Integrand size = 128, antiderivative size = 25 \[ \int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx=(-3+x)^2-e^{-50+x-\frac {2 x}{1+x \log (5)}} x \]
Time = 0.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx=x \left (-6-e^{x-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}}+x\right ) \]
Integrate[(E^x*(-1 + x + (-2*x - 2*x^2)*Log[5] + (-x^2 - x^3)*Log[5]^2) + E^((2*(25 + x + 25*x*Log[5]))/(1 + x*Log[5]))*(-6 + 2*x + (-12*x + 4*x^2)* Log[5] + (-6*x^2 + 2*x^3)*Log[5]^2))/(E^((2*(25 + x + 25*x*Log[5]))/(1 + x *Log[5]))*(1 + 2*x*Log[5] + x^2*Log[5]^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.
| \(\displaystyle \int \frac {e^{-\frac {2 (x+25 x \log (5)+25)}{x \log (5)+1}} \left (e^x \left (\left (-2 x^2-2 x\right ) \log (5)+\left (-x^3-x^2\right ) \log ^2(5)+x-1\right )+e^{\frac {2 (x+25 x \log (5)+25)}{x \log (5)+1}} \left (\left (4 x^2-12 x\right ) \log (5)+\left (2 x^3-6 x^2\right ) \log ^2(5)+2 x-6\right )\right )}{x^2 \log ^2(5)+2 x \log (5)+1} \, dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {e^{-\frac {2 (x+25 x \log (5)+25)}{x \log (5)+1}} \left (e^x \left (\left (-2 x^2-2 x\right ) \log (5)+\left (-x^3-x^2\right ) \log ^2(5)+x-1\right )+e^{\frac {2 (x+25 x \log (5)+25)}{x \log (5)+1}} \left (\left (4 x^2-12 x\right ) \log (5)+\left (2 x^3-6 x^2\right ) \log ^2(5)+2 x-6\right )\right )}{(x \log (5)+1)^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (e^x \left (\left (-2 x^2-2 x\right ) \log (5)+\left (-x^3-x^2\right ) \log ^2(5)+x-1\right )+e^{\frac {2 (x+25 x \log (5)+25)}{x \log (5)+1}} \left (\left (4 x^2-12 x\right ) \log (5)+\left (2 x^3-6 x^2\right ) \log ^2(5)+2 x-6\right )\right ) \exp \left (-\frac {2 (x (1+25 \log (5))+25)}{x \log (5)+1}\right )}{(x \log (5)+1)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {\left (x^3 \left (-\log ^2(5)\right )-x^2 \log (5) (2+\log (5))+x (1-\log (25))-1\right ) \exp \left (x-\frac {2 (x (1+25 \log (5))+25)}{x \log (5)+1}\right )}{(x \log (5)+1)^2}+2 (x-3) 5^{\frac {50 x}{x \log (5)+1}} \exp \left (\frac {2 (x+25)}{x \log (5)+1}-\frac {2 (x (1+25 \log (5))+25)}{x \log (5)+1}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \exp \left (x-\frac {2 ((1+25 \log (5)) x+25)}{\log (5) x+1}\right )dx-\int \exp \left (x-\frac {2 ((1+25 \log (5)) x+25)}{\log (5) x+1}\right ) xdx-\frac {2 \int \frac {\exp \left (x-\frac {2 ((1+25 \log (5)) x+25)}{\log (5) x+1}\right )}{(\log (5) x+1)^2}dx}{\log (5)}+\frac {2 \int \frac {\exp \left (x-\frac {2 ((1+25 \log (5)) x+25)}{\log (5) x+1}\right )}{\log (5) x+1}dx}{\log (5)}+(3-x)^2\) |
Int[(E^x*(-1 + x + (-2*x - 2*x^2)*Log[5] + (-x^2 - x^3)*Log[5]^2) + E^((2* (25 + x + 25*x*Log[5]))/(1 + x*Log[5]))*(-6 + 2*x + (-12*x + 4*x^2)*Log[5] + (-6*x^2 + 2*x^3)*Log[5]^2))/(E^((2*(25 + x + 25*x*Log[5]))/(1 + x*Log[5 ]))*(1 + 2*x*Log[5] + x^2*Log[5]^2)),x]
3.29.21.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 1.51 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48
| method | result | size |
