Integrand size = 160, antiderivative size = 33 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {x}{x+\frac {x}{e^3-x}}\right )}} \]
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}} \]
Integrate[(E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2)*(-25*x^2*Log[(E^3 - x)/(1 + E^3 - x)] + (25*E^6*x - 25*x^2 + 25*x^3 + E^3*(25*x - 50*x^2))*Log[(E^3 - x)/(1 + E^3 - x)]^2))/( 8*E^6 + E^3*(8 - 16*x) - 8*x + 8*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.
| \(\displaystyle \int \frac {\left (\left (25 x^3-25 x^2+e^3 \left (25 x-50 x^2\right )+25 e^6 x\right ) \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )-25 x^2 \log \left (\frac {e^3-x}{-x+e^3+1}\right )\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )}{8 x^2-8 x+e^3 (8-16 x)+8 e^6} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (\left (25 x^3-25 x^2+e^3 \left (25 x-50 x^2\right )+25 e^6 x\right ) \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )-25 x^2 \log \left (\frac {e^3-x}{-x+e^3+1}\right )\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )}{8 x^2-8 \left (1+2 e^3\right ) x+8 e^3 \left (1+e^3\right )}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {25}{8} x \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )-\frac {25 x^2 \log \left (\frac {e^3-x}{-x+e^3+1}\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )}{8 \left (e^3-x\right ) \left (-x+e^3+1\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {25}{8} \int \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right ) \log \left (\frac {e^3-x}{-x+e^3+1}\right )dx-\frac {25}{8} \int \frac {\exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}+6\right ) \log \left (\frac {e^3-x}{-x+e^3+1}\right )}{e^3-x}dx+\frac {25}{8} \left (1+e^3\right )^2 \int \frac {\exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right ) \log \left (\frac {e^3-x}{-x+e^3+1}\right )}{-x+e^3+1}dx+\frac {25}{8} \int \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right ) x \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )dx\) |
Int[(E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/( 1 + E^3 - x)]^2)*(-25*x^2*Log[(E^3 - x)/(1 + E^3 - x)] + (25*E^6*x - 25*x^ 2 + 25*x^3 + E^3*(25*x - 50*x^2))*Log[(E^3 - x)/(1 + E^3 - x)]^2))/(8*E^6 + E^3*(8 - 16*x) - 8*x + 8*x^2),x]
3.19.9.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 30.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88
| method | result | size |
| risch | \(\frac {{\mathrm e}^{{\mathrm e}^{25 x^{2} \ln \left (\frac {-x +{\mathrm e}^{3}}{{\mathrm e}^{3}-x +1}\right )^{2}}}}{16}\) | \(29\) |
| parallelrisch | \(\frac {{\mathrm e}^{{\mathrm e}^{25 x^{2} \ln \left (\frac {-x +{\mathrm e}^{3}}{{\mathrm e}^{3}-x +1}\right )^{2}}}}{16}\) | \(29\) |
int(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*ln((-x+exp(3))/(e xp(3)-x+1))^2-25*x^2*ln((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*ln((-x+exp(3 ))/(exp(3)-x+1))^2)*exp(exp(25*x^2*ln((-x+exp(3))/(exp(3)-x+1))^2))/(8*exp (3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )} \]
integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp (3))/(exp(3)-x+1))^2-25*x^2*log((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*log( (-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^ 2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm=\
Time = 3.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {e^{e^{25 x^{2} \log {\left (\frac {- x + e^{3}}{- x + 1 + e^{3}} \right )}^{2}}}}{16} \]
integrate(((25*x*exp(3)**2+(-50*x**2+25*x)*exp(3)+25*x**3-25*x**2)*ln((-x+ exp(3))/(exp(3)-x+1))**2-25*x**2*ln((-x+exp(3))/(exp(3)-x+1)))*exp(25*x**2 *ln((-x+exp(3))/(exp(3)-x+1))**2)*exp(exp(25*x**2*ln((-x+exp(3))/(exp(3)-x +1))**2))/(8*exp(3)**2+(-16*x+8)*exp(3)+8*x**2-8*x),x)
Time = 0.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (x - e^{3}\right )^{2} - 50 \, x^{2} \log \left (x - e^{3}\right ) \log \left (x - e^{3} - 1\right ) + 25 \, x^{2} \log \left (x - e^{3} - 1\right )^{2}\right )}\right )} \]
integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp (3))/(exp(3)-x+1))^2-25*x^2*log((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*log( (-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^ 2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm=\
1/16*e^(e^(25*x^2*log(x - e^3)^2 - 50*x^2*log(x - e^3)*log(x - e^3 - 1) + 25*x^2*log(x - e^3 - 1)^2))
\[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\int { -\frac {25 \, {\left (x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right ) - {\left (x^{3} - x^{2} + x e^{6} - {\left (2 \, x^{2} - x\right )} e^{3}\right )} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )} e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2} + e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )}}{8 \, {\left (x^{2} - {\left (2 \, x - 1\right )} e^{3} - x + e^{6}\right )}} \,d x } \]
integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp (3))/(exp(3)-x+1))^2-25*x^2*log((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*log( (-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^ 2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm=\
integrate(-25/8*(x^2*log((x - e^3)/(x - e^3 - 1)) - (x^3 - x^2 + x*e^6 - ( 2*x^2 - x)*e^3)*log((x - e^3)/(x - e^3 - 1))^2)*e^(25*x^2*log((x - e^3)/(x - e^3 - 1))^2 + e^(25*x^2*log((x - e^3)/(x - e^3 - 1))^2))/(x^2 - (2*x - 1)*e^3 - x + e^6), x)
Time = 14.48 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{25\,x^2\,{\ln \left (-\frac {x-{\mathrm {e}}^3}{{\mathrm {e}}^3-x+1}\right )}^2}}}{16} \]
int(-(exp(25*x^2*log(-(x - exp(3))/(exp(3) - x + 1))^2)*exp(exp(25*x^2*log (-(x - exp(3))/(exp(3) - x + 1))^2))*(log(-(x - exp(3))/(exp(3) - x + 1))^ 2*(exp(3)*(25*x - 50*x^2) + 25*x*exp(6) - 25*x^2 + 25*x^3) - 25*x^2*log(-( x - exp(3))/(exp(3) - x + 1))))/(8*x - 8*exp(6) - 8*x^2 + exp(3)*(16*x - 8 )),x)