Integrand size = 94, antiderivative size = 27 \[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=1+\frac {\log (25)}{4}-\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right ) \] Output:
1+1/2*ln(5)-ln(ln(ln((exp(1)^2+x+5)^2/(1+x)^2)))
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=-\log \left (\log \left (\log \left (\frac {\left (5+e^2+x\right )^2}{(1+x)^2}\right )\right )\right ) \] Input:
Integrate[(8 + 2*E^2)/((5 + 6*x + x^2 + E^2*(1 + x))*Log[(25 + E^4 + 10*x + x^2 + E^2*(10 + 2*x))/(1 + 2*x + x^2)]*Log[Log[(25 + E^4 + 10*x + x^2 + E^2*(10 + 2*x))/(1 + 2*x + x^2)]]),x]
Output:
-Log[Log[Log[(5 + E^2 + x)^2/(1 + x)^2]]]
Time = 0.58 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {27, 7239, 7235}
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.
\(\displaystyle \int \frac {8+2 e^2}{\left (x^2+6 x+e^2 (x+1)+5\right ) \log \left (\frac {x^2+10 x+e^2 (2 x+10)+e^4+25}{x^2+2 x+1}\right ) \log \left (\log \left (\frac {x^2+10 x+e^2 (2 x+10)+e^4+25}{x^2+2 x+1}\right )\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (4+e^2\right ) \int \frac {1}{\left (x^2+6 x+e^2 (x+1)+5\right ) \log \left (\frac {x^2+10 x+2 e^2 (x+5)+e^4+25}{x^2+2 x+1}\right ) \log \left (\log \left (\frac {x^2+10 x+2 e^2 (x+5)+e^4+25}{x^2+2 x+1}\right )\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle 2 \left (4+e^2\right ) \int \frac {1}{(x+1) \left (x+e^2+5\right ) \log \left (\frac {\left (x+e^2+5\right )^2}{(x+1)^2}\right ) \log \left (\log \left (\frac {\left (x+e^2+5\right )^2}{(x+1)^2}\right )\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle -\log \left (\log \left (\log \left (\frac {\left (x+e^2+5\right )^2}{(x+1)^2}\right )\right )\right )\) |
Input:
Int[(8 + 2*E^2)/((5 + 6*x + x^2 + E^2*(1 + x))*Log[(25 + E^4 + 10*x + x^2 + E^2*(10 + 2*x))/(1 + 2*x + x^2)]*Log[Log[(25 + E^4 + 10*x + x^2 + E^2*(1 0 + 2*x))/(1 + 2*x + x^2)]]),x]
Output:
-Log[Log[Log[(5 + E^2 + x)^2/(1 + x)^2]]]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.44 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44
method | result | size |
norman | \(-\ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4}+\left (2 x +10\right ) {\mathrm e}^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )\) | \(39\) |
default | \(-\frac {\left (2 \,{\mathrm e}^{2}+8\right ) \ln \left (\ln \left (\ln \left (\frac {2 \,{\mathrm e}^{2} x +x^{2}+10 \,{\mathrm e}^{2}+{\mathrm e}^{4}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )}{2 \left (4+{\mathrm e}^{2}\right )}\) | \(58\) |
parallelrisch | \(\frac {\left (2 \,{\mathrm e}^{2}+8\right ) \left (-25 \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4}+\left (2 x +10\right ) {\mathrm e}^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )-{\mathrm e}^{4} \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4}+\left (2 x +10\right ) {\mathrm e}^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )-10 \,{\mathrm e}^{2} \ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4}+\left (2 x +10\right ) {\mathrm e}^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right )\right )}{2 \left (4+{\mathrm e}^{2}\right ) \left ({\mathrm e}^{2}+5\right )^{2}}\) | \(150\) |
Input:
int((2*exp(1)^2+8)/((1+x)*exp(1)^2+x^2+6*x+5)/ln((exp(1)^4+(2*x+10)*exp(1) ^2+x^2+10*x+25)/(x^2+2*x+1))/ln(ln((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25 )/(x^2+2*x+1))),x,method=_RETURNVERBOSE)
Output:
-ln(ln(ln((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))))