| risch | \(x^{2}-6 x -x \,{\mathrm e}^{\frac {x^{2} \ln \left (5\right )-50 x \ln \left (5\right )-x -50}{x \ln \left (5\right )+1}}\) | \(37\) |
| parts | \(x^{2}-6 x +\frac {\left (-{\mathrm e}^{x} x -x^{2} \ln \left (5\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-\frac {2 \left (25 x \ln \left (5\right )+x +25\right )}{x \ln \left (5\right )+1}}}{x \ln \left (5\right )+1}\) | \(52\) |
| norman | \(\frac {\left (x^{3} \ln \left (5\right ) {\mathrm e}^{\frac {50 x \ln \left (5\right )+2 x +50}{x \ln \left (5\right )+1}}-6 \,{\mathrm e}^{\frac {50 x \ln \left (5\right )+2 x +50}{x \ln \left (5\right )+1}} x +\left (-6 \ln \left (5\right )+1\right ) x^{2} {\mathrm e}^{\frac {50 x \ln \left (5\right )+2 x +50}{x \ln \left (5\right )+1}}-{\mathrm e}^{x} x -x^{2} \ln \left (5\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-\frac {2 \left (25 x \ln \left (5\right )+x +25\right )}{x \ln \left (5\right )+1}}}{x \ln \left (5\right )+1}\) | \(124\) |
| parallelrisch | \(\frac {\left (x^{3} \ln \left (5\right ) {\mathrm e}^{\frac {50 x \ln \left (5\right )+2 x +50}{x \ln \left (5\right )+1}}-6 \ln \left (5\right ) {\mathrm e}^{\frac {50 x \ln \left (5\right )+2 x +50}{x \ln \left (5\right )+1}} x^{2}-x^{2} \ln \left (5\right ) {\mathrm e}^{x}+{\mathrm e}^{\frac {50 x \ln \left (5\right )+2 x +50}{x \ln \left (5\right )+1}} x^{2}-6 \,{\mathrm e}^{\frac {50 x \ln \left (5\right )+2 x +50}{x \ln \left (5\right )+1}} x -{\mathrm e}^{x} x \right ) {\mathrm e}^{-\frac {2 \left (25 x \ln \left (5\right )+x +25\right )}{x \ln \left (5\right )+1}}}{x \ln \left (5\right )+1}\) | \(145\) |
int((((-x^3-x^2)*ln(5)^2+(-2*x^2-2*x)*ln(5)+x-1)*exp(x)+((2*x^3-6*x^2)*ln( 5)^2+(4*x^2-12*x)*ln(5)+2*x-6)*exp((25*x*ln(5)+x+25)/(x*ln(5)+1))^2)/(x^2* ln(5)^2+2*x*ln(5)+1)/exp((25*x*ln(5)+x+25)/(x*ln(5)+1))^2,x,method=_RETURN VERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (24) = 48\).
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx=-{\left (x e^{x} - {\left (x^{2} - 6 \, x\right )} e^{\left (\frac {2 \, {\left (25 \, x \log \left (5\right ) + x + 25\right )}}{x \log \left (5\right ) + 1}\right )}\right )} e^{\left (-\frac {2 \, {\left (25 \, x \log \left (5\right ) + x + 25\right )}}{x \log \left (5\right ) + 1}\right )} \]
integrate((((-x^3-x^2)*log(5)^2+(-2*x^2-2*x)*log(5)+x-1)*exp(x)+((2*x^3-6* x^2)*log(5)^2+(4*x^2-12*x)*log(5)+2*x-6)*exp((25*x*log(5)+x+25)/(x*log(5)+ 1))^2)/(x^2*log(5)^2+2*x*log(5)+1)/exp((25*x*log(5)+x+25)/(x*log(5)+1))^2, x, algorithm=\
-(x*e^x - (x^2 - 6*x)*e^(2*(25*x*log(5) + x + 25)/(x*log(5) + 1)))*e^(-2*( 25*x*log(5) + x + 25)/(x*log(5) + 1))
Time = 6.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx=x^{2} - x e^{x} e^{- \frac {2 \left (x + 25 x \log {\left (5 \right )} + 25\right )}{x \log {\left (5 \right )} + 1}} - 6 x \]
integrate((((-x**3-x**2)*ln(5)**2+(-2*x**2-2*x)*ln(5)+x-1)*exp(x)+((2*x**3 -6*x**2)*ln(5)**2+(4*x**2-12*x)*ln(5)+2*x-6)*exp((25*x*ln(5)+x+25)/(x*ln(5 )+1))**2)/(x**2*ln(5)**2+2*x*ln(5)+1)/exp((25*x*ln(5)+x+25)/(x*ln(5)+1))** 2,x)
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (24) = 48\).
