Integrand size = 54, antiderivative size = 20 \[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=\frac {6 e^{e^{4 x}} \log (-1+x)}{5-x} \] Output:
6*exp(exp(x)^4)*ln(-1+x)/(5-x)
Time = 0.88 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=-\frac {6 e^{e^{4 x}} \log (-1+x)}{-5+x} \] Input:
Integrate[(E^E^(4*x)*(30 - 6*x + (-6 + 6*x + E^(4*x)*(-120 + 144*x - 24*x^ 2))*Log[-1 + x]))/(-25 + 35*x - 11*x^2 + x^3),x]
Output:
(-6*E^E^(4*x)*Log[-1 + x])/(-5 + 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.
\(\displaystyle \int \frac {e^{e^{4 x}} \left (\left (e^{4 x} \left (-24 x^2+144 x-120\right )+6 x-6\right ) \log (x-1)-6 x+30\right )}{x^3-11 x^2+35 x-25} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {e^{e^{4 x}} \left (\left (e^{4 x} \left (-24 x^2+144 x-120\right )+6 x-6\right ) \log (x-1)-6 x+30\right )}{16 (x-5)}+\frac {e^{e^{4 x}} \left (\left (e^{4 x} \left (-24 x^2+144 x-120\right )+6 x-6\right ) \log (x-1)-6 x+30\right )}{16 (x-1)}+\frac {e^{e^{4 x}} \left (\left (e^{4 x} \left (-24 x^2+144 x-120\right )+6 x-6\right ) \log (x-1)-6 x+30\right )}{4 (x-5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{2} \int \frac {e^{e^{4 x}}}{x-5}dx+\frac {3}{2} \int \frac {e^{e^{4 x}}}{x-1}dx-6 \int \frac {\int \frac {e^{e^{4 x}}}{(x-5)^2}dx}{x-1}dx+24 \int \frac {\int \frac {e^{4 x+e^{4 x}}}{x-5}dx}{x-1}dx+6 \log (x-1) \int \frac {e^{e^{4 x}}}{(x-5)^2}dx-24 \log (x-1) \int \frac {e^{4 x+e^{4 x}}}{x-5}dx\) |
Input:
Int[(E^E^(4*x)*(30 - 6*x + (-6 + 6*x + E^(4*x)*(-120 + 144*x - 24*x^2))*Lo g[-1 + x]))/(-25 + 35*x - 11*x^2 + x^3),x]
Output:
$Aborted
Time = 6.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-\frac {6 \ln \left (-1+x \right ) {\mathrm e}^{{\mathrm e}^{4 x}}}{-5+x}\) | \(17\) |
parallelrisch | \(-\frac {6 \ln \left (-1+x \right ) {\mathrm e}^{{\mathrm e}^{4 x}}}{-5+x}\) | \(17\) |
Input:
int((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*ln(-1+x)-6*x+30)*exp(exp(x)^4)/( x^3-11*x^2+35*x-25),x,method=_RETURNVERBOSE)
Output:
-6/(-5+x)*ln(-1+x)*exp(exp(4*x))
Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=-\frac {6 \, e^{\left (e^{\left (4 \, x\right )}\right )} \log \left (x - 1\right )}{x - 5} \] Input:
integrate((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*log(-1+x)-6*x+30)*exp(exp( x)^4)/(x^3-11*x^2+35*x-25),x, algorithm="fricas")
Output:
-6*e^(e^(4*x))*log(x - 1)/(x - 5)
Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=- \frac {6 e^{e^{4 x}} \log {\left (x - 1 \right )}}{x - 5} \] Input:
integrate((((-24*x**2+144*x-120)*exp(x)**4+6*x-6)*ln(-1+x)-6*x+30)*exp(exp (x)**4)/(x**3-11*x**2+35*x-25),x)
Output:
-6*exp(exp(4*x))*log(x - 1)/(x - 5)
Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=-\frac {6 \, e^{\left (e^{\left (4 \, x\right )}\right )} \log \left (x - 1\right )}{x - 5} \] Input:
integrate((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*log(-1+x)-6*x+30)*exp(exp( x)^4)/(x^3-11*x^2+35*x-25),x, algorithm="maxima")
Output:
-6*e^(e^(4*x))*log(x - 1)/(x - 5)
\[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=\int { -\frac {6 \, {\left ({\left (4 \, {\left (x^{2} - 6 \, x + 5\right )} e^{\left (4 \, x\right )} - x + 1\right )} \log \left (x - 1\right ) + x - 5\right )} e^{\left (e^{\left (4 \, x\right )}\right )}}{x^{3} - 11 \, x^{2} + 35 \, x - 25} \,d x } \] Input:
integrate((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*log(-1+x)-6*x+30)*exp(exp( x)^4)/(x^3-11*x^2+35*x-25),x, algorithm="giac")
Output:
integrate(-6*((4*(x^2 - 6*x + 5)*e^(4*x) - x + 1)*log(x - 1) + x - 5)*e^(e ^(4*x))/(x^3 - 11*x^2 + 35*x - 25), x)
Timed out. \[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=\int -\frac {{\mathrm {e}}^{{\mathrm {e}}^{4\,x}}\,\left (6\,x+\ln \left (x-1\right )\,\left ({\mathrm {e}}^{4\,x}\,\left (24\,x^2-144\,x+120\right )-6\,x+6\right )-30\right )}{x^3-11\,x^2+35\,x-25} \,d x \] Input:
int(-(exp(exp(4*x))*(6*x + log(x - 1)*(exp(4*x)*(24*x^2 - 144*x + 120) - 6 *x + 6) - 30))/(35*x - 11*x^2 + x^3 - 25),x)
Output:
int(-(exp(exp(4*x))*(6*x + log(x - 1)*(exp(4*x)*(24*x^2 - 144*x + 120) - 6 *x + 6) - 30))/(35*x - 11*x^2 + x^3 - 25), x)
Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{e^{4 x}} \left (30-6 x+\left (-6+6 x+e^{4 x} \left (-120+144 x-24 x^2\right )\right ) \log (-1+x)\right )}{-25+35 x-11 x^2+x^3} \, dx=-\frac {6 e^{e^{4 x}} \mathrm {log}\left (x -1\right )}{-5+x} \] Input:
int((((-24*x^2+144*x-120)*exp(x)^4+6*x-6)*log(-1+x)-6*x+30)*exp(exp(x)^4)/ (x^3-11*x^2+35*x-25),x)
Output:
( - 6*e**(e**(4*x))*log(x - 1))/(x - 5)