Integrand size = 91, antiderivative size = 28 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=\frac {1}{x}-\frac {e^{x^2}}{x}+\frac {1}{x^2 \log \left (\frac {8}{3}-16 x\right )} \] Output:
1/x-exp(x^2)/x+1/ln(-16*x+8/3)/x^2
Time = 0.50 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=\frac {x-e^{x^2} x+\frac {1}{\log \left (\frac {8}{3}-16 x\right )}}{x^2} \] Input:
Integrate[(-6*x + (2 - 12*x)*Log[(8 - 48*x)/3] + (x - 6*x^2 + E^x^2*(-x + 6*x^2 + 2*x^3 - 12*x^4))*Log[(8 - 48*x)/3]^2)/((-x^3 + 6*x^4)*Log[(8 - 48* x)/3]^2),x]
Output:
(x - E^x^2*x + Log[8/3 - 16*x]^(-1))/x^2
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.
\(\displaystyle \int \frac {\left (-6 x^2+e^{x^2} \left (-12 x^4+2 x^3+6 x^2-x\right )+x\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )}{\left (6 x^4-x^3\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-6 x^2+e^{x^2} \left (-12 x^4+2 x^3+6 x^2-x\right )+x\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )}{x^3 (6 x-1) \log ^2\left (\frac {1}{3} (8-48 x)\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (e^{x^2} \left (1-2 x^2\right )-1\right ) x-\frac {6 x}{(6 x-1) \log ^2\left (\frac {8}{3}-16 x\right )}-\frac {2}{\log \left (\frac {8}{3}-16 x\right )}}{x^3}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {-6 x^2 \log ^2\left (\frac {8}{3}-16 x\right )-6 x+x \log ^2\left (\frac {8}{3}-16 x\right )-12 x \log \left (\frac {8}{3}-16 x\right )+2 \log \left (\frac {8}{3}-16 x\right )}{x^3 (6 x-1) \log ^2\left (\frac {8}{3}-16 x\right )}-\frac {e^{x^2} \left (2 x^2-1\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )}dx+6 \int \frac {1}{x^2 \log ^2\left (\frac {8}{3}-16 x\right )}dx+36 \int \frac {1}{x \log ^2\left (\frac {8}{3}-16 x\right )}dx-\frac {e^{x^2}}{x}+\frac {1}{x}+\frac {36}{\log \left (\frac {8}{3}-16 x\right )}\) |
Input:
Int[(-6*x + (2 - 12*x)*Log[(8 - 48*x)/3] + (x - 6*x^2 + E^x^2*(-x + 6*x^2 + 2*x^3 - 12*x^4))*Log[(8 - 48*x)/3]^2)/((-x^3 + 6*x^4)*Log[(8 - 48*x)/3]^ 2),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(74\) vs. \(2(25)=50\).
Time = 1.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.68
\[\frac {1}{x}+\frac {3 \ln \left (2\right )-\ln \left (3\right )+\ln \left (-6 x +1\right )+6 x}{\left (3 \ln \left (2\right )-\ln \left (3\right )+\ln \left (-6 x +1\right )\right )^{2} x^{2}}-\frac {6}{\left (3 \ln \left (2\right )-\ln \left (3\right )+\ln \left (-6 x +1\right )\right )^{2} x}-\frac {{\mathrm e}^{x^{2}}}{x}\]
Input:
int((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*ln(-16*x+8/3)^2+(-12*x+2)* ln(-16*x+8/3)-6*x)/(6*x^4-x^3)/ln(-16*x+8/3)^2,x)
Output:
1/x+(3*ln(2)-ln(3)+ln(-6*x+1)+6*x)/(3*ln(2)-ln(3)+ln(-6*x+1))^2/x^2-6/(3*l n(2)-ln(3)+ln(-6*x+1))^2/x-exp(x^2)/x
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=-\frac {{\left (x e^{\left (x^{2}\right )} - x\right )} \log \left (-16 \, x + \frac {8}{3}\right ) - 1}{x^{2} \log \left (-16 \, x + \frac {8}{3}\right )} \] Input:
