Integrand size = 54, antiderivative size = 26 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=e^{\frac {x \left (e^{10}-\frac {6}{5 x}-\log (25 x)\right )}{-2+x}} \]
Time = 4.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=5^{-1+\frac {2+x}{2-x}} e^{\frac {6-5 e^{10} x}{10-5 x}} x^{-\frac {x}{-2+x}} \]
Integrate[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 5*x + 10*Log[25*x]))/(20 - 20*x + 5*x^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.
\(\displaystyle \int \frac {e^{\frac {5 e^{10} x-5 x \log (25 x)-6}{5 x-10}} \left (-5 x+10 \log (25 x)-10 e^{10}+16\right )}{5 x^2-20 x+20} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 20 \int \frac {25^{\frac {5 x}{10-5 x}-1} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {5 x}{10-5 x}} \left (-5 x+10 \log (25 x)+2 \left (8-5 e^{10}\right )\right )}{4 (2-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int \frac {25^{\frac {x}{2-x}-1} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {x}{2-x}} \left (-5 x+10 \log (25 x)+2 \left (8-5 e^{10}\right )\right )}{(2-x)^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle 5 \int \frac {25^{-\frac {2 (x-1)}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {x}{2-x}} \left (-5 x+10 \log (25 x)+2 \left (8-5 e^{10}\right )\right )}{(2-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 5 \int \left (\frac {2\ 5^{1-\frac {4 (x-1)}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} \log (25 x) x^{\frac {x}{2-x}}}{(x-2)^2}+\frac {25^{-\frac {2 (x-1)}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} \left (-5 x-10 e^{10}+16\right ) x^{\frac {x}{2-x}}}{(x-2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \left (2 \left (3-5 e^{10}\right ) \int \frac {25^{-\frac {2 (x-1)}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {x}{2-x}}}{(x-2)^2}dx-\int \frac {5^{\frac {2-3 x}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {x}{2-x}}}{x-2}dx-2 \int \frac {\int \frac {5^{\frac {2-3 x}{x-2}} e^{\frac {6-5 e^{10} x}{10-5 x}} x^{\frac {x}{2-x}}}{(x-2)^2}dx}{x}dx+2 \log (25 x) \int \frac {5^{\frac {2-3 x}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {x}{2-x}}}{(2-x)^2}dx\right )\) |
Int[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 5*x + 10*Log[25*x]))/(20 - 20*x + 5*x^2),x]
3.8.88.3.1 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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 3.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
risch | \({\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}\) | \(23\) |
parallelrisch | \({\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}\) | \(25\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}-2 \,{\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}}{-2+x}\) | \(62\) |
int((10*ln(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)^2-6)/(5 *x-10))/(5*x^2-20*x+20),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )} \]
integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5) ^2-6)/(5*x-10))/(5*x^2-20*x+20),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=e^{\frac {- 5 x \log {\left (25 x \right )} + 5 x e^{10} - 6}{5 x - 10}} \]
integrate((10*ln(25*x)-10*exp(5)**2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)* *2-6)/(5*x-10))/(5*x**2-20*x+20),x)
\[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=\int { -\frac {{\left (5 \, x + 10 \, e^{10} - 10 \, \log \left (25 \, x\right ) - 16\right )} e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )}}{5 \, {\left (x^{2} - 4 \, x + 4\right )}} \,d x } \]
integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5) ^2-6)/(5*x-10))/(5*x^2-20*x+20),x, algorithm=\
-1/5*integrate((5*x + 10*e^10 - 10*log(25*x) - 16)*e^(1/5*(5*x*e^10 - 5*x* log(25*x) - 6)/(x - 2))/(x^2 - 4*x + 4), x)
\[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=\int { -\frac {{\left (5 \, x + 10 \, e^{10} - 10 \, \log \left (25 \, x\right ) - 16\right )} e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )}}{5 \, {\left (x^{2} - 4 \, x + 4\right )}} \,d x } \]
integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5) ^2-6)/(5*x-10))/(5*x^2-20*x+20),x, algorithm=\
integrate(-1/5*(5*x + 10*e^10 - 10*log(25*x) - 16)*e^(1/5*(5*x*e^10 - 5*x* log(25*x) - 6)/(x - 2))/(x^2 - 4*x + 4), x)
Time = 9.74 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=\frac {{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{10}}{5\,x-10}}\,{\mathrm {e}}^{-\frac {6}{5\,x-10}}}{5^{\frac {10\,x}{5\,x-10}}\,x^{\frac {5\,x}{5\,x-10}}} \]
int(-(exp(-(5*x*log(25*x) - 5*x*exp(10) + 6)/(5*x - 10))*(5*x - 10*log(25* x) + 10*exp(10) - 16))/(5*x^2 - 20*x + 20),x)