Integrand size = 96, antiderivative size = 22 \[ \int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{e^4+4 e^6-x-\log (5)+\log (x)} \]
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{e^4+4 e^6-x+\log \left (\frac {x}{5}\right )} \]
Integrate[(-1 + E^4 + 4*E^6 - Log[5] + Log[x])/(E^8 + 16*E^12 + E^4*(8*E^6 - 2*x) - 8*E^6*x + x^2 + (-2*E^4 - 8*E^6 + 2*x)*Log[5] + Log[5]^2 + (2*E^ 4 + 8*E^6 - 2*x - 2*Log[5])*Log[x] + 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.
\(\displaystyle \int \frac {\log (x)+4 e^6+e^4-1-\log (5)}{x^2-8 e^6 x+e^4 \left (8 e^6-2 x\right )+\log ^2(x)+\left (-2 x+8 e^6+2 e^4-2 \log (5)\right ) \log (x)+\left (2 x-8 e^6-2 e^4\right ) \log (5)+16 e^{12}+e^8+\log ^2(5)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )-1}{\left (-x+\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )\right )^2}dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle 5 \int -\frac {-\log \left (\frac {x}{5}\right )-4 e^6-e^4+1}{\left (-x+\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )\right )^2}d\frac {x}{5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {-\log \left (\frac {x}{5}\right )-4 e^6-e^4+1}{\left (-x+\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )\right )^2}d\frac {x}{5}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -5 \int \frac {-\log \left (\frac {x}{5}\right )-e^4 \left (1+4 e^2\right )+1}{\left (-x+\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )\right )^2}d\frac {x}{5}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (\frac {1-x}{\left (-x+\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )\right )^2}+\frac {1}{x-\log \left (\frac {x}{5}\right )-e^4 \left (1+4 e^2\right )}\right )d\frac {x}{5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \left (-\int \frac {1}{x-\log \left (\frac {x}{5}\right )-e^4 \left (1+4 e^2\right )}d\frac {x}{5}-\int \frac {1}{\left (-x+\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )\right )^2}d\frac {x}{5}+5 \int \frac {x}{5 \left (-x+\log \left (\frac {x}{5}\right )+e^4 \left (1+4 e^2\right )\right )^2}d\frac {x}{5}\right )\) |
Int[(-1 + E^4 + 4*E^6 - Log[5] + Log[x])/(E^8 + 16*E^12 + E^4*(8*E^6 - 2*x ) - 8*E^6*x + x^2 + (-2*E^4 - 8*E^6 + 2*x)*Log[5] + Log[5]^2 + (2*E^4 + 8* E^6 - 2*x - 2*Log[5])*Log[x] + Log[x]^2),x]
3.27.98.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, 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 = 1.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {x}{4 \,{\mathrm e}^{6}+{\mathrm e}^{4}-\ln \left (5\right )-x +\ln \left (x \right )}\) | \(21\) |
default | \(\frac {x}{4 \,{\mathrm e}^{6}+{\mathrm e}^{4}-\ln \left (5\right )-x +\ln \left (x \right )}\) | \(23\) |
parallelrisch | \(\frac {x}{4 \,{\mathrm e}^{6}+{\mathrm e}^{4}-\ln \left (5\right )-x +\ln \left (x \right )}\) | \(23\) |
norman | \(\frac {\ln \left (x \right )+4 \,{\mathrm e}^{6}+{\mathrm e}^{4}-\ln \left (5\right )}{4 \,{\mathrm e}^{6}+{\mathrm e}^{4}-\ln \left (5\right )-x +\ln \left (x \right )}\) | \(37\) |
int((ln(x)-ln(5)+exp(4)+4*exp(3)^2-1)/(ln(x)^2+(-2*ln(5)+2*exp(4)+8*exp(3) ^2-2*x)*ln(x)+ln(5)^2+(-2*exp(4)-8*exp(3)^2+2*x)*ln(5)+exp(4)^2+(8*exp(3)^ 2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx=-\frac {x}{x - 4 \, e^{6} - e^{4} + \log \left (5\right ) - \log \left (x\right )} \]
integrate((log(x)-log(5)+exp(4)+4*exp(3)^2-1)/(log(x)^2+(-2*log(5)+2*exp(4 )+8*exp(3)^2-2*x)*log(x)+log(5)^2+(-2*exp(4)-8*exp(3)^2+2*x)*log(5)+exp(4) ^2+(8*exp(3)^2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x, algorithm=\
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{- x + \log {\left (x \right )} - \log {\left (5 \right )} + e^{4} + 4 e^{6}} \]
integrate((ln(x)-ln(5)+exp(4)+4*exp(3)**2-1)/(ln(x)**2+(-2*ln(5)+2*exp(4)+ 8*exp(3)**2-2*x)*ln(x)+ln(5)**2+(-2*exp(4)-8*exp(3)**2+2*x)*ln(5)+exp(4)** 2+(8*exp(3)**2-2*x)*exp(4)+16*exp(3)**4-8*x*exp(3)**2+x**2),x)
Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx=-\frac {x}{x - 4 \, e^{6} - e^{4} + \log \left (5\right ) - \log \left (x\right )} \]
integrate((log(x)-log(5)+exp(4)+4*exp(3)^2-1)/(log(x)^2+(-2*log(5)+2*exp(4 )+8*exp(3)^2-2*x)*log(x)+log(5)^2+(-2*exp(4)-8*exp(3)^2+2*x)*log(5)+exp(4) ^2+(8*exp(3)^2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx=-\frac {x}{x - 4 \, e^{6} - e^{4} + \log \left (5\right ) - \log \left (x\right )} \]
integrate((log(x)-log(5)+exp(4)+4*exp(3)^2-1)/(log(x)^2+(-2*log(5)+2*exp(4 )+8*exp(3)^2-2*x)*log(x)+log(5)^2+(-2*exp(4)-8*exp(3)^2+2*x)*log(5)+exp(4) ^2+(8*exp(3)^2-2*x)*exp(4)+16*exp(3)^4-8*x*exp(3)^2+x^2),x, algorithm=\
Time = 9.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-1+e^4+4 e^6-\log (5)+\log (x)}{e^8+16 e^{12}+e^4 \left (8 e^6-2 x\right )-8 e^6 x+x^2+\left (-2 e^4-8 e^6+2 x\right ) \log (5)+\log ^2(5)+\left (2 e^4+8 e^6-2 x-2 \log (5)\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{\ln \left (\frac {x}{5}\right )-x+{\mathrm {e}}^4+4\,{\mathrm {e}}^6} \]