Integrand size = 168, antiderivative size = 26 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=2-3 x+\frac {\log (5)}{\log \left (\log \left (2+\left (4+e^2-x\right )^2+\log (4)\right )\right )} \] Output:
ln(5)/ln(ln(2*ln(2)+(4-x+exp(2))^2+2))-3*x+2
Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=-3 x+\frac {\log (5)}{\log \left (\log \left (18+e^4-2 e^2 (-4+x)-8 x+x^2+\log (4)\right )\right )} \] Input:
Integrate[((8 + 2*E^2 - 2*x)*Log[5] + (-54 - 3*E^4 + 24*x - 3*x^2 + E^2*(- 24 + 6*x) - 3*Log[4])*Log[18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4]]*L og[Log[18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4]]]^2)/((18 + E^4 + E^2 *(8 - 2*x) - 8*x + x^2 + Log[4])*Log[18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4]]*Log[Log[18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4]]]^2),x]
Output:
-3*x + Log[5]/Log[Log[18 + E^4 - 2*E^2*(-4 + x) - 8*x + x^2 + Log[4]]]
Time = 2.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {7292, 7239, 2009}
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.
| \(\displaystyle \int \frac {\left (-3 x^2+24 x+e^2 (6 x-24)-3 e^4-54-3 \log (4)\right ) \log \left (x^2-8 x+e^2 (8-2 x)+e^4+18+\log (4)\right ) \log ^2\left (\log \left (x^2-8 x+e^2 (8-2 x)+e^4+18+\log (4)\right )\right )+\left (-2 x+2 e^2+8\right ) \log (5)}{\left (x^2-8 x+e^2 (8-2 x)+e^4+18+\log (4)\right ) \log \left (x^2-8 x+e^2 (8-2 x)+e^4+18+\log (4)\right ) \log ^2\left (\log \left (x^2-8 x+e^2 (8-2 x)+e^4+18+\log (4)\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (-3 x^2+24 x+e^2 (6 x-24)-3 e^4-54-3 \log (4)\right ) \log \left (x^2-8 x+e^2 (8-2 x)+e^4+18+\log (4)\right ) \log ^2\left (\log \left (x^2-8 x+e^2 (8-2 x)+e^4+18+\log (4)\right )\right )+\left (-2 x+2 e^2+8\right ) \log (5)}{\left (x^2-2 \left (4+e^2\right ) x+e^4+8 e^2+18+\log (4)\right ) \log \left (x^2-2 \left (4+e^2\right ) x+e^4+8 e^2+18+\log (4)\right ) \log ^2\left (\log \left (x^2-2 \left (4+e^2\right ) x+e^4+8 e^2+18+\log (4)\right )\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (\frac {2 \left (-x+e^2+4\right ) \log (5)}{\left (x^2-2 \left (4+e^2\right ) x+e^4+8 e^2+18+\log (4)\right ) \log \left (x^2-2 \left (4+e^2\right ) x+e^4+8 e^2+18+\log (4)\right ) \log ^2\left (\log \left (x^2-2 \left (4+e^2\right ) x+e^4+8 e^2+18+\log (4)\right )\right )}-3\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\log (5)}{\log \left (\log \left (x^2-2 \left (4+e^2\right ) x+e^4+8 e^2+18+\log (4)\right )\right )}-3 x\) |
Input:
Int[((8 + 2*E^2 - 2*x)*Log[5] + (-54 - 3*E^4 + 24*x - 3*x^2 + E^2*(-24 + 6 *x) - 3*Log[4])*Log[18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4]]*Log[Log [18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4]]]^2)/((18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4])*Log[18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[ 4]]*Log[Log[18 + E^4 + E^2*(8 - 2*x) - 8*x + x^2 + Log[4]]]^2),x]
Output:
-3*x + Log[5]/Log[Log[18 + 8*E^2 + E^4 - 2*(4 + E^2)*x + x^2 + Log[4]]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.70 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
| method | result | size |
| risch | \(-3 x +\frac {\ln \left (5\right )}{\ln \left (\ln \left (2 \ln \left (2\right )+{\mathrm e}^{4}+\left (-2 x +8\right ) {\mathrm e}^{2}+x^{2}-8 x +18\right )\right )}\) | \(34\) |
| default | \(-3 x +\frac {\ln \left (5\right )}{\ln \left (\ln \left (-2 \,{\mathrm e}^{2} x +x^{2}+8 \,{\mathrm e}^{2}+2 \ln \left (2\right )+{\mathrm e}^{4}-8 x +18\right )\right )}\) | \(37\) |
| parts | \(-3 x +\frac {\ln \left (5\right )}{\ln \left (\ln \left (-2 \,{\mathrm e}^{2} x +x^{2}+8 \,{\mathrm e}^{2}+2 \ln \left (2\right )+{\mathrm e}^{4}-8 x +18\right )\right )}\) | \(37\) |
