\frac {-2+x}{x}-x\right )^2\right )\right ) \] Output:
3/exp(4)*x/exp(x)*(1-ln(((-2+x)/x-5-x)^2))
Time = 0.59 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=-3 e^{-4-x} x \left (-1+\log \left (\frac {\left (2+4 x+x^2\right )^2}{x^2}\right )\right ) \] Input:
Integrate[(E^(-4 - x)*(18 + 6*x - 15*x^2 - 3*x^3 + (-6 - 6*x + 9*x^2 + 3*x ^3)*Log[(4 + 16*x + 20*x^2 + 8*x^3 + x^4)/x^2]))/(2 + 4*x + x^2),x]
Output:
-3*E^(-4 - x)*x*(-1 + Log[(2 + 4*x + x^2)^2/x^2])
Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(29)=58\).
Time = 0.89 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {7279, 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.
\(\displaystyle \int \frac {e^{-x-4} \left (-3 x^3-15 x^2+\left (3 x^3+9 x^2-6 x-6\right ) \log \left (\frac {x^4+8 x^3+20 x^2+16 x+4}{x^2}\right )+6 x+18\right )}{x^2+4 x+2} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (3 e^{-x-4} (x-1) \log \left (\frac {\left (x^2+4 x+2\right )^2}{x^2}\right )-\frac {3 e^{-x-4} \left (x^3+5 x^2-2 x-6\right )}{x^2+4 x+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 e^{-x-4} \log \left (\frac {\left (x^2+4 x+2\right )^2}{x^2}\right )+3 e^{-x-4} (1-x) \log \left (\frac {\left (x^2+4 x+2\right )^2}{x^2}\right )+3 e^{-x-4} x\) |
Input:
Int[(E^(-4 - x)*(18 + 6*x - 15*x^2 - 3*x^3 + (-6 - 6*x + 9*x^2 + 3*x^3)*Lo g[(4 + 16*x + 20*x^2 + 8*x^3 + x^4)/x^2]))/(2 + 4*x + x^2),x]
Output:
3*E^(-4 - x)*x - 3*E^(-4 - x)*Log[(2 + 4*x + x^2)^2/x^2] + 3*E^(-4 - x)*(1 - x)*Log[(2 + 4*x + x^2)^2/x^2]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.87 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{-4} \left (-24 \ln \left (\frac {x^{4}+8 x^{3}+20 x^{2}+16 x +4}{x^{2}}\right ) x +24 x \right ) {\mathrm e}^{-x}}{8}\) | \(41\) |
norman | \(\left (3 \,{\mathrm e}^{-4} x -3 x \,{\mathrm e}^{-4} \ln \left (\frac {x^{4}+8 x^{3}+20 x^{2}+16 x +4}{x^{2}}\right )\right ) {\mathrm e}^{-x}\) | \(44\) |
risch | \(-6 \ln \left (x^{2}+4 x +2\right ) x \,{\mathrm e}^{-4-x}+\frac {3 \left (i \pi {\operatorname {csgn}\left (i \left (x^{2}+4 x +2\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+4 x +2\right )^{2}\right )-2 i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +2\right )\right ) {\operatorname {csgn}\left (i \left (x^{2}+4 x +2\right )^{2}\right )}^{2}+i \pi {\operatorname {csgn}\left (i \left (x^{2}+4 x +2\right )^{2}\right )}^{3}-i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +2\right )^{2}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+4 x +2\right )^{2}}{x^{2}}\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left (x^{2}+4 x +2\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+4 x +2\right )^{2}}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+i \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+4 x +2\right )^{2}}{x^{2}}\right )}^{3}-i \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+4 x +2\right )^{2}}{x^{2}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 \ln \left (x \right )+2\right ) x \,{\mathrm e}^{-4-x}}{2}\) | \(300\) |
orering | \(-\frac {x \left (2 x^{5}+7 x^{4}-8 x^{3}-24 x^{2}-8 x +4\right ) \left (\left (3 x^{3}+9 x^{2}-6 x -6\right ) \ln \left (\frac {x^{4}+8 x^{3}+20 x^{2}+16 x +4}{x^{2}}\right )-3 x^{3}-15 x^{2}+6 x +18\right ) {\mathrm e}^{-4} {\mathrm e}^{-x}}{\left (x^{6}+3 x^{5}-7 x^{4}-8 x^{3}+4 x^{2}+4 x -4\right ) \left (x^{2}+4 x +2\right )}-\frac {\left (x^{2}+4 x +2\right ) \left (x^{2}-2\right ) x^{2} \left (\frac {\left (\left (9 x^{2}+18 x -6\right ) \ln \left (\frac {x^{4}+8 x^{3}+20 x^{2}+16 x +4}{x^{2}}\right )+\frac {\left (3 x^{3}+9 x^{2}-6 x -6\right ) \left (\frac {4 x^{3}+24 