Integrand size = 101, antiderivative size = 26 \[ \int \frac {e^{\sqrt [4]{e}} (-4+x)-4 x-2 x^2+3 x^3}{4 x^2+8 x^3+4 x^4+e^{\sqrt [4]{e}} \left (4 x+4 x^2\right )+\left (e^{\sqrt [4]{e}} x^2+x^3+x^4\right ) \log \left (\frac {1}{2} \left (e^{\sqrt [4]{e}} x+x^2+x^3\right )\right )} \, dx=\log \left (4+\frac {4}{x}+\log \left (\frac {1}{2} x \left (e^{\sqrt [4]{e}}+x+x^2\right )\right )\right ) \]
Time = 5.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\sqrt [4]{e}} (-4+x)-4 x-2 x^2+3 x^3}{4 x^2+8 x^3+4 x^4+e^{\sqrt [4]{e}} \left (4 x+4 x^2\right )+\left (e^{\sqrt [4]{e}} x^2+x^3+x^4\right ) \log \left (\frac {1}{2} \left (e^{\sqrt [4]{e}} x+x^2+x^3\right )\right )} \, dx=-\log (x)+\log \left (4+4 x+x \log \left (\frac {1}{2} x \left (e^{\sqrt [4]{e}}+x+x^2\right )\right )\right ) \]
Integrate[(E^E^(1/4)*(-4 + x) - 4*x - 2*x^2 + 3*x^3)/(4*x^2 + 8*x^3 + 4*x^ 4 + E^E^(1/4)*(4*x + 4*x^2) + (E^E^(1/4)*x^2 + x^3 + x^4)*Log[(E^E^(1/4)*x + x^2 + x^3)/2]),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.
\(\displaystyle \int \frac {3 x^3-2 x^2-4 x+e^{\sqrt [4]{e}} (x-4)}{4 x^4+8 x^3+4 x^2+e^{\sqrt [4]{e}} \left (4 x^2+4 x\right )+\left (x^4+x^3+e^{\sqrt [4]{e}} x^2\right ) \log \left (\frac {1}{2} \left (x^3+x^2+e^{\sqrt [4]{e}} x\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 x^3-2 x^2-\left (4-e^{\sqrt [4]{e}}\right ) x-4 e^{\sqrt [4]{e}}}{x \left (x^2+x+e^{\sqrt [4]{e}}\right ) \left (x \log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right )+4 x+4\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (-\frac {x+2 e^{\sqrt [4]{e}}}{\left (x^2+x+e^{\sqrt [4]{e}}\right ) \left (x \log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right )+4 x+4\right )}-\frac {4}{x \left (x \log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right )+4 x+4\right )}+\frac {3}{x \log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right )+4 x+4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \int \frac {1}{\log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right ) x+4 x+4}dx-\frac {4 i e^{\sqrt [4]{e}} \int \frac {1}{\left (-2 x+i \sqrt {-1+4 e^{\sqrt [4]{e}}}-1\right ) \left (\log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right ) x+4 x+4\right )}dx}{\sqrt {4 e^{\sqrt [4]{e}}-1}}-4 \int \frac {1}{x \left (\log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right ) x+4 x+4\right )}dx-\left (1+\frac {i}{\sqrt {4 e^{\sqrt [4]{e}}-1}}\right ) \int \frac {1}{\left (2 x-i \sqrt {-1+4 e^{\sqrt [4]{e}}}+1\right ) \left (\log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right ) x+4 x+4\right )}dx-\left (1-\frac {i}{\sqrt {4 e^{\sqrt [4]{e}}-1}}\right ) \int \frac {1}{\left (2 x+i \sqrt {-1+4 e^{\sqrt [4]{e}}}+1\right ) \left (\log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right ) x+4 x+4\right )}dx-\frac {4 i e^{\sqrt [4]{e}} \int \frac {1}{\left (2 x+i \sqrt {-1+4 e^{\sqrt [4]{e}}}+1\right ) \left (\log \left (\frac {1}{2} x \left (x^2+x+e^{\sqrt [4]{e}}\right )\right ) x+4 x+4\right )}dx}{\sqrt {4 e^{\sqrt [4]{e}}-1}}\) |
Int[(E^E^(1/4)*(-4 + x) - 4*x - 2*x^2 + 3*x^3)/(4*x^2 + 8*x^3 + 4*x^4 + E^ E^(1/4)*(4*x + 4*x^2) + (E^E^(1/4)*x^2 + x^3 + x^4)*Log[(E^E^(1/4)*x + x^2 + x^3)/2]),x]
3.7.66.3.1 Defintions of rubi rules used
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 = 1.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(-\ln \left (x \right )+\ln \left (x \ln \left (\frac {x \left ({\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}+x +x^{2}\right )}{2}\right )+4 x +4\right )\) | \(26\) |
