Integrand size = 130, antiderivative size = 26 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=1+e^{x-x^2 \log ^2\left (\frac {1}{(4+\log (x-x \log (4)))^2}\right )} \]
Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{x-x^2 \log ^2\left (\frac {1}{(4+\log (x-x \log (4)))^2}\right )} \]
Integrate[(E^(x - x^2*Log[(16 + 8*Log[x - x*Log[4]] + Log[x - x*Log[4]]^2) ^(-1)]^2)*(4 + Log[x - x*Log[4]] + 4*x*Log[(16 + 8*Log[x - x*Log[4]] + Log [x - x*Log[4]]^2)^(-1)] + (-8*x - 2*x*Log[x - x*Log[4]])*Log[(16 + 8*Log[x - x*Log[4]] + Log[x - x*Log[4]]^2)^(-1)]^2))/(4 + Log[x - x*Log[4]]),x]
Time = 0.72 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2894, 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.
\(\displaystyle \int \frac {\left ((-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{\log ^2(x-x \log (4))+8 \log (x-x \log (4))+16}\right )+4 x \log \left (\frac {1}{\log ^2(x-x \log (4))+8 \log (x-x \log (4))+16}\right )+\log (x-x \log (4))+4\right ) \exp \left (x-x^2 \log ^2\left (\frac {1}{\log ^2(x-x \log (4))+8 \log (x-x \log (4))+16}\right )\right )}{\log (x-x \log (4))+4} \, dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \int \frac {\left ((-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{\log ^2(x-x \log (4))+8 \log (x-x \log (4))+16}\right )+4 x \log \left (\frac {1}{\log ^2(x-x \log (4))+8 \log (x-x \log (4))+16}\right )+\log (x-x \log (4))+4\right ) \exp \left (x-x^2 \log ^2\left (\frac {1}{\log ^2(x-x \log (4))+8 \log (x-x \log (4))+16}\right )\right )}{\log (x (1-\log (4)))+4}dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle \exp \left (x-x^2 \log ^2\left (\frac {1}{\log ^2(x (1-\log (4)))+8 \log (x (1-\log (4)))+16}\right )\right )\) |
Int[(E^(x - x^2*Log[(16 + 8*Log[x - x*Log[4]] + Log[x - x*Log[4]]^2)^(-1)] ^2)*(4 + Log[x - x*Log[4]] + 4*x*Log[(16 + 8*Log[x - x*Log[4]] + Log[x - x *Log[4]]^2)^(-1)] + (-8*x - 2*x*Log[x - x*Log[4]])*Log[(16 + 8*Log[x - x*L og[4]] + Log[x - x*Log[4]]^2)^(-1)]^2))/(4 + Log[x - x*Log[4]]),x]
3.26.59.3.1 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]
Time = 2.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38
method | result | size |
parallelrisch | \({\mathrm e}^{-x^{2} \ln \left (\frac {1}{\ln \left (x -2 x \ln \left (2\right )\right )^{2}+8 \ln \left (x -2 x \ln \left (2\right )\right )+16}\right )^{2}+x}\) | \(36\) |
risch | \(\left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{4 i \pi \,\operatorname {csgn}\left (i \left (\ln \left (-\left (2 \ln \left (2\right )-1\right ) x \right )+4\right )^{2}\right ) x^{2}} \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{-4 i \pi \,\operatorname {csgn}\left (i \left (\ln \left (-x \right )+\ln \left (2 \ln \left (2\right )-1\right )+4\right )\right ) x^{2}} {\mathrm e}^{-\frac {x \left (-x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{6}+4 x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{5} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )-6 x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{4} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )^{2}+4 x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{3} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )^{3}-x \,\pi ^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )\right )^{4}+16 x \ln \left (\ln \left (x -2 x \ln \left (2\right )\right )+4\right )^{2}-4\right )}{4}}\) | \(280\) |