| parallelrisch | \(-\frac {12 \,{\mathrm e}^{2} \ln \left (\ln \left (2 \ln \left (2\right )+{\mathrm e}^{4}+\left (-2 x +8\right ) {\mathrm e}^{2}+x^{2}-8 x +18\right )\right )+3 x \ln \left (\ln \left (2 \ln \left (2\right )+{\mathrm e}^{4}+\left (-2 x +8\right ) {\mathrm e}^{2}+x^{2}-8 x +18\right )\right )-\ln \left (5\right )+48 \ln \left (\ln \left (2 \ln \left (2\right )+{\mathrm e}^{4}+\left (-2 x +8\right ) {\mathrm e}^{2}+x^{2}-8 x +18\right )\right )}{\ln \left (\ln \left (2 \ln \left (2\right )+{\mathrm e}^{4}+\left (-2 x +8\right ) {\mathrm e}^{2}+x^{2}-8 x +18\right )\right )}\) | \(123\) |
Input:
int(((-6*ln(2)-3*exp(2)^2+(6*x-24)*exp(2)-3*x^2+24*x-54)*ln(2*ln(2)+exp(2) ^2+(-2*x+8)*exp(2)+x^2-8*x+18)*ln(ln(2*ln(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2- 8*x+18))^2+(2*exp(2)-2*x+8)*ln(5))/(2*ln(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8 *x+18)/ln(2*ln(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)/ln(ln(2*ln(2)+exp(2 )^2+(-2*x+8)*exp(2)+x^2-8*x+18))^2,x,method=_RETURNVERBOSE)
Output:
-3*x+ln(5)/ln(ln(2*ln(2)+exp(4)+(-2*x+8)*exp(2)+x^2-8*x+18))
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.23 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=-\frac {3 \, x \log \left (\log \left (x^{2} - 2 \, {\left (x - 4\right )} e^{2} - 8 \, x + e^{4} + 2 \, \log \left (2\right ) + 18\right )\right ) - \log \left (5\right )}{\log \left (\log \left (x^{2} - 2 \, {\left (x - 4\right )} e^{2} - 8 \, x + e^{4} + 2 \, \log \left (2\right ) + 18\right )\right )} \] Input:
integrate(((-6*log(2)-3*exp(2)^2+(6*x-24)*exp(2)-3*x^2+24*x-54)*log(2*log( 2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)*log(log(2*log(2)+exp(2)^2+(-2*x+8) *exp(2)+x^2-8*x+18))^2+(2*exp(2)-2*x+8)*log(5))/(2*log(2)+exp(2)^2+(-2*x+8 )*exp(2)+x^2-8*x+18)/log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)/log (log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18))^2,x, algorithm="fricas ")
Output:
-(3*x*log(log(x^2 - 2*(x - 4)*e^2 - 8*x + e^4 + 2*log(2) + 18)) - log(5))/ log(log(x^2 - 2*(x - 4)*e^2 - 8*x + e^4 + 2*log(2) + 18))
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=- 3 x + \frac {\log {\left (5 \right )}}{\log {\left (\log {\left (x^{2} - 8 x + \left (8 - 2 x\right ) e^{2} + 2 \log {\left (2 \right )} + 18 + e^{4} \right )} \right )}} \] Input:
integrate(((-6*ln(2)-3*exp(2)**2+(6*x-24)*exp(2)-3*x**2+24*x-54)*ln(2*ln(2 )+exp(2)**2+(-2*x+8)*exp(2)+x**2-8*x+18)*ln(ln(2*ln(2)+exp(2)**2+(-2*x+8)* exp(2)+x**2-8*x+18))**2+(2*exp(2)-2*x+8)*ln(5))/(2*ln(2)+exp(2)**2+(-2*x+8 )*exp(2)+x**2-8*x+18)/ln(2*ln(2)+exp(2)**2+(-2*x+8)*exp(2)+x**2-8*x+18)/ln (ln(2*ln(2)+exp(2)**2+(-2*x+8)*exp(2)+x**2-8*x+18))**2,x)
Output:
-3*x + log(5)/log(log(x**2 - 8*x + (8 - 2*x)*exp(2) + 2*log(2) + 18 + exp( 4)))
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=-\frac {3 \, x \log \left (\log \left (x^{2} - 2 \, x {\left (e^{2} + 4\right )} + e^{4} + 8 \, e^{2} + 2 \, \log \left (2\right ) + 18\right )\right ) - \log \left (5\right )}{\log \left (\log \left (x^{2} - 2 \, x {\left (e^{2} + 4\right )} + e^{4} + 8 \, e^{2} + 2 \, \log \left (2\right ) + 18\right )\right )} \] Input:
integrate(((-6*log(2)-3*exp(2)^2+(6*x-24)*exp(2)-3*x^2+24*x-54)*log(2*log( 2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)*log(log(2*log(2)+exp(2)^2+(-2*x+8) *exp(2)+x^2-8*x+18))^2+(2*exp(2)-2*x+8)*log(5))/(2*log(2)+exp(2)^2+(-2*x+8 )*exp(2)+x^2-8*x+18)/log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)/log (log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18))^2,x, algorithm="maxima ")
Output:
-(3*x*log(log(x^2 - 2*x*(e^2 + 4) + e^4 + 8*e^2 + 2*log(2) + 18)) - log(5) )/log(log(x^2 - 2*x*(e^2 + 4) + e^4 + 8*e^2 + 2*log(2) + 18))
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (27) = 54\).
