Integrand size = 56, antiderivative size = 27 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=\log \left (6-4 e^{-\frac {1}{3} x \left (5-e^4+3 x-\log (x)\right )} x\right ) \] Output:
ln(6-4/exp(1/3*x*(3*x+5-exp(4)-ln(x)))*x)
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(27)=54\).
Time = 5.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=-\frac {1}{3} x \left (6-e^4+3 x-\log (x)\right )+\frac {1}{3} \left (x+3 \log \left (-2 x+3 e^{-\frac {1}{3} \left (-5+e^4-3 x\right ) x} x^{-x/3}\right )\right ) \] Input:
Integrate[(-6 + 8*x - 2*E^4*x + 12*x^2 - 2*x*Log[x])/(9*E^((5*x - E^4*x + 3*x^2 - x*Log[x])/3) - 6*x),x]
Output:
-1/3*(x*(6 - E^4 + 3*x - Log[x])) + (x + 3*Log[-2*x + 3/(E^(((-5 + E^4 - 3 *x)*x)/3)*x^(x/3))])/3
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.
\(\displaystyle \int \frac {12 x^2-2 e^4 x+8 x-2 x \log (x)-6}{9 e^{\frac {1}{3} \left (3 x^2-e^4 x+5 x-x \log (x)\right )}-6 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {12 x^2+\left (8-2 e^4\right ) x-2 x \log (x)-6}{9 e^{\frac {1}{3} \left (3 x^2-e^4 x+5 x-x \log (x)\right )}-6 x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (6 x^2+4 \left (1-\frac {e^4}{4}\right ) x-x \log (x)-3\right )}{9 e^{\frac {1}{3} \left (3 x^2-e^4 x+5 x-x \log (x)\right )}-6 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {-6 x^2+\log (x) x-\left (4-e^4\right ) x+3}{3 \left (2 x-3 e^{\frac {1}{3} \left (3 x^2-e^4 x+5 x\right )} x^{-x/3}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \int \frac {-6 x^2+\log (x) x-\left (4-e^4\right ) x+3}{2 x-3 e^{\frac {1}{3} \left (3 x^2-e^4 x+5 x\right )} x^{-x/3}}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{3} \int \left (\frac {3 e^{x^2+\frac {5 x}{3}} \left (6 x^2-\log (x) x+4 \left (1-\frac {e^4}{4}\right ) x-3\right )}{2 x \left (3 e^{x^2+\frac {5 x}{3}}-2 e^{\frac {e^4 x}{3}} x^{\frac {x}{3}+1}\right )}+\frac {-6 x^2+\log (x) x-4 \left (1-\frac {e^4}{4}\right ) x+3}{2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{3} \left (\frac {3}{2} \left (4-e^4\right ) \int \frac {e^{x^2+\frac {5 x}{3}}}{3 e^{x^2+\frac {5 x}{3}}-2 e^{\frac {e^4 x}{3}} x^{\frac {x}{3}+1}}dx+\frac {9}{2} \int \frac {e^{x^2+\frac {5 x}{3}}}{x \left (2 e^{\frac {e^4 x}{3}} x^{\frac {x}{3}+1}-3 e^{x^2+\frac {5 x}{3}}\right )}dx-9 \int \frac {e^{x^2+\frac {5 x}{3}} x}{2 e^{\frac {e^4 x}{3}} x^{\frac {x}{3}+1}-3 e^{x^2+\frac {5 x}{3}}}dx+\frac {3}{2} \int \frac {\int \frac {e^{x^2+\frac {5 x}{3}}}{3 e^{x^2+\frac {5 x}{3}}-2 e^{\frac {e^4 x}{3}} x^{\frac {x}{3}+1}}dx}{x}dx-\frac {3}{2} \log (x) \int \frac {e^{x^2+\frac {5 x}{3}}}{3 e^{x^2+\frac {5 x}{3}}-2 e^{\frac {e^4 x}{3}} x^{\frac {x}{3}+1}}dx-\frac {3 x^2}{2}-\frac {1}{2} \left (4-e^4\right ) x-\frac {x}{2}+\frac {1}{2} x \log (x)+\frac {3 \log (x)}{2}\right )\) |
Input:
Int[(-6 + 8*x - 2*E^4*x + 12*x^2 - 2*x*Log[x])/(9*E^((5*x - E^4*x + 3*x^2 - x*Log[x])/3) - 6*x),x]
Output:
$Aborted
Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41
method | result | size |
parallelrisch | \(\frac {x \,{\mathrm e}^{4}}{3}-x^{2}+\frac {x \ln \left (x \right )}{3}-\frac {5 x}{3}+\ln \left (-\frac {3 \,{\mathrm e}^{-\frac {x \left (\ln \left (x \right )+{\mathrm e}^{4}-3 x -5\right )}{3}}}{2}+x \right )\) | \(38\) |
risch | \(\frac {x \ln \left (x \right )}{3}+\frac {x \,{\mathrm e}^{4}}{3}-x^{2}-\frac {5 x}{3}+\ln \left (x^{-\frac {x}{3}} {\mathrm e}^{-\frac {x \left ({\mathrm e}^{4}-3 x -5\right )}{3}}-\frac {2 x}{3}\right )\) | \(42\) |