int(((-2*x*ln(x-2*x*ln(2))-8*x)*ln(1/(ln(x-2*x*ln(2))^2+8*ln(x-2*x*ln(2))+ 16))^2+4*x*ln(1/(ln(x-2*x*ln(2))^2+8*ln(x-2*x*ln(2))+16))+ln(x-2*x*ln(2))+ 4)*exp(-x^2*ln(1/(ln(x-2*x*ln(2))^2+8*ln(x-2*x*ln(2))+16))^2+x)/(ln(x-2*x* ln(2))+4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{\left (-x^{2} \log \left (\frac {1}{\log \left (-2 \, x \log \left (2\right ) + x\right )^{2} + 8 \, \log \left (-2 \, x \log \left (2\right ) + x\right ) + 16}\right )^{2} + x\right )} \]
integrate(((-2*x*log(x-2*x*log(2))-8*x)*log(1/(log(x-2*x*log(2))^2+8*log(x -2*x*log(2))+16))^2+4*x*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log(2))+16) )+log(x-2*x*log(2))+4)*exp(-x^2*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log (2))+16))^2+x)/(log(x-2*x*log(2))+4),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{- x^{2} \log {\left (\frac {1}{\log {\left (- 2 x \log {\left (2 \right )} + x \right )}^{2} + 8 \log {\left (- 2 x \log {\left (2 \right )} + x \right )} + 16} \right )}^{2} + x} \]
integrate(((-2*x*ln(x-2*x*ln(2))-8*x)*ln(1/(ln(x-2*x*ln(2))**2+8*ln(x-2*x* ln(2))+16))**2+4*x*ln(1/(ln(x-2*x*ln(2))**2+8*ln(x-2*x*ln(2))+16))+ln(x-2* x*ln(2))+4)*exp(-x**2*ln(1/(ln(x-2*x*ln(2))**2+8*ln(x-2*x*ln(2))+16))**2+x )/(ln(x-2*x*ln(2))+4),x)
Time = 0.41 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{\left (-4 \, x^{2} \log \left (\log \left (x\right ) + \log \left (-2 \, \log \left (2\right ) + 1\right ) + 4\right )^{2} + x\right )} \]
integrate(((-2*x*log(x-2*x*log(2))-8*x)*log(1/(log(x-2*x*log(2))^2+8*log(x -2*x*log(2))+16))^2+4*x*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log(2))+16) )+log(x-2*x*log(2))+4)*exp(-x^2*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log (2))+16))^2+x)/(log(x-2*x*log(2))+4),x, algorithm=\
Time = 1.44 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx=e^{\left (-x^{2} \log \left (\log \left (-2 \, x \log \left (2\right ) + x\right )^{2} + 8 \, \log \left (-2 \, x \log \left (2\right ) + x\right ) + 16\right )^{2} + x\right )} \]
integrate(((-2*x*log(x-2*x*log(2))-8*x)*log(1/(log(x-2*x*log(2))^2+8*log(x -2*x*log(2))+16))^2+4*x*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log(2))+16) )+log(x-2*x*log(2))+4)*exp(-x^2*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log (2))+16))^2+x)/(log(x-2*x*log(2))+4),x, algorithm=\
Time = 14.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )} \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x-x \log (4))} \, dx={\mathrm {e}}^x\,{\mathrm {e}}^{-x^2\,{\ln \left (\frac {1}{{\ln \left (x-2\,x\,\ln \left (2\right )\right )}^2+8\,\ln \left (x-2\,x\,\ln \left (2\right )\right )+16}\right )}^2} \]
int((exp(x - x^2*log(1/(8*log(x - 2*x*log(2)) + log(x - 2*x*log(2))^2 + 16 ))^2)*(log(x - 2*x*log(2)) - log(1/(8*log(x - 2*x*log(2)) + log(x - 2*x*lo g(2))^2 + 16))^2*(8*x + 2*x*log(x - 2*x*log(2))) + 4*x*log(1/(8*log(x - 2* x*log(2)) + log(x - 2*x*log(2))^2 + 16)) + 4))/(log(x - 2*x*log(2)) + 4),x )