Integrand size = 105, antiderivative size = 26 \[ \int \frac {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx=x \left (-1-4 x^2+\frac {1}{5 x^2 \left (e^x+x\right )}+\log (\log (5))\right ) \]
Time = 0.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx=\frac {1}{5} \left (-20 x^3+\frac {1}{x \left (e^x+x\right )}+5 x (-1+\log (\log (5)))\right ) \]
Integrate[(-2*x - 5*x^4 - 60*x^6 + E^(2*x)*(-5*x^2 - 60*x^4) + E^x*(-1 - x - 10*x^3 - 120*x^5) + (5*E^(2*x)*x^2 + 10*E^x*x^3 + 5*x^4)*Log[Log[5]])/( 5*E^(2*x)*x^2 + 10*E^x*x^3 + 5*x^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.
\(\displaystyle \int \frac {-60 x^6-5 x^4+e^x \left (-120 x^5-10 x^3-x-1\right )+e^{2 x} \left (-60 x^4-5 x^2\right )+\left (5 x^4+10 e^x x^3+5 e^{2 x} x^2\right ) \log (\log (5))-2 x}{5 x^4+10 e^x x^3+5 e^{2 x} x^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-60 x^6-5 x^4+e^x \left (-120 x^5-10 x^3-x-1\right )+e^{2 x} \left (-60 x^4-5 x^2\right )+\left (5 x^4+10 e^x x^3+5 e^{2 x} x^2\right ) \log (\log (5))-2 x}{5 x^2 \left (x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {60 x^6+5 x^4+2 x+5 e^{2 x} \left (12 x^4+x^2\right )+e^x \left (120 x^5+10 x^3+x+1\right )-5 \left (x^4+2 e^x x^3+e^{2 x} x^2\right ) \log (\log (5))}{x^2 \left (x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {60 x^6+5 x^4+2 x+5 e^{2 x} \left (12 x^4+x^2\right )+e^x \left (120 x^5+10 x^3+x+1\right )-5 \left (x^4+2 e^x x^3+e^{2 x} x^2\right ) \log (\log (5))}{x^2 \left (x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (-\frac {x-1}{x \left (x+e^x\right )^2}+5 \left (12 x^2-\log (\log (5))+1\right )+\frac {x+1}{x^2 \left (x+e^x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-\int \frac {1}{x^2 \left (x+e^x\right )}dx+\int \frac {1}{\left (x+e^x\right )^2}dx-\int \frac {1}{x \left (x+e^x\right )^2}dx-\int \frac {1}{x \left (x+e^x\right )}dx-20 x^3-5 x (1-\log (\log (5)))\right )\) |
Int[(-2*x - 5*x^4 - 60*x^6 + E^(2*x)*(-5*x^2 - 60*x^4) + E^x*(-1 - x - 10* x^3 - 120*x^5) + (5*E^(2*x)*x^2 + 10*E^x*x^3 + 5*x^4)*Log[Log[5]])/(5*E^(2 *x)*x^2 + 10*E^x*x^3 + 5*x^4),x]
3.11.65.3.1 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]]
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-4 x^{3}+x \ln \left (\ln \left (5\right )\right )-x +\frac {1}{5 \left ({\mathrm e}^{x}+x \right ) x}\) | \(26\) |
norman | \(\frac {\frac {1}{5}+\left (-1+\ln \left (\ln \left (5\right )\right )\right ) x^{3}+\left (-1+\ln \left (\ln \left (5\right )\right )\right ) x^{2} {\mathrm e}^{x}-4 x^{5}-4 \,{\mathrm e}^{x} x^{4}}{\left ({\mathrm e}^{x}+x \right ) x}\) | \(45\) |
parallelrisch | \(\frac {-20 \,{\mathrm e}^{x} x^{4}-20 x^{5}+5 \ln \left (\ln \left (5\right )\right ) x^{2} {\mathrm e}^{x}+5 \ln \left (\ln \left (5\right )\right ) x^{3}-5 \,{\mathrm e}^{x} x^{2}-5 x^{3}+1}{5 \left ({\mathrm e}^{x}+x \right ) x}\) | \(56\) |
