Integrand size = 123, antiderivative size = 24 \[ \int \frac {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx=\frac {2 x}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \]
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx=\frac {2 x}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \]
Integrate[(2*x + E^x*(-2 + 4*x) + 2*Log[-3 + E^(2*E^4)])/(-15*E^(3*x) - 45 *E^(2*x)*x - 45*E^x*x^2 - 15*x^3 + (45*E^(2*x) + 90*E^x*x + 45*x^2)*Log[-3 + E^(2*E^4)] + (-45*E^x - 45*x)*Log[-3 + E^(2*E^4)]^2 + 15*Log[-3 + E^(2* E^4)]^3),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.
\(\displaystyle \int \frac {2 x+e^x (4 x-2)+2 \log \left (e^{2 e^4}-3\right )}{-15 x^3-45 e^x x^2+\left (45 x^2+90 e^x x+45 e^{2 x}\right ) \log \left (e^{2 e^4}-3\right )-45 e^{2 x} x-15 e^{3 x}+\left (-45 x-45 e^x\right ) \log ^2\left (e^{2 e^4}-3\right )+15 \log ^3\left (e^{2 e^4}-3\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (-x-e^x (2 x-1)-\log \left (e^{2 e^4}-3\right )\right )}{15 \left (x+e^x-\log \left (e^{2 e^4}-3\right )\right )^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{15} \int \frac {e^x (1-2 x)-x-\log \left (-3+e^{2 e^4}\right )}{\left (x+e^x-\log \left (-3+e^{2 e^4}\right )\right )^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{15} \int \left (\frac {2 x \left (x-\log \left (-3+e^{2 e^4}\right )-1\right )}{\left (x+e^x-\log \left (-3+e^{2 e^4}\right )\right )^3}-\frac {2 x-1}{\left (x+e^x-\log \left (-3+e^{2 e^4}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{15} \left (2 \int \frac {x^2}{\left (x+e^x-\log \left (-3+e^{2 e^4}\right )\right )^3}dx-2 \left (1+\log \left (e^{2 e^4}-3\right )\right ) \int \frac {x}{\left (x+e^x-\log \left (-3+e^{2 e^4}\right )\right )^3}dx+\int \frac {1}{\left (x+e^x-\log \left (-3+e^{2 e^4}\right )\right )^2}dx-2 \int \frac {x}{\left (x+e^x-\log \left (-3+e^{2 e^4}\right )\right )^2}dx\right )\) |
Int[(2*x + E^x*(-2 + 4*x) + 2*Log[-3 + E^(2*E^4)])/(-15*E^(3*x) - 45*E^(2* x)*x - 45*E^x*x^2 - 15*x^3 + (45*E^(2*x) + 90*E^x*x + 45*x^2)*Log[-3 + E^( 2*E^4)] + (-45*E^x - 45*x)*Log[-3 + E^(2*E^4)]^2 + 15*Log[-3 + E^(2*E^4)]^ 3),x]
3.6.68.3.1 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 = 2.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
norman | \(\frac {2 x}{15 {\left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )-x -{\mathrm e}^{x}\right )}^{2}}\) | \(22\) |
risch | \(\frac {2 x}{15 {\left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )-x -{\mathrm e}^{x}\right )}^{2}}\) | \(22\) |
parallelrisch | \(\frac {2 x}{15 \left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )^{2}-2 \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right ) x -2 \,{\mathrm e}^{x} \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )+x^{2}+2 \,{\mathrm e}^{x} x +{\mathrm e}^{2 x}\right )}\) | \(52\) |
int((2*ln(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*ln(exp(exp(4))^2-3)^3+( -45*exp(x)-45*x)*ln(exp(exp(4))^2-3)^2+(45*exp(x)^2+90*exp(x)*x+45*x^2)*ln (exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x^3),x,method =_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx=\frac {2 \, x}{15 \, {\left (x^{2} + 2 \, x e^{x} - 2 \, {\left (x + e^{x}\right )} \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \]
integrate((2*log(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*log(exp(exp(4))^ 2-3)^3+(-45*exp(x)-45*x)*log(exp(exp(4))^2-3)^2+(45*exp(x)^2+90*exp(x)*x+4 5*x^2)*log(exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x^3 ),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50 \[ \int \frac {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx=\frac {2 x}{15 x^{2} - 30 x \log {\left (-3 + e^{2 e^{4}} \right )} + \left (30 x - 30 \log {\left (-3 + e^{2 e^{4}} \right )}\right ) e^{x} + 15 e^{2 x} + 15 \log {\left (-3 + e^{2 e^{4}} \right )}^{2}} \]
integrate((2*ln(exp(exp(4))**2-3)+(4*x-2)*exp(x)+2*x)/(15*ln(exp(exp(4))** 2-3)**3+(-45*exp(x)-45*x)*ln(exp(exp(4))**2-3)**2+(45*exp(x)**2+90*exp(x)* x+45*x**2)*ln(exp(exp(4))**2-3)-15*exp(x)**3-45*x*exp(x)**2-45*exp(x)*x**2 -15*x**3),x)
2*x/(15*x**2 - 30*x*log(-3 + exp(2*exp(4))) + (30*x - 30*log(-3 + exp(2*ex p(4))))*exp(x) + 15*exp(2*x) + 15*log(-3 + exp(2*exp(4)))**2)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (19) = 38\).
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx=\frac {2 \, x}{15 \, {\left (x^{2} + 2 \, {\left (x - \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )\right )} e^{x} - 2 \, x \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \]
integrate((2*log(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*log(exp(exp(4))^ 2-3)^3+(-45*exp(x)-45*x)*log(exp(exp(4))^2-3)^2+(45*exp(x)^2+90*exp(x)*x+4 5*x^2)*log(exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x^3 ),x, algorithm=\
2/15*x/(x^2 + 2*(x - log(e^(2*e^4) - 3))*e^x - 2*x*log(e^(2*e^4) - 3) + lo g(e^(2*e^4) - 3)^2 + e^(2*x))
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (19) = 38\).
Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx=\frac {2 \, x}{15 \, {\left (x^{2} + 2 \, x e^{x} - 2 \, x \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) - 2 \, e^{x} \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \]
integrate((2*log(exp(exp(4))^2-3)+(4*x-2)*exp(x)+2*x)/(15*log(exp(exp(4))^ 2-3)^3+(-45*exp(x)-45*x)*log(exp(exp(4))^2-3)^2+(45*exp(x)^2+90*exp(x)*x+4 5*x^2)*log(exp(exp(4))^2-3)-15*exp(x)^3-45*x*exp(x)^2-45*exp(x)*x^2-15*x^3 ),x, algorithm=\
2/15*x/(x^2 + 2*x*e^x - 2*x*log(e^(2*e^4) - 3) - 2*e^x*log(e^(2*e^4) - 3) + log(e^(2*e^4) - 3)^2 + e^(2*x))
Time = 0.49 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.38 \[ \int \frac {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx=\frac {2\,x}{15\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )+{\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )}^2+2\,x\,{\mathrm {e}}^x-2\,x\,\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )+x^2\right )} \]
int(-(2*x + 2*log(exp(2*exp(4)) - 3) + exp(x)*(4*x - 2))/(15*exp(3*x) + 45 *x*exp(2*x) + 45*x^2*exp(x) + log(exp(2*exp(4)) - 3)^2*(45*x + 45*exp(x)) - log(exp(2*exp(4)) - 3)*(45*exp(2*x) + 90*x*exp(x) + 45*x^2) - 15*log(exp (2*exp(4)) - 3)^3 + 15*x^3),x)