norman | \(\left (-\frac {5}{3}+\frac {{\mathrm e}^{4}}{3}\right ) x -x^{2}+\frac {x \ln \left (x \right )}{3}+\ln \left (-9 \,{\mathrm e}^{-\frac {x \ln \left (x \right )}{3}-\frac {x \,{\mathrm e}^{4}}{3}+x^{2}+\frac {5 x}{3}}+6 x \right )\) | \(45\) |
Input:
int((-2*x*ln(x)-2*x*exp(4)+12*x^2+8*x-6)/(9*exp(-1/3*x*ln(x)-1/3*x*exp(4)+ x^2+5/3*x)-6*x),x,method=_RETURNVERBOSE)
Output:
1/3*x*exp(4)-x^2+1/3*x*ln(x)-5/3*x+ln(-3/2*exp(-1/3*x*(ln(x)+exp(4)-3*x-5) )+x)
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=-x^{2} + \frac {1}{3} \, x e^{4} + \frac {1}{3} \, x \log \left (x\right ) - \frac {5}{3} \, x + \log \left (-2 \, x + 3 \, e^{\left (x^{2} - \frac {1}{3} \, x e^{4} - \frac {1}{3} \, x \log \left (x\right ) + \frac {5}{3} \, x\right )}\right ) \] Input:
integrate((-2*x*log(x)-2*x*exp(4)+12*x^2+8*x-6)/(9*exp(-1/3*x*log(x)-1/3*x *exp(4)+x^2+5/3*x)-6*x),x, algorithm="fricas")
Output:
-x^2 + 1/3*x*e^4 + 1/3*x*log(x) - 5/3*x + log(-2*x + 3*e^(x^2 - 1/3*x*e^4 - 1/3*x*log(x) + 5/3*x))
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=- x^{2} + \frac {x \log {\left (x \right )}}{3} + x \left (- \frac {5}{3} + \frac {e^{4}}{3}\right ) + \log {\left (- \frac {2 x}{3} + e^{x^{2} - \frac {x \log {\left (x \right )}}{3} - \frac {x e^{4}}{3} + \frac {5 x}{3}} \right )} \] Input:
integrate((-2*x*ln(x)-2*x*exp(4)+12*x**2+8*x-6)/(9*exp(-1/3*x*ln(x)-1/3*x* exp(4)+x**2+5/3*x)-6*x),x)
Output:
-x**2 + x*log(x)/3 + x*(-5/3 + exp(4)/3) + log(-2*x/3 + exp(x**2 - x*log(x )/3 - x*exp(4)/3 + 5*x/3))
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=-x^{2} + \frac {1}{3} \, {\left (x + 3\right )} \log \left (x\right ) - \frac {5}{3} \, x + \log \left (\frac {2 \, x e^{\left (\frac {1}{3} \, x e^{4} + \frac {1}{3} \, x \log \left (x\right )\right )} - 3 \, e^{\left (x^{2} + \frac {5}{3} \, x\right )}}{2 \, x x^{\frac {1}{3} \, x}}\right ) \] Input:
integrate((-2*x*log(x)-2*x*exp(4)+12*x^2+8*x-6)/(9*exp(-1/3*x*log(x)-1/3*x *exp(4)+x^2+5/3*x)-6*x),x, algorithm="maxima")
Output:
-x^2 + 1/3*(x + 3)*log(x) - 5/3*x + log(1/2*(2*x*e^(1/3*x*e^4 + 1/3*x*log( x)) - 3*e^(x^2 + 5/3*x))/(x*x^(1/3*x)))
Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=-x^{2} + \frac {1}{3} \, x e^{4} + \frac {1}{3} \, x \log \left (x\right ) - \frac {5}{3} \, x + \log \left (2 \, x - 3 \, e^{\left (x^{2} - \frac {1}{3} \, x e^{4} - \frac {1}{3} \, x \log \left (x\right ) + \frac {5}{3} \, x\right )}\right ) \] Input:
integrate((-2*x*log(x)-2*x*exp(4)+12*x^2+8*x-6)/(9*exp(-1/3*x*log(x)-1/3*x *exp(4)+x^2+5/3*x)-6*x),x, algorithm="giac")
Output:
-x^2 + 1/3*x*e^4 + 1/3*x*log(x) - 5/3*x + log(2*x - 3*e^(x^2 - 1/3*x*e^4 - 1/3*x*log(x) + 5/3*x))
Time = 3.75 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=\ln \left (x-\frac {3\,{\mathrm {e}}^{x^2}\,{\left ({\mathrm {e}}^x\right )}^{5/3}}{2\,x^{x/3}\,{\left ({\mathrm {e}}^{x\,{\mathrm {e}}^4}\right )}^{1/3}}\right )+\frac {x\,\ln \left (x\right )}{3}-x^2+x\,\left (\frac {{\mathrm {e}}^4}{3}-\frac {5}{3}\right ) \] Input:
int((2*x*exp(4) - 8*x + 2*x*log(x) - 12*x^2 + 6)/(6*x - 9*exp((5*x)/3 - (x *exp(4))/3 - (x*log(x))/3 + x^2)),x)
Output:
log(x - (3*exp(x^2)*exp(x)^(5/3))/(2*x^(x/3)*exp(x*exp(4))^(1/3))) + (x*lo g(x))/3 - x^2 + x*(exp(4)/3 - 5/3)
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {-6+8 x-2 e^4 x+12 x^2-2 x \log (x)}{9 e^{\frac {1}{3} \left (5 x-e^4 x+3 x^2-x \log (x)\right )}-6 x} \, dx=\mathrm {log}\left (3 e^{x^{2}+\frac {5}{3} x}-2 x^{\frac {x}{3}} e^{\frac {e^{4} x}{3}} x \right )-x^{2}-\frac {5 x}{3} \] Input:
int((-2*x*log(x)-2*x*exp(4)+12*x^2+8*x-6)/(9*exp(-1/3*x*log(x)-1/3*x*exp(4 )+x^2+5/3*x)-6*x),x)
Output:
(3*log(3*e**((3*x**2 + 5*x)/3) - 2*x**(x/3)*e**((e**4*x)/3)*x) - 3*x**2 - 5*x)/3