Integrand size = 135, antiderivative size = 30 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=\frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{3 x^2 \left (e^x+x\right )} \] Output:
1/3/x^2*ln(x/(5*ln(x)-10*x^2))/(exp(x)+x)
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=\frac {\log \left (\frac {x}{5 \left (-2 x^2+\log (x)\right )}\right )}{3 x^2 \left (e^x+x\right )} \] Input:
Integrate[(-x + 2*x^3 + E^x*(-1 + 2*x^2) + (E^x + x)*Log[x] + (6*x^3 + E^x *(4*x^2 + 2*x^3) + (E^x*(-2 - x) - 3*x)*Log[x])*Log[x/(-10*x^2 + 5*Log[x]) ])/(-6*E^(2*x)*x^5 - 12*E^x*x^6 - 6*x^7 + (3*E^(2*x)*x^3 + 6*E^x*x^4 + 3*x ^5)*Log[x]),x]
Output:
Log[x/(5*(-2*x^2 + Log[x]))]/(3*x^2*(E^x + 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.
| \(\displaystyle \int \frac {2 x^3+e^x \left (2 x^2-1\right )+\left (6 x^3+e^x \left (2 x^3+4 x^2\right )+\left (e^x (-x-2)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{5 \log (x)-10 x^2}\right )-x+\left (x+e^x\right ) \log (x)}{-6 x^7-12 e^x x^6-6 e^{2 x} x^5+\left (3 x^5+6 e^x x^4+3 e^{2 x} x^3\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-2 x^3-e^x \left (2 x^2-1\right )-\left (6 x^3+e^x \left (2 x^3+4 x^2\right )+\left (e^x (-x-2)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{5 \log (x)-10 x^2}\right )+x-\left (x+e^x\right ) \log (x)}{3 x^3 \left (x+e^x\right )^2 \left (2 x^2-\log (x)\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {-2 x^3+x+e^x \left (1-2 x^2\right )-\left (x+e^x\right ) \log (x)-\left (6 x^3+2 e^x \left (x^3+2 x^2\right )-\left (3 x+e^x (x+2)\right ) \log (x)\right ) \log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right )}{x^3 \left (x+e^x\right )^2 \left (2 x^2-\log (x)\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {(x-1) \log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right )}{x^2 \left (x+e^x\right )^2}-\frac {2 \log \left (-\frac {x}{10 x^2-5 \log (x)}\right ) x^3+4 \log \left (-\frac {x}{10 x^2-5 \log (x)}\right ) x^2+2 x^2-\log (x) \log \left (-\frac {x}{10 x^2-5 \log (x)}\right ) x+\log (x)-2 \log (x) \log \left (-\frac {x}{10 x^2-5 \log (x)}\right )-1}{x^3 \left (x+e^x\right ) \left (2 x^2-\log (x)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-\log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right ) \int \frac {1}{x^2 \left (x+e^x\right )^2}dx+\log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right ) \int \frac {1}{x \left (x+e^x\right )^2}dx-2 \int \frac {1}{x \left (x+e^x\right ) \left (2 x^2-\log (x)\right )}dx-2 \int \frac {\log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right )}{\left (x+e^x\right ) \left (2 x^2-\log (x)\right )}dx-4 \int \frac {\log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right )}{x \left (x+e^x\right ) \left (2 x^2-\log (x)\right )}dx-\int \frac {\log (x) \log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right )}{\left (-x-e^x\right ) x^2 \left (2 x^2-\log (x)\right )}dx+\int \frac {\int \frac {1}{x^2 \left (x+e^x\right )^2}dx}{x \left (2 x^2-\log (x)\right )}dx-2 \int \frac {x \int \frac {1}{x^2 \left (x+e^x\right )^2}dx}{2 x^2-\log (x)}dx+\int \frac {\log (x) \int \frac {1}{x^2 \left (x+e^x\right )^2}dx}{x \left (\log (x)-2 x^2\right )}dx-\int \frac {\int \frac {1}{x \left (x+e^x\right )^2}dx}{x \left (2 x^2-\log (x)\right )}dx+2 \int \frac {x \int \frac {1}{x \left (x+e^x\right )^2}dx}{2 x^2-\log (x)}dx-\int \frac {\log (x) \int \frac {1}{x \left (x+e^x\right )^2}dx}{x \left (\log (x)-2 x^2\right )}dx+\int \frac {1}{x^3 \left (x+e^x\right ) \left (2 x^2-\log (x)\right )}dx-\int \frac {\log (x)}{x^3 \left (x+e^x\right ) \left (2 x^2-\log (x)\right )}dx-2 \int \frac {\log (x) \log \left (-\frac {x}{5 \left (2 x^2-\log (x)\right )}\right )}{\left (-x-e^x\right ) x^3 \left (2 x^2-\log (x)\right )}dx\right )\) |
Input:
Int[(-x + 2*x^3 + E^x*(-1 + 2*x^2) + (E^x + x)*Log[x] + (6*x^3 + E^x*(4*x^ 2 + 2*x^3) + (E^x*(-2 - x) - 3*x)*Log[x])*Log[x/(-10*x^2 + 5*Log[x])])/(-6 *E^(2*x)*x^5 - 12*E^x*x^6 - 6*x^7 + (3*E^(2*x)*x^3 + 6*E^x*x^4 + 3*x^5)*Lo g[x]),x]
Output:
$Aborted
Time = 20.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87
| method | result | size |
| parallelrisch | \(\frac {\ln \left (\frac {x}{5 \ln \left (x \right )-10 x^{2}}\right )}{3 x^{2} \left ({\mathrm e}^{x}+x \right )}\) | \(26\) |
| risch | \(-\frac {\ln \left (x^{2}-\frac {\ln \left (x \right )}{2}\right )}{3 x^{2} \left ({\mathrm e}^{x}+x \right )}-\frac {2 i \pi \operatorname {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \left (x \right )}{2}}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{-x^{2}+\frac {\ln \left (x \right )}{2}}\right ) \operatorname {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \left (x \right )}{2}}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{-x^{2}+\frac {\ln \left (x \right )}{2}}\right ) \operatorname {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \left (x \right )}{2}}\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \left (x \right )}{2}}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (\frac {i x}{-x^{2}+\frac {\ln \left (x \right )}{2}}\right )^{3}-2 i \pi +2 \ln \left (2\right )+2 \ln \left (5\right )-2 \ln \left (x \right )}{6 x^{2} \left ({\mathrm e}^{x}+x \right )}\) | \(205\) |
Input:
int(((((-2-x)*exp(x)-3*x)*ln(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*ln(x/(5*ln(x)- 10*x^2))+(exp(x)+x)*ln(x)+(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2*x^3+6*exp (x)*x^4+3*x^5)*ln(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x,method=_RETURNV ERBOSE)
Output:
1/3*ln(1/5*x/(-2*x^2+ln(x)))/x^2/(exp(x)+x)
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=\frac {\log \left (-\frac {x}{5 \, {\left (2 \, x^{2} - \log \left (x\right )\right )}}\right )}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \] Input:
integrate(((((-2-x)*exp(x)-3*x)*log(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*log(x/( 5*log(x)-10*x^2))+(exp(x)+x)*log(x)+(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2 *x^3+6*exp(x)*x^4+3*x^5)*log(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x, alg orithm="fricas")
Output:
1/3*log(-1/5*x/(2*x^2 - log(x)))/(x^3 + x^2*e^x)
Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=\frac {\log {\left (\frac {x}{- 10 x^{2} + 5 \log {\left (x \right )}} \right )}}{3 x^{3} + 3 x^{2} e^{x}} \] Input:
integrate(((((-2-x)*exp(x)-3*x)*ln(x)+(2*x**3+4*x**2)*exp(x)+6*x**3)*ln(x/ (5*ln(x)-10*x**2))+(exp(x)+x)*ln(x)+(2*x**2-1)*exp(x)+2*x**3-x)/((3*exp(x) **2*x**3+6*exp(x)*x**4+3*x**5)*ln(x)-6*x**5*exp(x)**2-12*x**6*exp(x)-6*x** 7),x)
Output:
log(x/(-10*x**2 + 5*log(x)))/(3*x**3 + 3*x**2*exp(x))
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=-\frac {\log \left (5\right ) + \log \left (-2 \, x^{2} + \log \left (x\right )\right ) - \log \left (x\right )}{3 \, {\left (x^{3} + x^{2} e^{x}\right )}} \] Input:
integrate(((((-2-x)*exp(x)-3*x)*log(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*log(x/( 5*log(x)-10*x^2))+(exp(x)+x)*log(x)+(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2 *x^3+6*exp(x)*x^4+3*x^5)*log(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x, alg orithm="maxima")
Output:
-1/3*(log(5) + log(-2*x^2 + log(x)) - log(x))/(x^3 + x^2*e^x)
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=-\frac {\log \left (100 \, x^{4} - 100 \, x^{2} \log \left ({\left | x \right |}\right ) - \frac {25}{2} \, \pi ^{2} \mathrm {sgn}\left (x\right ) + \frac {25}{2} \, \pi ^{2} + 25 \, \log \left ({\left | x \right |}\right )^{2}\right ) - 2 \, \log \left ({\left | x \right |}\right )}{6 \, {\left (x^{3} + x^{2} e^{x}\right )}} \] Input:
integrate(((((-2-x)*exp(x)-3*x)*log(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*log(x/( 5*log(x)-10*x^2))+(exp(x)+x)*log(x)+(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2 *x^3+6*exp(x)*x^4+3*x^5)*log(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x, alg orithm="giac")
Output:
-1/6*(log(100*x^4 - 100*x^2*log(abs(x)) - 25/2*pi^2*sgn(x) + 25/2*pi^2 + 2 5*log(abs(x))^2) - 2*log(abs(x)))/(x^3 + x^2*e^x)
Time = 2.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=\frac {\ln \left (\frac {x}{5\,\left (\ln \left (x\right )-2\,x^2\right )}\right )}{3\,x^2\,\left (x+{\mathrm {e}}^x\right )} \] Input:
int(-(log(x/(5*log(x) - 10*x^2))*(exp(x)*(4*x^2 + 2*x^3) - log(x)*(3*x + e xp(x)*(x + 2)) + 6*x^3) - x + exp(x)*(2*x^2 - 1) + 2*x^3 + log(x)*(x + exp (x)))/(12*x^6*exp(x) - log(x)*(6*x^4*exp(x) + 3*x^3*exp(2*x) + 3*x^5) + 6* x^5*exp(2*x) + 6*x^7),x)
Output:
log(x/(5*(log(x) - 2*x^2)))/(3*x^2*(x + exp(x)))
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-x+2 x^3+e^x \left (-1+2 x^2\right )+\left (e^x+x\right ) \log (x)+\left (6 x^3+e^x \left (4 x^2+2 x^3\right )+\left (e^x (-2-x)-3 x\right ) \log (x)\right ) \log \left (\frac {x}{-10 x^2+5 \log (x)}\right )}{-6 e^{2 x} x^5-12 e^x x^6-6 x^7+\left (3 e^{2 x} x^3+6 e^x x^4+3 x^5\right ) \log (x)} \, dx=\frac {\mathrm {log}\left (\frac {x}{5 \,\mathrm {log}\left (x \right )-10 x^{2}}\right )}{3 x^{2} \left (e^{x}+x \right )} \] Input:
int(((((-2-x)*exp(x)-3*x)*log(x)+(2*x^3+4*x^2)*exp(x)+6*x^3)*log(x/(5*log( x)-10*x^2))+(exp(x)+x)*log(x)+(2*x^2-1)*exp(x)+2*x^3-x)/((3*exp(x)^2*x^3+6 *exp(x)*x^4+3*x^5)*log(x)-6*x^5*exp(x)^2-12*x^6*exp(x)-6*x^7),x)
Output:
log(x/(5*log(x) - 10*x**2))/(3*x**2*(e**x + x))