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=-\log \left (\log \left (\log \left (\frac {x^{2} + 2 \, {\left (x + 5\right )} e^{2} + 10 \, x + e^{4} + 25}{x^{2} + 2 \, x + 1}\right )\right )\right ) \] Input:
integrate((2*exp(1)^2+8)/((1+x)*exp(1)^2+x^2+6*x+5)/log((exp(1)^4+(2*x+10) *exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))/log(log((exp(1)^4+(2*x+10)*exp(1)^2+x^ 2+10*x+25)/(x^2+2*x+1))),x, algorithm="fricas")
Output:
-log(log(log((x^2 + 2*(x + 5)*e^2 + 10*x + e^4 + 25)/(x^2 + 2*x + 1))))
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=- \log {\left (\log {\left (\log {\left (\frac {x^{2} + 10 x + \left (2 x + 10\right ) e^{2} + 25 + e^{4}}{x^{2} + 2 x + 1} \right )} \right )} \right )} \] Input:
integrate((2*exp(1)**2+8)/((1+x)*exp(1)**2+x**2+6*x+5)/ln((exp(1)**4+(2*x+ 10)*exp(1)**2+x**2+10*x+25)/(x**2+2*x+1))/ln(ln((exp(1)**4+(2*x+10)*exp(1) **2+x**2+10*x+25)/(x**2+2*x+1))),x)
Output:
-log(log(log((x**2 + 10*x + (2*x + 10)*exp(2) + 25 + exp(4))/(x**2 + 2*x + 1))))
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=-\log \left (\log \left (2\right ) + \log \left (\log \left (x + e^{2} + 5\right ) - \log \left (x + 1\right )\right )\right ) \] Input:
integrate((2*exp(1)^2+8)/((1+x)*exp(1)^2+x^2+6*x+5)/log((exp(1)^4+(2*x+10) *exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))/log(log((exp(1)^4+(2*x+10)*exp(1)^2+x^ 2+10*x+25)/(x^2+2*x+1))),x, algorithm="maxima")
Output:
-log(log(2) + log(log(x + e^2 + 5) - log(x + 1)))
\[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=\int { \frac {2 \, {\left (e^{2} + 4\right )}}{{\left (x^{2} + {\left (x + 1\right )} e^{2} + 6 \, x + 5\right )} \log \left (\frac {x^{2} + 2 \, {\left (x + 5\right )} e^{2} + 10 \, x + e^{4} + 25}{x^{2} + 2 \, x + 1}\right ) \log \left (\log \left (\frac {x^{2} + 2 \, {\left (x + 5\right )} e^{2} + 10 \, x + e^{4} + 25}{x^{2} + 2 \, x + 1}\right )\right )} \,d x } \] Input:
integrate((2*exp(1)^2+8)/((1+x)*exp(1)^2+x^2+6*x+5)/log((exp(1)^4+(2*x+10) *exp(1)^2+x^2+10*x+25)/(x^2+2*x+1))/log(log((exp(1)^4+(2*x+10)*exp(1)^2+x^ 2+10*x+25)/(x^2+2*x+1))),x, algorithm="giac")
Output:
integrate(2*(e^2 + 4)/((x^2 + (x + 1)*e^2 + 6*x + 5)*log((x^2 + 2*(x + 5)* e^2 + 10*x + e^4 + 25)/(x^2 + 2*x + 1))*log(log((x^2 + 2*(x + 5)*e^2 + 10* x + e^4 + 25)/(x^2 + 2*x + 1)))), x)
Time = 16.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=-\ln \left (\ln \left (\ln \left (\frac {10\,x+{\mathrm {e}}^4+x^2+{\mathrm {e}}^2\,\left (2\,x+10\right )+25}{x^2+2\,x+1}\right )\right )\right ) \] Input:
int((2*exp(2) + 8)/(log(log((10*x + exp(4) + x^2 + exp(2)*(2*x + 10) + 25) /(2*x + x^2 + 1)))*log((10*x + exp(4) + x^2 + exp(2)*(2*x + 10) + 25)/(2*x + x^2 + 1))*(6*x + exp(2)*(x + 1) + x^2 + 5)),x)
Output:
-log(log(log((10*x + exp(4) + x^2 + exp(2)*(2*x + 10) + 25)/(2*x + x^2 + 1 ))))
Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {8+2 e^2}{\left (5+6 x+x^2+e^2 (1+x)\right ) \log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right ) \log \left (\log \left (\frac {25+e^4+10 x+x^2+e^2 (10+2 x)}{1+2 x+x^2}\right )\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {e^{4}+2 e^{2} x +10 e^{2}+x^{2}+10 x +25}{x^{2}+2 x +1}\right )\right )\right ) \] Input:
int((2*exp(1)^2+8)/((1+x)*exp(1)^2+x^2+6*x+5)/log((exp(1)^4+(2*x+10)*exp(1 )^2+x^2+10*x+25)/(x^2+2*x+1))/log(log((exp(1)^4+(2*x+10)*exp(1)^2+x^2+10*x +25)/(x^2+2*x+1))),x)
Output:
- log(log(log((e**4 + 2*e**2*x + 10*e**2 + x**2 + 10*x + 25)/(x**2 + 2*x + 1))))