Time = 0.37 (sec) , antiderivative size = 225, normalized size of antiderivative = 9.00 \[ \int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx={\left (\frac {2}{x \log \left (5\right )^{5} + \log \left (5\right )^{4}} + \frac {x^{2} \log \left (5\right ) - 4 \, x}{\log \left (5\right )^{3}} + \frac {6 \, \log \left (x \log \left (5\right ) + 1\right )}{\log \left (5\right )^{4}}\right )} \log \left (5\right )^{2} + 6 \, {\left (\frac {1}{x \log \left (5\right )^{4} + \log \left (5\right )^{3}} - \frac {x}{\log \left (5\right )^{2}} + \frac {2 \, \log \left (x \log \left (5\right ) + 1\right )}{\log \left (5\right )^{3}}\right )} \log \left (5\right )^{2} - x e^{\left (x + \frac {2}{x \log \left (5\right )^{2} + \log \left (5\right )} - \frac {2}{\log \left (5\right )} - 50\right )} - 4 \, {\left (\frac {1}{x \log \left (5\right )^{4} + \log \left (5\right )^{3}} - \frac {x}{\log \left (5\right )^{2}} + \frac {2 \, \log \left (x \log \left (5\right ) + 1\right )}{\log \left (5\right )^{3}}\right )} \log \left (5\right ) - 12 \, {\left (\frac {1}{x \log \left (5\right )^{3} + \log \left (5\right )^{2}} + \frac {\log \left (x \log \left (5\right ) + 1\right )}{\log \left (5\right )^{2}}\right )} \log \left (5\right ) + \frac {2}{x \log \left (5\right )^{3} + \log \left (5\right )^{2}} + \frac {6}{x \log \left (5\right )^{2} + \log \left (5\right )} + \frac {2 \, \log \left (x \log \left (5\right ) + 1\right )}{\log \left (5\right )^{2}} \]
integrate((((-x^3-x^2)*log(5)^2+(-2*x^2-2*x)*log(5)+x-1)*exp(x)+((2*x^3-6* x^2)*log(5)^2+(4*x^2-12*x)*log(5)+2*x-6)*exp((25*x*log(5)+x+25)/(x*log(5)+ 1))^2)/(x^2*log(5)^2+2*x*log(5)+1)/exp((25*x*log(5)+x+25)/(x*log(5)+1))^2, x, algorithm=\
(2/(x*log(5)^5 + log(5)^4) + (x^2*log(5) - 4*x)/log(5)^3 + 6*log(x*log(5) + 1)/log(5)^4)*log(5)^2 + 6*(1/(x*log(5)^4 + log(5)^3) - x/log(5)^2 + 2*lo g(x*log(5) + 1)/log(5)^3)*log(5)^2 - x*e^(x + 2/(x*log(5)^2 + log(5)) - 2/ log(5) - 50) - 4*(1/(x*log(5)^4 + log(5)^3) - x/log(5)^2 + 2*log(x*log(5) + 1)/log(5)^3)*log(5) - 12*(1/(x*log(5)^3 + log(5)^2) + log(x*log(5) + 1)/ log(5)^2)*log(5) + 2/(x*log(5)^3 + log(5)^2) + 6/(x*log(5)^2 + log(5)) + 2 *log(x*log(5) + 1)/log(5)^2
Time = 0.82 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx={\left (x^{2} e^{50} - 6 \, x e^{50} - x e^{\left (\frac {x^{2} \log \left (5\right ) - x}{x \log \left (5\right ) + 1}\right )}\right )} e^{\left (-50\right )} \]
integrate((((-x^3-x^2)*log(5)^2+(-2*x^2-2*x)*log(5)+x-1)*exp(x)+((2*x^3-6* x^2)*log(5)^2+(4*x^2-12*x)*log(5)+2*x-6)*exp((25*x*log(5)+x+25)/(x*log(5)+ 1))^2)/(x^2*log(5)^2+2*x*log(5)+1)/exp((25*x*log(5)+x+25)/(x*log(5)+1))^2, x, algorithm=\
Timed out. \[ \int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx=\int -\frac {{\mathrm {e}}^{-\frac {2\,\left (x+25\,x\,\ln \left (5\right )+25\right )}{x\,\ln \left (5\right )+1}}\,\left ({\mathrm {e}}^{\frac {2\,\left (x+25\,x\,\ln \left (5\right )+25\right )}{x\,\ln \left (5\right )+1}}\,\left (\ln \left (5\right )\,\left (12\,x-4\,x^2\right )-2\,x+{\ln \left (5\right )}^2\,\left (6\,x^2-2\,x^3\right )+6\right )+{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (2\,x^2+2\,x\right )-x+{\ln \left (5\right )}^2\,\left (x^3+x^2\right )+1\right )\right )}{{\ln \left (5\right )}^2\,x^2+2\,\ln \left (5\right )\,x+1} \,d x \]
int(-(exp(-(2*(x + 25*x*log(5) + 25))/(x*log(5) + 1))*(exp((2*(x + 25*x*lo g(5) + 25))/(x*log(5) + 1))*(log(5)*(12*x - 4*x^2) - 2*x + log(5)^2*(6*x^2 - 2*x^3) + 6) + exp(x)*(log(5)*(2*x + 2*x^2) - x + log(5)^2*(x^2 + x^3) + 1)))/(x^2*log(5)^2 + 2*x*log(5) + 1),x)