Integrand size = 170, antiderivative size = 28 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x+\log (x)}{5 \left (4+e^5 \left (-5+2 \left (-2+e^x\right )+x+\log (2)\right )\right )} \]
Time = 1.70 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x+\log (x)}{5 \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \]
Integrate[(4 + E^5*(-9 - 8*x) + 4*x + E^(5 + x)*(2 + 2*x - 2*x^2) + E^5*(1 + x)*Log[2] + (-(E^5*x) - 2*E^(5 + x)*x)*Log[x])/(80*x + 20*E^(10 + 2*x)* x + E^5*(-360*x + 40*x^2) + E^10*(405*x - 90*x^2 + 5*x^3) + (40*E^5*x + E^ 10*(-90*x + 10*x^2))*Log[2] + 5*E^10*x*Log[2]^2 + E^x*(80*E^5*x + E^10*(-1 80*x + 20*x^2) + 20*E^10*x*Log[2])),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.
| \(\displaystyle \int \frac {e^{x+5} \left (-2 x^2+2 x+2\right )+e^5 (-8 x-9)+4 x+\left (-2 e^{x+5} x-e^5 x\right ) \log (x)+e^5 (x+1) \log (2)+4}{e^5 \left (40 x^2-360 x\right )+e^x \left (e^{10} \left (20 x^2-180 x\right )+80 e^5 x+20 e^{10} x \log (2)\right )+\left (e^{10} \left (10 x^2-90 x\right )+40 e^5 x\right ) \log (2)+e^{10} \left (5 x^3-90 x^2+405 x\right )+20 e^{2 x+10} x+80 x+5 e^{10} x \log ^2(2)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{x+5} \left (-2 x^2+2 x+2\right )+e^5 (-8 x-9)+4 x+\left (-2 e^{x+5} x-e^5 x\right ) \log (x)+e^5 (x+1) \log (2)+4}{e^5 \left (40 x^2-360 x\right )+e^x \left (e^{10} \left (20 x^2-180 x\right )+80 e^5 x+20 e^{10} x \log (2)\right )+\left (e^{10} \left (10 x^2-90 x\right )+40 e^5 x\right ) \log (2)+e^{10} \left (5 x^3-90 x^2+405 x\right )+20 e^{2 x+10} x+x \left (80+5 e^{10} \log ^2(2)\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{x+5} \left (-2 x^2+2 x+2\right )+4 (x+1)+e^5 (x (\log (2)-8)-9+\log (2))-e^5 \left (2 e^x+1\right ) x \log (x)}{5 x \left (2 e^{x+5}+e^5 (x-9+\log (2))+4\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {4 (x+1)+2 e^{x+5} \left (-x^2+x+1\right )-e^5 ((8-\log (2)) x-\log (2)+9)-e^5 \left (1+2 e^x\right ) x \log (x)}{x \left (-e^5 (-x-\log (2)+9)+2 e^{x+5}+4\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{5} \int \left (\frac {\left (e^5 x+e^5 \log (2)-10 e^5+4\right ) (x+\log (x))}{\left (e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}+\frac {-x^2-\log (x) x+x+1}{x \left (e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (e^5 \int \frac {x^2}{\left (e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}dx+\log (x) \int \frac {1}{-e^5 x-2 e^{x+5}-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}dx+\int \frac {x}{-e^5 x-2 e^{x+5}-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}dx+\left (4-e^5 (10-\log (2))\right ) \log (x) \int \frac {1}{\left (e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}dx+e^5 \log (x) \int \frac {x}{\left (e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}dx+\left (4-e^5 (10-\log (2))\right ) \int \frac {x}{\left (e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}dx+\int \frac {1}{e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}dx+\int \frac {1}{x \left (e^5 x+2 e^{x+5}+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )}dx-\left (4-e^5 (10-\log (2))\right ) \int \frac {\int \frac {1}{\left (e^5 (x+\log (2)-9)+2 e^{x+5}+4\right )^2}dx}{x}dx-e^5 \int \frac {\int \frac {x}{\left (e^5 (x+\log (2)-9)+2 e^{x+5}+4\right )^2}dx}{x}dx-\int \frac {\int -\frac {1}{e^5 (x+\log (2)-9)+2 e^{x+5}+4}dx}{x}dx\right )\) |
Int[(4 + E^5*(-9 - 8*x) + 4*x + E^(5 + x)*(2 + 2*x - 2*x^2) + E^5*(1 + x)* Log[2] + (-(E^5*x) - 2*E^(5 + x)*x)*Log[x])/(80*x + 20*E^(10 + 2*x)*x + E^ 5*(-360*x + 40*x^2) + E^10*(405*x - 90*x^2 + 5*x^3) + (40*E^5*x + E^10*(-9 0*x + 10*x^2))*Log[2] + 5*E^10*x*Log[2]^2 + E^x*(80*E^5*x + E^10*(-180*x + 20*x^2) + 20*E^10*x*Log[2])),x]
3.21.59.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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 = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50
| method | result | size |
| parallelrisch | \(\frac {\left (2 x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{5} \ln \left (x \right )\right ) {\mathrm e}^{-5}}{20 \,{\mathrm e}^{5} {\mathrm e}^{x}+10 \,{\mathrm e}^{5} \ln \left (2\right )+10 x \,{\mathrm e}^{5}-90 \,{\mathrm e}^{5}+40}\) | \(42\) |
| risch | \(\frac {\ln \left (x \right )}{10 \,{\mathrm e}^{5+x}+5 \,{\mathrm e}^{5} \ln \left (2\right )+5 x \,{\mathrm e}^{5}-45 \,{\mathrm e}^{5}+20}+\frac {x}{10 \,{\mathrm e}^{5+x}+5 \,{\mathrm e}^{5} \ln \left (2\right )+5 x \,{\mathrm e}^{5}-45 \,{\mathrm e}^{5}+20}\) | \(55\) |
int(((-2*x*exp(5)*exp(x)-x*exp(5))*ln(x)+(-2*x^2+2*x+2)*exp(5)*exp(x)+(1+x )*exp(5)*ln(2)+(-8*x-9)*exp(5)+4*x+4)/(20*x*exp(5)^2*exp(x)^2+(20*x*exp(5) ^2*ln(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)^2*ln(2)^2+ ((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*ln(2)+(5*x^3-90*x^2+405*x)*exp(5)^2+( 40*x^2-360*x)*exp(5)+80*x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log \left (x\right )}{5 \, {\left ({\left (x - 9\right )} e^{5} + e^{5} \log \left (2\right ) + 2 \, e^{\left (x + 5\right )} + 4\right )}} \]
integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*log(x)+(-2*x^2+2*x+2)*exp(5)*exp( x)+(1+x)*exp(5)*log(2)+(-8*x-9)*exp(5)+4*x+4)/(20*x*exp(5)^2*exp(x)^2+(20* x*exp(5)^2*log(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)^2 *log(2)^2+((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*log(2)+(5*x^3-90*x^2+405*x) *exp(5)^2+(40*x^2-360*x)*exp(5)+80*x),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log {\left (x \right )}}{5 x e^{5} + 10 e^{5} e^{x} - 45 e^{5} + 20 + 5 e^{5} \log {\left (2 \right )}} \]
integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*ln(x)+(-2*x**2+2*x+2)*exp(5)*exp( x)+(1+x)*exp(5)*ln(2)+(-8*x-9)*exp(5)+4*x+4)/(20*x*exp(5)**2*exp(x)**2+(20 *x*exp(5)**2*ln(2)+(20*x**2-180*x)*exp(5)**2+80*x*exp(5))*exp(x)+5*x*exp(5 )**2*ln(2)**2+((10*x**2-90*x)*exp(5)**2+40*x*exp(5))*ln(2)+(5*x**3-90*x**2 +405*x)*exp(5)**2+(40*x**2-360*x)*exp(5)+80*x),x)
Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log \left (x\right )}{5 \, {\left (x e^{5} + {\left (\log \left (2\right ) - 9\right )} e^{5} + 2 \, e^{\left (x + 5\right )} + 4\right )}} \]
integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*log(x)+(-2*x^2+2*x+2)*exp(5)*exp( x)+(1+x)*exp(5)*log(2)+(-8*x-9)*exp(5)+4*x+4)/(20*x*exp(5)^2*exp(x)^2+(20* x*exp(5)^2*log(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)^2 *log(2)^2+((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*log(2)+(5*x^3-90*x^2+405*x) *exp(5)^2+(40*x^2-360*x)*exp(5)+80*x),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\frac {x + \log \left (x\right )}{5 \, {\left (x e^{5} + e^{5} \log \left (2\right ) - 9 \, e^{5} + 2 \, e^{\left (x + 5\right )} + 4\right )}} \]
integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*log(x)+(-2*x^2+2*x+2)*exp(5)*exp( x)+(1+x)*exp(5)*log(2)+(-8*x-9)*exp(5)+4*x+4)/(20*x*exp(5)^2*exp(x)^2+(20* x*exp(5)^2*log(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)^2 *log(2)^2+((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*log(2)+(5*x^3-90*x^2+405*x) *exp(5)^2+(40*x^2-360*x)*exp(5)+80*x),x, algorithm=\
Timed out. \[ \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx=\int \frac {4\,x+{\mathrm {e}}^{x+5}\,\left (-2\,x^2+2\,x+2\right )-\ln \left (x\right )\,\left (2\,x\,{\mathrm {e}}^{x+5}+x\,{\mathrm {e}}^5\right )-{\mathrm {e}}^5\,\left (8\,x+9\right )+{\mathrm {e}}^5\,\ln \left (2\right )\,\left (x+1\right )+4}{80\,x-{\mathrm {e}}^5\,\left (360\,x-40\,x^2\right )+{\mathrm {e}}^{10}\,\left (5\,x^3-90\,x^2+405\,x\right )+20\,x\,{\mathrm {e}}^{2\,x+10}-\ln \left (2\right )\,\left ({\mathrm {e}}^{10}\,\left (90\,x-10\,x^2\right )-40\,x\,{\mathrm {e}}^5\right )+{\mathrm {e}}^x\,\left (80\,x\,{\mathrm {e}}^5-{\mathrm {e}}^{10}\,\left (180\,x-20\,x^2\right )+20\,x\,{\mathrm {e}}^{10}\,\ln \left (2\right )\right )+5\,x\,{\mathrm {e}}^{10}\,{\ln \left (2\right )}^2} \,d x \]
int((4*x - log(x)*(x*exp(5) + 2*x*exp(5)*exp(x)) - exp(5)*(8*x + 9) + exp( 5)*log(2)*(x + 1) + exp(5)*exp(x)*(2*x - 2*x^2 + 2) + 4)/(80*x - exp(5)*(3 60*x - 40*x^2) + exp(10)*(405*x - 90*x^2 + 5*x^3) - log(2)*(exp(10)*(90*x - 10*x^2) - 40*x*exp(5)) + exp(x)*(80*x*exp(5) - exp(10)*(180*x - 20*x^2) + 20*x*exp(10)*log(2)) + 5*x*exp(10)*log(2)^2 + 20*x*exp(2*x)*exp(10)),x)
int((4*x + exp(x + 5)*(2*x - 2*x^2 + 2) - log(x)*(2*x*exp(x + 5) + x*exp(5 )) - exp(5)*(8*x + 9) + exp(5)*log(2)*(x + 1) + 4)/(80*x - exp(5)*(360*x - 40*x^2) + exp(10)*(405*x - 90*x^2 + 5*x^3) + 20*x*exp(2*x + 10) - log(2)* (exp(10)*(90*x - 10*x^2) - 40*x*exp(5)) + exp(x)*(80*x*exp(5) - exp(10)*(1 80*x - 20*x^2) + 20*x*exp(10)*log(2)) + 5*x*exp(10)*log(2)^2), x)