int(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*ln(ln(5))+(-60*x^4-5*x^2)*exp(x) ^2+(-120*x^5-10*x^3-x-1)*exp(x)-60*x^6-5*x^4-2*x)/(5*exp(x)^2*x^2+10*exp(x )*x^3+5*x^4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx=-\frac {20 \, x^{5} + 5 \, x^{3} + 5 \, {\left (4 \, x^{4} + x^{2}\right )} e^{x} - 5 \, {\left (x^{3} + x^{2} e^{x}\right )} \log \left (\log \left (5\right )\right ) - 1}{5 \, {\left (x^{2} + x e^{x}\right )}} \]
integrate(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*log(log(5))+(-60*x^4-5*x^2 )*exp(x)^2+(-120*x^5-10*x^3-x-1)*exp(x)-60*x^6-5*x^4-2*x)/(5*exp(x)^2*x^2+ 10*exp(x)*x^3+5*x^4),x, algorithm=\
-1/5*(20*x^5 + 5*x^3 + 5*(4*x^4 + x^2)*e^x - 5*(x^3 + x^2*e^x)*log(log(5)) - 1)/(x^2 + x*e^x)
Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx=- 4 x^{3} + x \left (-1 + \log {\left (\log {\left (5 \right )} \right )}\right ) + \frac {1}{5 x^{2} + 5 x e^{x}} \]
integrate(((5*exp(x)**2*x**2+10*exp(x)*x**3+5*x**4)*ln(ln(5))+(-60*x**4-5* x**2)*exp(x)**2+(-120*x**5-10*x**3-x-1)*exp(x)-60*x**6-5*x**4-2*x)/(5*exp( x)**2*x**2+10*exp(x)*x**3+5*x**4),x)
Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx=-\frac {20 \, x^{5} - 5 \, x^{3} {\left (\log \left (\log \left (5\right )\right ) - 1\right )} + 5 \, {\left (4 \, x^{4} - x^{2} {\left (\log \left (\log \left (5\right )\right ) - 1\right )}\right )} e^{x} - 1}{5 \, {\left (x^{2} + x e^{x}\right )}} \]
integrate(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*log(log(5))+(-60*x^4-5*x^2 )*exp(x)^2+(-120*x^5-10*x^3-x-1)*exp(x)-60*x^6-5*x^4-2*x)/(5*exp(x)^2*x^2+ 10*exp(x)*x^3+5*x^4),x, algorithm=\
-1/5*(20*x^5 - 5*x^3*(log(log(5)) - 1) + 5*(4*x^4 - x^2*(log(log(5)) - 1)) *e^x - 1)/(x^2 + x*e^x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (26) = 52\).
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx=-\frac {20 \, x^{5} + 20 \, x^{4} e^{x} - 5 \, x^{3} \log \left (\log \left (5\right )\right ) - 5 \, x^{2} e^{x} \log \left (\log \left (5\right )\right ) + 5 \, x^{3} + 5 \, x^{2} e^{x} - 1}{5 \, {\left (x^{2} + x e^{x}\right )}} \]
integrate(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*log(log(5))+(-60*x^4-5*x^2 )*exp(x)^2+(-120*x^5-10*x^3-x-1)*exp(x)-60*x^6-5*x^4-2*x)/(5*exp(x)^2*x^2+ 10*exp(x)*x^3+5*x^4),x, algorithm=\
-1/5*(20*x^5 + 20*x^4*e^x - 5*x^3*log(log(5)) - 5*x^2*e^x*log(log(5)) + 5* x^3 + 5*x^2*e^x - 1)/(x^2 + x*e^x)
Time = 8.40 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx=\frac {x^3\,\left (\ln \left ({\ln \left (5\right )}^5\right )-5\right )-20\,x^4\,{\mathrm {e}}^x-20\,x^5+x^2\,{\mathrm {e}}^x\,\left (\ln \left ({\ln \left (5\right )}^5\right )-5\right )+1}{5\,\left (x\,{\mathrm {e}}^x+x^2\right )} \]
int(-(2*x - log(log(5))*(10*x^3*exp(x) + 5*x^2*exp(2*x) + 5*x^4) + exp(2*x )*(5*x^2 + 60*x^4) + exp(x)*(x + 10*x^3 + 120*x^5 + 1) + 5*x^4 + 60*x^6)/( 10*x^3*exp(x) + 5*x^2*exp(2*x) + 5*x^4),x)