Integrand size = 108, antiderivative size = 26 \[ \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx=2+\frac {9 \log \left (e^x-\log (\log (4))\right )}{x^2 \left (-e^3+x\right )} \]
Time = 0.64 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx=-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2} \]
Integrate[(E^x*(9*E^3*x - 9*x^2) + (E^x*(-18*E^3 + 27*x) + (18*E^3 - 27*x) *Log[Log[4]])*Log[E^x - Log[Log[4]]])/(E^x*(-(E^6*x^3) + 2*E^3*x^4 - x^5) + (E^6*x^3 - 2*E^3*x^4 + x^5)*Log[Log[4]]),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.
\(\displaystyle \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (27 x-18 e^3\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-x^5+2 e^3 x^4-e^6 x^3\right )+\left (x^5-2 e^3 x^4+e^6 x^3\right ) \log (\log (4))} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {9 \left (\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )-\frac {e^x \left (e^3-x\right ) x}{e^x-\log (\log (4))}\right )}{\left (e^3-x\right )^2 x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 9 \int -\frac {\frac {e^x \left (e^3-x\right ) x}{e^x-\log (\log (4))}-\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -9 \int \frac {\frac {e^x \left (e^3-x\right ) x}{e^x-\log (\log (4))}-\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -9 \int \left (\frac {-x^2+3 \log \left (e^x-\log (\log (4))\right ) x+e^3 x-2 e^3 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3}+\frac {\log (\log (4))}{\left (e^3-x\right ) x^2 \left (e^x-\log (\log (4))\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -9 \left (\frac {\log (\log (4)) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )}dx}{e^3}-\frac {\int \frac {e^x}{x^2 \left (e^x-\log (\log (4))\right )}dx}{e^3}+\frac {\log (\log (4)) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )}dx}{e^6}-\frac {\int \frac {e^x}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )}dx}{e^6}+\frac {\log (\log (4)) \int \frac {1}{x \left (e^x-\log (\log (4))\right )}dx}{e^6}-\frac {\int \frac {e^x}{x \left (e^x-\log (\log (4))\right )}dx}{e^6}+\frac {\log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}-\frac {1}{e^3 x}-\frac {\log \left (e^3-x\right )}{e^6}+\frac {\log (x)}{e^6}\right )\) |
Int[(E^x*(9*E^3*x - 9*x^2) + (E^x*(-18*E^3 + 27*x) + (18*E^3 - 27*x)*Log[L og[4]])*Log[E^x - Log[Log[4]]])/(E^x*(-(E^6*x^3) + 2*E^3*x^4 - x^5) + (E^6 *x^3 - 2*E^3*x^4 + x^5)*Log[Log[4]]),x]
3.22.98.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.91 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\frac {9 \ln \left (-\ln \left (2 \ln \left (2\right )\right )+{\mathrm e}^{x}\right )}{x^{2} \left (-x +{\mathrm e}^{3}\right )}\) | \(25\) |
risch | \(-\frac {9 \ln \left (-\ln \left (2\right )-\ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )}{x^{2} \left (-x +{\mathrm e}^{3}\right )}\) | \(27\) |
int((((18*exp(3)-27*x)*ln(2*ln(2))+(-18*exp(3)+27*x)*exp(x))*ln(-ln(2*ln(2 ))+exp(x))+(9*x*exp(3)-9*x^2)*exp(x))/((x^3*exp(3)^2-2*x^4*exp(3)+x^5)*ln( 2*ln(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx=\frac {9 \, \log \left (e^{x} - \log \left (2 \, \log \left (2\right )\right )\right )}{x^{3} - x^{2} e^{3}} \]
integrate((((18*exp(3)-27*x)*log(2*log(2))+(-18*exp(3)+27*x)*exp(x))*log(- log(2*log(2))+exp(x))+(9*x*exp(3)-9*x^2)*exp(x))/((x^3*exp(3)^2-2*x^4*exp( 3)+x^5)*log(2*log(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x, algorith m=\
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx=\frac {9 \log {\left (e^{x} - \log {\left (2 \log {\left (2 \right )} \right )} \right )}}{x^{3} - x^{2} e^{3}} \]
integrate((((18*exp(3)-27*x)*ln(2*ln(2))+(-18*exp(3)+27*x)*exp(x))*ln(-ln( 2*ln(2))+exp(x))+(9*x*exp(3)-9*x**2)*exp(x))/((x**3*exp(3)**2-2*x**4*exp(3 )+x**5)*ln(2*ln(2))+(-x**3*exp(3)**2+2*x**4*exp(3)-x**5)*exp(x)),x)
Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx=\frac {9 \, \log \left (e^{x} - \log \left (2\right ) - \log \left (\log \left (2\right )\right )\right )}{x^{3} - x^{2} e^{3}} \]
integrate((((18*exp(3)-27*x)*log(2*log(2))+(-18*exp(3)+27*x)*exp(x))*log(- log(2*log(2))+exp(x))+(9*x*exp(3)-9*x^2)*exp(x))/((x^3*exp(3)^2-2*x^4*exp( 3)+x^5)*log(2*log(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x, algorith m=\
Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx=\frac {9 \, \log \left (e^{x} - \log \left (2\right ) - \log \left (\log \left (2\right )\right )\right )}{x^{3} - x^{2} e^{3}} \]
integrate((((18*exp(3)-27*x)*log(2*log(2))+(-18*exp(3)+27*x)*exp(x))*log(- log(2*log(2))+exp(x))+(9*x*exp(3)-9*x^2)*exp(x))/((x^3*exp(3)^2-2*x^4*exp( 3)+x^5)*log(2*log(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x, algorith m=\
Time = 0.85 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx=-\frac {9\,\ln \left ({\mathrm {e}}^x-\ln \left (2\,\ln \left (2\right )\right )\right )}{x^2\,{\mathrm {e}}^3-x^3} \]