integrate((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*log(-16*x+8/3)^2+(-1 2*x+2)*log(-16*x+8/3)-6*x)/(6*x^4-x^3)/log(-16*x+8/3)^2,x, algorithm="fric as")
Output:
-((x*e^(x^2) - x)*log(-16*x + 8/3) - 1)/(x^2*log(-16*x + 8/3))
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=- \frac {e^{x^{2}}}{x} + \frac {1}{x} + \frac {1}{x^{2} \log {\left (\frac {8}{3} - 16 x \right )}} \] Input:
integrate((((-12*x**4+2*x**3+6*x**2-x)*exp(x**2)-6*x**2+x)*ln(-16*x+8/3)** 2+(-12*x+2)*ln(-16*x+8/3)-6*x)/(6*x**4-x**3)/ln(-16*x+8/3)**2,x)
Output:
-exp(x**2)/x + 1/x + 1/(x**2*log(8/3 - 16*x))
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.86 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=-\frac {{\left (i \, \pi - \log \left (3\right ) + 3 \, \log \left (2\right )\right )} x e^{\left (x^{2}\right )} + {\left (-i \, \pi + \log \left (3\right ) - 3 \, \log \left (2\right )\right )} x + {\left (x e^{\left (x^{2}\right )} - x\right )} \log \left (6 \, x - 1\right ) - 1}{{\left (i \, \pi - \log \left (3\right ) + 3 \, \log \left (2\right )\right )} x^{2} + x^{2} \log \left (6 \, x - 1\right )} \] Input:
integrate((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*log(-16*x+8/3)^2+(-1 2*x+2)*log(-16*x+8/3)-6*x)/(6*x^4-x^3)/log(-16*x+8/3)^2,x, algorithm="maxi ma")
Output:
-((I*pi - log(3) + 3*log(2))*x*e^(x^2) + (-I*pi + log(3) - 3*log(2))*x + ( x*e^(x^2) - x)*log(6*x - 1) - 1)/((I*pi - log(3) + 3*log(2))*x^2 + x^2*log (6*x - 1))
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=-\frac {x e^{\left (x^{2}\right )} \log \left (-16 \, x + \frac {8}{3}\right ) - x \log \left (-16 \, x + \frac {8}{3}\right ) - 1}{x^{2} \log \left (-16 \, x + \frac {8}{3}\right )} \] Input:
integrate((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*log(-16*x+8/3)^2+(-1 2*x+2)*log(-16*x+8/3)-6*x)/(6*x^4-x^3)/log(-16*x+8/3)^2,x, algorithm="giac ")
Output:
-(x*e^(x^2)*log(-16*x + 8/3) - x*log(-16*x + 8/3) - 1)/(x^2*log(-16*x + 8/ 3))
Time = 1.43 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=\frac {1}{x^2\,\ln \left (\frac {8}{3}-16\,x\right )}-\frac {{\mathrm {e}}^{x^2}}{x}+\frac {1}{x} \] Input:
int((6*x + log(8/3 - 16*x)*(12*x - 2) + log(8/3 - 16*x)^2*(exp(x^2)*(x - 6 *x^2 - 2*x^3 + 12*x^4) - x + 6*x^2))/(log(8/3 - 16*x)^2*(x^3 - 6*x^4)),x)
Output:
1/(x^2*log(8/3 - 16*x)) - exp(x^2)/x + 1/x
Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx=\frac {-e^{x^{2}} \mathrm {log}\left (-16 x +\frac {8}{3}\right ) x +\mathrm {log}\left (-16 x +\frac {8}{3}\right ) x +1}{\mathrm {log}\left (-16 x +\frac {8}{3}\right ) x^{2}} \] Input:
int((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*log(-16*x+8/3)^2+(-12*x+2) *log(-16*x+8/3)-6*x)/(6*x^4-x^3)/log(-16*x+8/3)^2,x)
Output:
( - e**(x**2)*log(( - 48*x + 8)/3)*x + log(( - 48*x + 8)/3)*x + 1)/(log(( - 48*x + 8)/3)*x**2)