risch | \(\ln \left (\ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}}{2}+\frac {x^{3}}{2}+\frac {x^{2}}{2}\right )+\frac {4+4 x}{x}\right )\) | \(29\) |
norman | \(-\ln \left (x \right )+\ln \left (x \ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{\frac {1}{4}}}}{2}+\frac {x^{3}}{2}+\frac {x^{2}}{2}\right )+4 x +4\right )\) | \(32\) |
int(((x-4)*exp(exp(1/4))+3*x^3-2*x^2-4*x)/((x^2*exp(exp(1/4))+x^4+x^3)*ln( 1/2*x*exp(exp(1/4))+1/2*x^3+1/2*x^2)+(4*x^2+4*x)*exp(exp(1/4))+4*x^4+8*x^3 +4*x^2),x,method=_RETURNVERBOSE)
Time = 0.35 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\sqrt [4]{e}} (-4+x)-4 x-2 x^2+3 x^3}{4 x^2+8 x^3+4 x^4+e^{\sqrt [4]{e}} \left (4 x+4 x^2\right )+\left (e^{\sqrt [4]{e}} x^2+x^3+x^4\right ) \log \left (\frac {1}{2} \left (e^{\sqrt [4]{e}} x+x^2+x^3\right )\right )} \, dx=\log \left (\frac {x \log \left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x^{2} + \frac {1}{2} \, x e^{\left (e^{\frac {1}{4}}\right )}\right ) + 4 \, x + 4}{x}\right ) \]
integrate(((x-4)*exp(exp(1/4))+3*x^3-2*x^2-4*x)/((x^2*exp(exp(1/4))+x^4+x^ 3)*log(1/2*x*exp(exp(1/4))+1/2*x^3+1/2*x^2)+(4*x^2+4*x)*exp(exp(1/4))+4*x^ 4+8*x^3+4*x^2),x, algorithm=\
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\sqrt [4]{e}} (-4+x)-4 x-2 x^2+3 x^3}{4 x^2+8 x^3+4 x^4+e^{\sqrt [4]{e}} \left (4 x+4 x^2\right )+\left (e^{\sqrt [4]{e}} x^2+x^3+x^4\right ) \log \left (\frac {1}{2} \left (e^{\sqrt [4]{e}} x+x^2+x^3\right )\right )} \, dx=\log {\left (\log {\left (\frac {x^{3}}{2} + \frac {x^{2}}{2} + \frac {x e^{e^{\frac {1}{4}}}}{2} \right )} + \frac {4 x + 4}{x} \right )} \]
integrate(((x-4)*exp(exp(1/4))+3*x**3-2*x**2-4*x)/((x**2*exp(exp(1/4))+x** 4+x**3)*ln(1/2*x*exp(exp(1/4))+1/2*x**3+1/2*x**2)+(4*x**2+4*x)*exp(exp(1/4 ))+4*x**4+8*x**3+4*x**2),x)
Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\sqrt [4]{e}} (-4+x)-4 x-2 x^2+3 x^3}{4 x^2+8 x^3+4 x^4+e^{\sqrt [4]{e}} \left (4 x+4 x^2\right )+\left (e^{\sqrt [4]{e}} x^2+x^3+x^4\right ) \log \left (\frac {1}{2} \left (e^{\sqrt [4]{e}} x+x^2+x^3\right )\right )} \, dx=\log \left (-\frac {x {\left (\log \left (2\right ) - 4\right )} - x \log \left (x^{2} + x + e^{\left (e^{\frac {1}{4}}\right )}\right ) - x \log \left (x\right ) - 4}{x}\right ) \]
integrate(((x-4)*exp(exp(1/4))+3*x^3-2*x^2-4*x)/((x^2*exp(exp(1/4))+x^4+x^ 3)*log(1/2*x*exp(exp(1/4))+1/2*x^3+1/2*x^2)+(4*x^2+4*x)*exp(exp(1/4))+4*x^ 4+8*x^3+4*x^2),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\sqrt [4]{e}} (-4+x)-4 x-2 x^2+3 x^3}{4 x^2+8 x^3+4 x^4+e^{\sqrt [4]{e}} \left (4 x+4 x^2\right )+\left (e^{\sqrt [4]{e}} x^2+x^3+x^4\right ) \log \left (\frac {1}{2} \left (e^{\sqrt [4]{e}} x+x^2+x^3\right )\right )} \, dx=\log \left (x \log \left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x^{2} + \frac {1}{2} \, x e^{\left (e^{\frac {1}{4}}\right )}\right ) + 4 \, x + 4\right ) - \log \left (x\right ) \]
integrate(((x-4)*exp(exp(1/4))+3*x^3-2*x^2-4*x)/((x^2*exp(exp(1/4))+x^4+x^ 3)*log(1/2*x*exp(exp(1/4))+1/2*x^3+1/2*x^2)+(4*x^2+4*x)*exp(exp(1/4))+4*x^ 4+8*x^3+4*x^2),x, algorithm=\
Time = 15.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\sqrt [4]{e}} (-4+x)-4 x-2 x^2+3 x^3}{4 x^2+8 x^3+4 x^4+e^{\sqrt [4]{e}} \left (4 x+4 x^2\right )+\left (e^{\sqrt [4]{e}} x^2+x^3+x^4\right ) \log \left (\frac {1}{2} \left (e^{\sqrt [4]{e}} x+x^2+x^3\right )\right )} \, dx=\ln \left (\ln \left (\frac {x^3}{2}+\frac {x^2}{2}+\frac {{\mathrm {e}}^{{\mathrm {e}}^{1/4}}\,x}{2}\right )+\frac {4}{x}+4\right ) \]