Integrand size = 115, antiderivative size = 33 \[ \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x+2 x^2+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )+\left (-1+x+x^2\right ) \log ^2\left (\frac {x^2}{4}\right )\right )}{\left (9 x^2-18 x^3+9 x^4\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx=\frac {-4+e^{-x+\frac {x}{\log \left (\frac {x^2}{4}\right )}}}{9 \left (x-x^2\right )} \]
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x+2 x^2+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )+\left (-1+x+x^2\right ) \log ^2\left (\frac {x^2}{4}\right )\right )}{\left (9 x^2-18 x^3+9 x^4\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx=\frac {e^{-x} \left (4 e^x-e^{\frac {x}{\log \left (\frac {x^2}{4}\right )}}\right )}{9 (-1+x) x} \]
Integrate[((4 - 8*x)*Log[x^2/4]^2 + E^((x - x*Log[x^2/4])/Log[x^2/4])*(-2* x + 2*x^2 + (x - x^2)*Log[x^2/4] + (-1 + x + x^2)*Log[x^2/4]^2))/((9*x^2 - 18*x^3 + 9*x^4)*Log[x^2/4]^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.
\(\displaystyle \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (2 x^2+\left (x^2+x-1\right ) \log ^2\left (\frac {x^2}{4}\right )+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )-2 x\right )}{\left (9 x^4-18 x^3+9 x^2\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (2 x^2+\left (x^2+x-1\right ) \log ^2\left (\frac {x^2}{4}\right )+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )-2 x\right )}{x^2 \left (9 x^2-18 x+9\right ) \log ^2\left (\frac {x^2}{4}\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 36 \int \frac {4 (1-2 x) \log ^2\left (\frac {x^2}{4}\right )-(4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x^2+2 x+\left (-x^2-x+1\right ) \log ^2\left (\frac {x^2}{4}\right )-\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )\right )}{324 (1-x)^2 x^2 \log ^2\left (\frac {x^2}{4}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {4 (1-2 x) \log ^2\left (\frac {x^2}{4}\right )-(4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x^2+2 x+\left (-x^2-x+1\right ) \log ^2\left (\frac {x^2}{4}\right )-\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )\right )}{(1-x)^2 x^2 \log ^2\left (\frac {x^2}{4}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{9} \int \left (\frac {(4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1} \left (\log ^2\left (\frac {x^2}{4}\right ) x^2-\log \left (\frac {x^2}{4}\right ) x^2+2 x^2+\log ^2\left (\frac {x^2}{4}\right ) x+\log \left (\frac {x^2}{4}\right ) x-2 x-\log ^2\left (\frac {x^2}{4}\right )\right )}{(x-1)^2 \log ^2\left (\frac {x^2}{4}\right )}-\frac {4 (2 x-1)}{(x-1)^2 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (\int \frac {2^{\frac {2 x}{\log \left (\frac {x^2}{4}\right )}+1} e^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1}}{\log ^2\left (\frac {x^2}{4}\right )}dx+\int \frac {2^{\frac {2 x}{\log \left (\frac {x^2}{4}\right )}+1} e^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1}}{(x-1) \log ^2\left (\frac {x^2}{4}\right )}dx+\int (4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1}dx+\int \frac {(4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1}}{(x-1)^2}dx+3 \int \frac {(4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1}}{x-1}dx-\int \frac {(4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1}}{\log \left (\frac {x^2}{4}\right )}dx-\int \frac {(4 e)^{\frac {x}{\log \left (\frac {x^2}{4}\right )}} \left (x^2\right )^{-\frac {x}{\log \left (\frac {x^2}{4}\right )}-1}}{(x-1) \log \left (\frac {x^2}{4}\right )}dx-\frac {4}{(1-x) x}\right )\) |
Int[((4 - 8*x)*Log[x^2/4]^2 + E^((x - x*Log[x^2/4])/Log[x^2/4])*(-2*x + 2* x^2 + (x - x^2)*Log[x^2/4] + (-1 + x + x^2)*Log[x^2/4]^2))/((9*x^2 - 18*x^ 3 + 9*x^4)*Log[x^2/4]^2),x]
3.8.57.3.1 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[(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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 6.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {4-{\mathrm e}^{\frac {-x \ln \left (\frac {x^{2}}{4}\right )+x}{\ln \left (\frac {x^{2}}{4}\right )}}}{9 x \left (-1+x \right )}\) | \(36\) |
