Integrand size = 77, antiderivative size = 22 \[ \int \frac {e^{\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx=e^{9-x+\frac {x}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \]
Time = 1.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx=e^{9-x+\frac {x}{\log \left (\frac {1}{4} x (1+\log (2))\right )}} \]
Integrate[(E^((x + (9 - x)*Log[(x + x*Log[2])/4])/Log[(x + x*Log[2])/4])*( -1 + Log[(x + x*Log[2])/4] - Log[(x + x*Log[2])/4]^2))/Log[(x + x*Log[2])/ 4]^2,x]
Time = 0.87 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {2894, 7281, 25, 7257}
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.
\(\displaystyle \int \frac {\left (-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )+\log \left (\frac {1}{4} (x+x \log (2))\right )-1\right ) \exp \left (\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \int \frac {\left (-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )+\log \left (\frac {1}{4} (x+x \log (2))\right )-1\right ) \exp \left (\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}\right )}{\log ^2\left (\frac {1}{4} x (1+\log (2))\right )}dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle 4 \int -\frac {e^{\frac {x}{\log \left (\frac {1}{4} x (1+\log (2))\right )}-x+9} \left (\log ^2\left (\frac {1}{4} x (1+\log (2))\right )-\log \left (\frac {1}{4} x (1+\log (2))\right )+1\right )}{\log ^2\left (\frac {1}{4} x (1+\log (2))\right )}d\frac {x}{4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {e^{\frac {x}{\log \left (\frac {1}{4} x (1+\log (2))\right )}-x+9} \left (\log ^2\left (\frac {1}{4} x (1+\log (2))\right )-\log \left (\frac {1}{4} x (1+\log (2))\right )+1\right )}{\log ^2\left (\frac {1}{4} x (1+\log (2))\right )}d\frac {x}{4}\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{-x+\frac {x}{\log \left (\frac {1}{4} x (1+\log (2))\right )}+9}\) |
Int[(E^((x + (9 - x)*Log[(x + x*Log[2])/4])/Log[(x + x*Log[2])/4])*(-1 + L og[(x + x*Log[2])/4] - Log[(x + x*Log[2])/4]^2))/Log[(x + x*Log[2])/4]^2,x ]
3.8.77.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*( a + b*Log[c*ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p}, x] && Lin earQ[v, x] && !LinearMatchQ[v, x] && !(EqQ[n, 1] && MatchQ[c*v, (e_.)*((f _) + (g_.)*x) /; FreeQ[{e, f, g}, x]])
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Time = 0.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (9-x \right ) \ln \left (\frac {\left (1+\ln \left (2\right )\right ) x}{4}\right )+x}{\ln \left (\frac {\left (1+\ln \left (2\right )\right ) x}{4}\right )}}\) | \(29\) |
norman | \({\mathrm e}^{\frac {\left (9-x \right ) \ln \left (\frac {x \ln \left (2\right )}{4}+\frac {x}{4}\right )+x}{\ln \left (\frac {x \ln \left (2\right )}{4}+\frac {x}{4}\right )}}\) | \(33\) |
risch | \({\mathrm e}^{-\frac {\ln \left (\frac {x \ln \left (2\right )}{4}+\frac {x}{4}\right ) x -9 \ln \left (\frac {x \ln \left (2\right )}{4}+\frac {x}{4}\right )-x}{\ln \left (\frac {x \ln \left (2\right )}{4}+\frac {x}{4}\right )}}\) | \(44\) |
int((-ln(1/4*x*ln(2)+1/4*x)^2+ln(1/4*x*ln(2)+1/4*x)-1)*exp(((9-x)*ln(1/4*x *ln(2)+1/4*x)+x)/ln(1/4*x*ln(2)+1/4*x))/ln(1/4*x*ln(2)+1/4*x)^2,x,method=_ RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx=e^{\left (-\frac {{\left (x - 9\right )} \log \left (\frac {1}{4} \, x \log \left (2\right ) + \frac {1}{4} \, x\right ) - x}{\log \left (\frac {1}{4} \, x \log \left (2\right ) + \frac {1}{4} \, x\right )}\right )} \]
integrate((-log(1/4*x*log(2)+1/4*x)^2+log(1/4*x*log(2)+1/4*x)-1)*exp(((9-x )*log(1/4*x*log(2)+1/4*x)+x)/log(1/4*x*log(2)+1/4*x))/log(1/4*x*log(2)+1/4 *x)^2,x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx=e^{\frac {x + \left (9 - x\right ) \log {\left (\frac {x \log {\left (2 \right )}}{4} + \frac {x}{4} \right )}}{\log {\left (\frac {x \log {\left (2 \right )}}{4} + \frac {x}{4} \right )}}} \]
integrate((-ln(1/4*x*ln(2)+1/4*x)**2+ln(1/4*x*ln(2)+1/4*x)-1)*exp(((9-x)*l n(1/4*x*ln(2)+1/4*x)+x)/ln(1/4*x*ln(2)+1/4*x))/ln(1/4*x*ln(2)+1/4*x)**2,x)
Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx=e^{\left (-x - \frac {x}{2 \, \log \left (2\right ) - \log \left (x\right ) - \log \left (\log \left (2\right ) + 1\right )} + 9\right )} \]
integrate((-log(1/4*x*log(2)+1/4*x)^2+log(1/4*x*log(2)+1/4*x)-1)*exp(((9-x )*log(1/4*x*log(2)+1/4*x)+x)/log(1/4*x*log(2)+1/4*x))/log(1/4*x*log(2)+1/4 *x)^2,x, algorithm=\
Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx=e^{\left (-x + \frac {x}{\log \left (\frac {1}{4} \, x \log \left (2\right ) + \frac {1}{4} \, x\right )} + 9\right )} \]
integrate((-log(1/4*x*log(2)+1/4*x)^2+log(1/4*x*log(2)+1/4*x)-1)*exp(((9-x )*log(1/4*x*log(2)+1/4*x)+x)/log(1/4*x*log(2)+1/4*x))/log(1/4*x*log(2)+1/4 *x)^2,x, algorithm=\
Time = 9.55 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} (x+x \log (2))\right )} \, dx={\mathrm {e}}^{-x}\,{\mathrm {e}}^9\,{\mathrm {e}}^{\frac {x}{\ln \left (\frac {x}{4}+\frac {x\,\ln \left (2\right )}{4}\right )}} \]