Time = 1.52 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.38 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=-\frac {3 \, x \log \left (\log \left (x^{2} - 2 \, x e^{2} - 8 \, x + e^{4} + 8 \, e^{2} + 2 \, \log \left (2\right ) + 18\right )\right ) - \log \left (5\right )}{\log \left (\log \left (x^{2} - 2 \, x e^{2} - 8 \, x + e^{4} + 8 \, e^{2} + 2 \, \log \left (2\right ) + 18\right )\right )} \] Input:
integrate(((-6*log(2)-3*exp(2)^2+(6*x-24)*exp(2)-3*x^2+24*x-54)*log(2*log( 2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)*log(log(2*log(2)+exp(2)^2+(-2*x+8) *exp(2)+x^2-8*x+18))^2+(2*exp(2)-2*x+8)*log(5))/(2*log(2)+exp(2)^2+(-2*x+8 )*exp(2)+x^2-8*x+18)/log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)/log (log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18))^2,x, algorithm="giac")
Output:
-(3*x*log(log(x^2 - 2*x*e^2 - 8*x + e^4 + 8*e^2 + 2*log(2) + 18)) - log(5) )/log(log(x^2 - 2*x*e^2 - 8*x + e^4 + 8*e^2 + 2*log(2) + 18))
Time = 3.76 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=\frac {\ln \left (5\right )}{\ln \left (\ln \left (8\,{\mathrm {e}}^2-8\,x+{\mathrm {e}}^4+2\,\ln \left (2\right )-2\,x\,{\mathrm {e}}^2+x^2+18\right )\right )}-3\,x \] Input:
int((log(5)*(2*exp(2) - 2*x + 8) - log(log(exp(4) - 8*x + 2*log(2) + x^2 - exp(2)*(2*x - 8) + 18))^2*log(exp(4) - 8*x + 2*log(2) + x^2 - exp(2)*(2*x - 8) + 18)*(3*exp(4) - 24*x + 6*log(2) + 3*x^2 - exp(2)*(6*x - 24) + 54)) /(log(log(exp(4) - 8*x + 2*log(2) + x^2 - exp(2)*(2*x - 8) + 18))^2*log(ex p(4) - 8*x + 2*log(2) + x^2 - exp(2)*(2*x - 8) + 18)*(exp(4) - 8*x + 2*log (2) + x^2 - exp(2)*(2*x - 8) + 18)),x)
Output:
log(5)/log(log(8*exp(2) - 8*x + exp(4) + 2*log(2) - 2*x*exp(2) + x^2 + 18) ) - 3*x
Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.50 \[ \int \frac {\left (8+2 e^2-2 x\right ) \log (5)+\left (-54-3 e^4+24 x-3 x^2+e^2 (-24+6 x)-3 \log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )}{\left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right ) \log ^2\left (\log \left (18+e^4+e^2 (8-2 x)-8 x+x^2+\log (4)\right )\right )} \, dx=\frac {-3 \,\mathrm {log}\left (\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )+e^{4}-2 e^{2} x +8 e^{2}+x^{2}-8 x +18\right )\right ) x +\mathrm {log}\left (5\right )}{\mathrm {log}\left (\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )+e^{4}-2 e^{2} x +8 e^{2}+x^{2}-8 x +18\right )\right )} \] Input:
int(((-6*log(2)-3*exp(2)^2+(6*x-24)*exp(2)-3*x^2+24*x-54)*log(2*log(2)+exp (2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)*log(log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2 )+x^2-8*x+18))^2+(2*exp(2)-2*x+8)*log(5))/(2*log(2)+exp(2)^2+(-2*x+8)*exp( 2)+x^2-8*x+18)/log(2*log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18)/log(log(2 *log(2)+exp(2)^2+(-2*x+8)*exp(2)+x^2-8*x+18))^2,x)
Output:
( - 3*log(log(2*log(2) + e**4 - 2*e**2*x + 8*e**2 + x**2 - 8*x + 18))*x + log(5))/log(log(2*log(2) + e**4 - 2*e**2*x + 8*e**2 + x**2 - 8*x + 18))