Integrand size = 83, antiderivative size = 37 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=4-e^{e^{4 \left (-2 x-x^2-\log (4)+\frac {x}{4 \log (x)}\right )}}-\frac {x}{4} \]
Time = 3.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-e^{\frac {1}{256} e^{-8 x-4 x^2+\frac {x}{\log (x)}}}-\frac {x}{4} \]
Integrate[(-Log[x]^2 + E^(E^((x + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x] ) + (x + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x])*(4 - 4*Log[x] + (32 + 3 2*x)*Log[x]^2))/(4*Log[x]^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.
\(\displaystyle \int \frac {\left ((32 x+32) \log ^2(x)-4 \log (x)+4\right ) \exp \left (\exp \left (\frac {\left (-4 x^2-8 x-4 \log (4)\right ) \log (x)+x}{\log (x)}\right )+\frac {\left (-4 x^2-8 x-4 \log (4)\right ) \log (x)+x}{\log (x)}\right )-\log ^2(x)}{4 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {\log ^2(x)-4 \exp \left (\frac {x-4 \left (x^2+2 x+\log (4)\right ) \log (x)}{\log (x)}+\frac {1}{256} e^{-4 x^2+\frac {x}{\log (x)}-8 x}\right ) \left (8 (x+1) \log ^2(x)-\log (x)+1\right )}{\log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {\log ^2(x)-4 \exp \left (\frac {x-4 \left (x^2+2 x+\log (4)\right ) \log (x)}{\log (x)}+\frac {1}{256} e^{-4 x^2+\frac {x}{\log (x)}-8 x}\right ) \left (8 (x+1) \log ^2(x)-\log (x)+1\right )}{\log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (1-\frac {\exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right ) \left (8 x \log ^2(x)+8 \log ^2(x)-\log (x)+1\right )}{64 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{64} \int \frac {\exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right )}{\log ^2(x)}dx+\frac {1}{8} \int \exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right )dx+\frac {1}{8} \int \exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right ) xdx-\frac {1}{64} \int \frac {\exp \left (-4 (x+2) x+\frac {x}{\log (x)}+\frac {1}{256} e^{x \left (\frac {1}{\log (x)}-4 (x+2)\right )}\right )}{\log (x)}dx-x\right )\) |
Int[(-Log[x]^2 + E^(E^((x + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x]) + (x + (-8*x - 4*x^2 - 4*Log[4])*Log[x])/Log[x])*(4 - 4*Log[x] + (32 + 32*x)*L og[x]^2))/(4*Log[x]^2),x]
3.23.51.3.1 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]]
Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {x}{4}-{\mathrm e}^{\frac {{\mathrm e}^{-\frac {x \left (4 x \ln \left (x \right )+8 \ln \left (x \right )-1\right )}{\ln \left (x \right )}}}{256}}\) | \(29\) |
parallelrisch | \(-\frac {x}{4}-{\mathrm e}^{{\mathrm e}^{\frac {\left (-8 \ln \left (2\right )-4 x^{2}-8 x \right ) \ln \left (x \right )+x}{\ln \left (x \right )}}}\) | \(32\) |
int(1/4*(((32*x+32)*ln(x)^2-4*ln(x)+4)*exp(((-8*ln(2)-4*x^2-8*x)*ln(x)+x)/ ln(x))*exp(exp(((-8*ln(2)-4*x^2-8*x)*ln(x)+x)/ln(x)))-ln(x)^2)/ln(x)^2,x,m ethod=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (28) = 56\).
Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.08 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-\frac {1}{4} \, {\left (x e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )} + 4 \, e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )}\right )} e^{\left (\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )} \]
integrate(1/4*(((32*x+32)*log(x)^2-4*log(x)+4)*exp(((-8*log(2)-4*x^2-8*x)* log(x)+x)/log(x))*exp(exp(((-8*log(2)-4*x^2-8*x)*log(x)+x)/log(x)))-log(x) ^2)/log(x)^2,x, algorithm=\
-1/4*(x*e^(-(4*(x^2 + 2*x + 2*log(2))*log(x) - x)/log(x)) + 4*e^(-(4*(x^2 + 2*x + 2*log(2))*log(x) - e^(-(4*(x^2 + 2*x + 2*log(2))*log(x) - x)/log(x ))*log(x) - x)/log(x)))*e^((4*(x^2 + 2*x + 2*log(2))*log(x) - x)/log(x))
Time = 0.47 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=- \frac {x}{4} - e^{e^{\frac {x + \left (- 4 x^{2} - 8 x - 8 \log {\left (2 \right )}\right ) \log {\left (x \right )}}{\log {\left (x \right )}}}} \]
integrate(1/4*(((32*x+32)*ln(x)**2-4*ln(x)+4)*exp(((-8*ln(2)-4*x**2-8*x)*l n(x)+x)/ln(x))*exp(exp(((-8*ln(2)-4*x**2-8*x)*ln(x)+x)/ln(x)))-ln(x)**2)/l n(x)**2,x)
Time = 0.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-\frac {1}{4} \, x - e^{\left (\frac {1}{256} \, e^{\left (-4 \, x^{2} - 8 \, x + \frac {x}{\log \left (x\right )}\right )}\right )} \]
integrate(1/4*(((32*x+32)*log(x)^2-4*log(x)+4)*exp(((-8*log(2)-4*x^2-8*x)* log(x)+x)/log(x))*exp(exp(((-8*log(2)-4*x^2-8*x)*log(x)+x)/log(x)))-log(x) ^2)/log(x)^2,x, algorithm=\
\[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=\int { \frac {4 \, {\left (8 \, {\left (x + 1\right )} \log \left (x\right )^{2} - \log \left (x\right ) + 1\right )} e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )} + e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) - x}{\log \left (x\right )}\right )}\right )} - \log \left (x\right )^{2}}{4 \, \log \left (x\right )^{2}} \,d x } \]
integrate(1/4*(((32*x+32)*log(x)^2-4*log(x)+4)*exp(((-8*log(2)-4*x^2-8*x)* log(x)+x)/log(x))*exp(exp(((-8*log(2)-4*x^2-8*x)*log(x)+x)/log(x)))-log(x) ^2)/log(x)^2,x, algorithm=\
Time = 11.66 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {-\log ^2(x)+e^{e^{\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}}+\frac {x+\left (-8 x-4 x^2-4 \log (4)\right ) \log (x)}{\log (x)}} \left (4-4 \log (x)+(32+32 x) \log ^2(x)\right )}{4 \log ^2(x)} \, dx=-\frac {x}{4}-{\mathrm {e}}^{\frac {{\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^{\frac {x}{\ln \left (x\right )}}\,{\mathrm {e}}^{-4\,x^2}}{256}} \]