risch | \(\frac {4}{9 x \left (-1+x \right )}-\frac {{\mathrm e}^{-\frac {x \left (\ln \left (\frac {x^{2}}{4}\right )-1\right )}{\ln \left (\frac {x^{2}}{4}\right )}}}{9 x \left (-1+x \right )}\) | \(42\) |
int((((x^2+x-1)*ln(1/4*x^2)^2+(-x^2+x)*ln(1/4*x^2)+2*x^2-2*x)*exp((-x*ln(1 /4*x^2)+x)/ln(1/4*x^2))+(-8*x+4)*ln(1/4*x^2)^2)/(9*x^4-18*x^3+9*x^2)/ln(1/ 4*x^2)^2,x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x+2 x^2+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )+\left (-1+x+x^2\right ) \log ^2\left (\frac {x^2}{4}\right )\right )}{\left (9 x^2-18 x^3+9 x^4\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx=-\frac {e^{\left (-\frac {x \log \left (\frac {1}{4} \, x^{2}\right ) - x}{\log \left (\frac {1}{4} \, x^{2}\right )}\right )} - 4}{9 \, {\left (x^{2} - x\right )}} \]
integrate((((x^2+x-1)*log(1/4*x^2)^2+(-x^2+x)*log(1/4*x^2)+2*x^2-2*x)*exp( (-x*log(1/4*x^2)+x)/log(1/4*x^2))+(-8*x+4)*log(1/4*x^2)^2)/(9*x^4-18*x^3+9 *x^2)/log(1/4*x^2)^2,x, algorithm=\
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x+2 x^2+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )+\left (-1+x+x^2\right ) \log ^2\left (\frac {x^2}{4}\right )\right )}{\left (9 x^2-18 x^3+9 x^4\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx=- \frac {e^{\frac {- x \log {\left (\frac {x^{2}}{4} \right )} + x}{\log {\left (\frac {x^{2}}{4} \right )}}}}{9 x^{2} - 9 x} + \frac {4}{9 x^{2} - 9 x} \]
integrate((((x**2+x-1)*ln(1/4*x**2)**2+(-x**2+x)*ln(1/4*x**2)+2*x**2-2*x)* exp((-x*ln(1/4*x**2)+x)/ln(1/4*x**2))+(-8*x+4)*ln(1/4*x**2)**2)/(9*x**4-18 *x**3+9*x**2)/ln(1/4*x**2)**2,x)
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x+2 x^2+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )+\left (-1+x+x^2\right ) \log ^2\left (\frac {x^2}{4}\right )\right )}{\left (9 x^2-18 x^3+9 x^4\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx=\frac {{\left (4 \, e^{x} - e^{\left (-\frac {x}{2 \, {\left (\log \left (2\right ) - \log \left (x\right )\right )}}\right )}\right )} e^{\left (-x\right )}}{9 \, {\left (x^{2} - x\right )}} \]
integrate((((x^2+x-1)*log(1/4*x^2)^2+(-x^2+x)*log(1/4*x^2)+2*x^2-2*x)*exp( (-x*log(1/4*x^2)+x)/log(1/4*x^2))+(-8*x+4)*log(1/4*x^2)^2)/(9*x^4-18*x^3+9 *x^2)/log(1/4*x^2)^2,x, algorithm=\
Time = 1.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x+2 x^2+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )+\left (-1+x+x^2\right ) \log ^2\left (\frac {x^2}{4}\right )\right )}{\left (9 x^2-18 x^3+9 x^4\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx=-\frac {e^{\left (-\frac {x \log \left (\frac {1}{4} \, x^{2}\right ) - x}{\log \left (\frac {1}{4} \, x^{2}\right )}\right )} - 4}{9 \, {\left (x^{2} - x\right )}} \]
integrate((((x^2+x-1)*log(1/4*x^2)^2+(-x^2+x)*log(1/4*x^2)+2*x^2-2*x)*exp( (-x*log(1/4*x^2)+x)/log(1/4*x^2))+(-8*x+4)*log(1/4*x^2)^2)/(9*x^4-18*x^3+9 *x^2)/log(1/4*x^2)^2,x, algorithm=\
Time = 8.94 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.97 \[ \int \frac {(4-8 x) \log ^2\left (\frac {x^2}{4}\right )+e^{\frac {x-x \log \left (\frac {x^2}{4}\right )}{\log \left (\frac {x^2}{4}\right )}} \left (-2 x+2 x^2+\left (x-x^2\right ) \log \left (\frac {x^2}{4}\right )+\left (-1+x+x^2\right ) \log ^2\left (\frac {x^2}{4}\right )\right )}{\left (9 x^2-18 x^3+9 x^4\right ) \log ^2\left (\frac {x^2}{4}\right )} \, dx=\frac {2^{\frac {2\,x}{\ln \left (\frac {x^2}{4}\right )}}\,{\mathrm {e}}^{\frac {x}{\ln \left (\frac {x^2}{4}\right )}}}{9\,\left (x-x^2\right )\,{\left (x^2\right )}^{\frac {x}{\ln \left (\frac {x^2}{4}\right )}}}-\frac {4}{9\,\left (x-x^2\right )} \]
int((exp((x - x*log(x^2/4))/log(x^2/4))*(log(x^2/4)*(x - x^2) - 2*x + log( x^2/4)^2*(x + x^2 - 1) + 2*x^2) - log(x^2/4)^2*(8*x - 4))/(log(x^2/4)^2*(9 *x^2 - 18*x^3 + 9*x^4)),x)