x^{2}+40 x +16}{x^{2}}-\frac {2 \left (x^{4}+8 x^{3}+20 x^{2}+16 x +4\right )}{x^{3}}\right ) x^{2}}{x^{4}+8 x^{3}+20 x^{2}+16 x +4}-9 x^{2}-30 x +6\right ) {\mathrm e}^{-4} {\mathrm e}^{-x}}{x^{2}+4 x +2}-\frac {\left (\left (3 x^{3}+9 x^{2}-6 x -6\right ) \ln \left (\frac {x^{4}+8 x^{3}+20 x^{2}+16 x +4}{x^{2}}\right )-3 x^{3}-15 x^{2}+6 x +18\right ) {\mathrm e}^{-4} {\mathrm e}^{-x} \left (4+2 x \right )}{\left (x^{2}+4 x +2\right )^{2}}-\frac {\left (\left (3 x^{3}+9 x^{2}-6 x -6\right ) \ln \left (\frac {x^{4}+8 x^{3}+20 x^{2}+16 x +4}{x^{2}}\right )-3 x^{3}-15 x^{2}+6 x +18\right ) {\mathrm e}^{-4} {\mathrm e}^{-x}}{x^{2}+4 x +2}\right )}{x^{6}+3 x^{5}-7 x^{4}-8 x^{3}+4 x^{2}+4 x -4}\) | \(479\) |
Input:
int(((3*x^3+9*x^2-6*x-6)*ln((x^4+8*x^3+20*x^2+16*x+4)/x^2)-3*x^3-15*x^2+6* x+18)/(x^2+4*x+2)/exp(4)/exp(x),x,method=_RETURNVERBOSE)
Output:
1/8/exp(4)*(-24*ln((x^4+8*x^3+20*x^2+16*x+4)/x^2)*x+24*x)/exp(x)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=-3 \, x e^{\left (-x - 4\right )} \log \left (\frac {x^{4} + 8 \, x^{3} + 20 \, x^{2} + 16 \, x + 4}{x^{2}}\right ) + 3 \, x e^{\left (-x - 4\right )} \] Input:
integrate(((3*x^3+9*x^2-6*x-6)*log((x^4+8*x^3+20*x^2+16*x+4)/x^2)-3*x^3-15 *x^2+6*x+18)/(x^2+4*x+2)/exp(4)/exp(x),x, algorithm="fricas")
Output:
-3*x*e^(-x - 4)*log((x^4 + 8*x^3 + 20*x^2 + 16*x + 4)/x^2) + 3*x*e^(-x - 4 )
Time = 0.43 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=\frac {\left (- 3 x \log {\left (\frac {x^{4} + 8 x^{3} + 20 x^{2} + 16 x + 4}{x^{2}} \right )} + 3 x\right ) e^{- x}}{e^{4}} \] Input:
integrate(((3*x**3+9*x**2-6*x-6)*ln((x**4+8*x**3+20*x**2+16*x+4)/x**2)-3*x **3-15*x**2+6*x+18)/(x**2+4*x+2)/exp(4)/exp(x),x)
Output:
(-3*x*log((x**4 + 8*x**3 + 20*x**2 + 16*x + 4)/x**2) + 3*x)*exp(-4)*exp(-x )
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=-3 \, {\left (2 \, x e^{\left (-x\right )} \log \left (x^{2} + 4 \, x + 2\right ) - {\left (2 \, x \log \left (x\right ) + x\right )} e^{\left (-x\right )}\right )} e^{\left (-4\right )} \] Input:
integrate(((3*x^3+9*x^2-6*x-6)*log((x^4+8*x^3+20*x^2+16*x+4)/x^2)-3*x^3-15 *x^2+6*x+18)/(x^2+4*x+2)/exp(4)/exp(x),x, algorithm="maxima")
Output:
-3*(2*x*e^(-x)*log(x^2 + 4*x + 2) - (2*x*log(x) + x)*e^(-x))*e^(-4)
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=-3 \, {\left (x e^{\left (-x\right )} \log \left (\frac {x^{4} + 8 \, x^{3} + 20 \, x^{2} + 16 \, x + 4}{x^{2}}\right ) - x e^{\left (-x\right )}\right )} e^{\left (-4\right )} \] Input:
integrate(((3*x^3+9*x^2-6*x-6)*log((x^4+8*x^3+20*x^2+16*x+4)/x^2)-3*x^3-15 *x^2+6*x+18)/(x^2+4*x+2)/exp(4)/exp(x),x, algorithm="giac")
Output:
-3*(x*e^(-x)*log((x^4 + 8*x^3 + 20*x^2 + 16*x + 4)/x^2) - x*e^(-x))*e^(-4)
Time = 4.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=-3\,x\,{\mathrm {e}}^{-x-4}\,\left (\ln \left (\frac {x^4+8\,x^3+20\,x^2+16\,x+4}{x^2}\right )-1\right ) \] Input:
int(-(exp(-x)*exp(-4)*(log((16*x + 20*x^2 + 8*x^3 + x^4 + 4)/x^2)*(6*x - 9 *x^2 - 3*x^3 + 6) - 6*x + 15*x^2 + 3*x^3 - 18))/(4*x + x^2 + 2),x)
Output:
-3*x*exp(- x - 4)*(log((16*x + 20*x^2 + 8*x^3 + x^4 + 4)/x^2) - 1)
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-4-x} \left (18+6 x-15 x^2-3 x^3+\left (-6-6 x+9 x^2+3 x^3\right ) \log \left (\frac {4+16 x+20 x^2+8 x^3+x^4}{x^2}\right )\right )}{2+4 x+x^2} \, dx=\frac {3 x \left (-\mathrm {log}\left (\frac {x^{4}+8 x^{3}+20 x^{2}+16 x +4}{x^{2}}\right )+1\right )}{e^{x} e^{4}} \] Input:
int(((3*x^3+9*x^2-6*x-6)*log((x^4+8*x^3+20*x^2+16*x+4)/x^2)-3*x^3-15*x^2+6 *x+18)/(x^2+4*x+2)/exp(4)/exp(x),x)
Output:
(3*x*( - log((x**4 + 8*x**3 + 20*x**2 + 16*x + 4)/x**2) + 1))/(e**x*e**4)