Integrand size = 66, antiderivative size = 32 \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\frac {1}{3} \left (2+x-\frac {e^e}{\log \left (\frac {x^2}{9}\right )}+5 \log (5-x+\log (x))\right ) \] Output:
5/3*ln(ln(x)-x+5)+2/3-1/3*exp(exp(1))/ln(1/9*x^2)+1/3*x
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\frac {1}{3} \left (x-\frac {e^e}{\log \left (\frac {x^2}{9}\right )}+5 \log (5-x+\log (x))\right ) \] Input:
Integrate[(E^E*(10 - 2*x) + 2*E^E*Log[x] + (5 - x^2 + x*Log[x])*Log[x^2/9] ^2)/((15*x - 3*x^2 + 3*x*Log[x])*Log[x^2/9]^2),x]
Output:
(x - E^E/Log[x^2/9] + 5*Log[5 - x + Log[x]])/3
Time = 1.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {7292, 27, 7293, 2009}
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.
\(\displaystyle \int \frac {\left (-x^2+x \log (x)+5\right ) \log ^2\left (\frac {x^2}{9}\right )+e^e (10-2 x)+2 e^e \log (x)}{\left (-3 x^2+15 x+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-x^2+x \log (x)+5\right ) \log ^2\left (\frac {x^2}{9}\right )+e^e (10-2 x)+2 e^e \log (x)}{3 x (-x+\log (x)+5) \log ^2\left (\frac {x^2}{9}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {\left (-x^2+\log (x) x+5\right ) \log ^2\left (\frac {x^2}{9}\right )+2 e^e (5-x)+2 e^e \log (x)}{x (-x+\log (x)+5) \log ^2\left (\frac {x^2}{9}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (\frac {x^2-\log (x) x-5}{x (x-\log (x)-5)}+\frac {2 e^e}{x \log ^2\left (\frac {x^2}{9}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {e^e}{\log \left (\frac {x^2}{9}\right )}+x+5 \log (-x+\log (x)+5)\right )\) |
Input:
Int[(E^E*(10 - 2*x) + 2*E^E*Log[x] + (5 - x^2 + x*Log[x])*Log[x^2/9]^2)/(( 15*x - 3*x^2 + 3*x*Log[x])*Log[x^2/9]^2),x]
Output:
(x - E^E/Log[x^2/9] + 5*Log[5 - x + Log[x]])/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25
method | result | size |
parallelrisch | \(-\frac {-5 \ln \left (x -\ln \left (x \right )-5\right ) \ln \left (\frac {x^{2}}{9}\right )-x \ln \left (\frac {x^{2}}{9}\right )+{\mathrm e}^{{\mathrm e}}}{3 \ln \left (\frac {x^{2}}{9}\right )}\) | \(40\) |
default | \(\frac {\left (2 \ln \left (3\right )+2 \ln \left (x \right )-\ln \left (x^{2}\right )\right ) x -2 x \ln \left (x \right )+{\mathrm e}^{{\mathrm e}}}{6 \ln \left (3\right )-3 \ln \left (x^{2}\right )}+\frac {5 \ln \left (\ln \left (x \right )-x +5\right )}{3}\) | \(53\) |
risch | \(\frac {x}{3}-\frac {2 i {\mathrm e}^{{\mathrm e}}}{3 \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-4 i \ln \left (3\right )+4 i \ln \left (x \right )\right )}+\frac {5 \ln \left (\ln \left (x \right )-x +5\right )}{3}\) | \(78\) |
Input:
int(((x*ln(x)-x^2+5)*ln(1/9*x^2)^2+2*exp(exp(1))*ln(x)+(-2*x+10)*exp(exp(1 )))/(3*x*ln(x)-3*x^2+15*x)/ln(1/9*x^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/3*(-5*ln(x-ln(x)-5)*ln(1/9*x^2)-x*ln(1/9*x^2)+exp(exp(1)))/ln(1/9*x^2)
Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\frac {2 \, x \log \left (3\right ) - 2 \, x \log \left (x\right ) + 10 \, {\left (\log \left (3\right ) - \log \left (x\right )\right )} \log \left (-x + \log \left (x\right ) + 5\right ) + e^{e}}{6 \, {\left (\log \left (3\right ) - \log \left (x\right )\right )}} \] Input:
integrate(((x*log(x)-x^2+5)*log(1/9*x^2)^2+2*exp(exp(1))*log(x)+(-2*x+10)* exp(exp(1)))/(3*x*log(x)-3*x^2+15*x)/log(1/9*x^2)^2,x, algorithm="fricas")
Output:
1/6*(2*x*log(3) - 2*x*log(x) + 10*(log(3) - log(x))*log(-x + log(x) + 5) + e^e)/(log(3) - log(x))
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\frac {x}{3} + \frac {5 \log {\left (- x + \log {\left (x \right )} + 5 \right )}}{3} - \frac {e^{e}}{6 \log {\left (x \right )} - 6 \log {\left (3 \right )}} \] Input:
integrate(((x*ln(x)-x**2+5)*ln(1/9*x**2)**2+2*exp(exp(1))*ln(x)+(-2*x+10)* exp(exp(1)))/(3*x*ln(x)-3*x**2+15*x)/ln(1/9*x**2)**2,x)
Output:
x/3 + 5*log(-x + log(x) + 5)/3 - exp(E)/(6*log(x) - 6*log(3))
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\frac {2 \, x \log \left (3\right ) - 2 \, x \log \left (x\right ) + e^{e}}{6 \, {\left (\log \left (3\right ) - \log \left (x\right )\right )}} + \frac {5}{3} \, \log \left (-x + \log \left (x\right ) + 5\right ) \] Input:
integrate(((x*log(x)-x^2+5)*log(1/9*x^2)^2+2*exp(exp(1))*log(x)+(-2*x+10)* exp(exp(1)))/(3*x*log(x)-3*x^2+15*x)/log(1/9*x^2)^2,x, algorithm="maxima")
Output:
1/6*(2*x*log(3) - 2*x*log(x) + e^e)/(log(3) - log(x)) + 5/3*log(-x + log(x ) + 5)
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\frac {2 \, x \log \left (3\right ) - 2 \, x \log \left (x\right ) + 10 \, \log \left (3\right ) \log \left (-x + \log \left (x\right ) + 5\right ) - 10 \, \log \left (x\right ) \log \left (-x + \log \left (x\right ) + 5\right ) + e^{e}}{6 \, {\left (\log \left (3\right ) - \log \left (x\right )\right )}} \] Input:
integrate(((x*log(x)-x^2+5)*log(1/9*x^2)^2+2*exp(exp(1))*log(x)+(-2*x+10)* exp(exp(1)))/(3*x*log(x)-3*x^2+15*x)/log(1/9*x^2)^2,x, algorithm="giac")
Output:
1/6*(2*x*log(3) - 2*x*log(x) + 10*log(3)*log(-x + log(x) + 5) - 10*log(x)* log(-x + log(x) + 5) + e^e)/(log(3) - log(x))
Timed out. \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\int \frac {\left (x\,\ln \left (x\right )-x^2+5\right )\,{\ln \left (\frac {x^2}{9}\right )}^2+2\,{\mathrm {e}}^{\mathrm {e}}\,\ln \left (x\right )-{\mathrm {e}}^{\mathrm {e}}\,\left (2\,x-10\right )}{{\ln \left (\frac {x^2}{9}\right )}^2\,\left (15\,x+3\,x\,\ln \left (x\right )-3\,x^2\right )} \,d x \] Input:
int((2*exp(exp(1))*log(x) - exp(exp(1))*(2*x - 10) + log(x^2/9)^2*(x*log(x ) - x^2 + 5))/(log(x^2/9)^2*(15*x + 3*x*log(x) - 3*x^2)),x)
Output:
int((2*exp(exp(1))*log(x) - exp(exp(1))*(2*x - 10) + log(x^2/9)^2*(x*log(x ) - x^2 + 5))/(log(x^2/9)^2*(15*x + 3*x*log(x) - 3*x^2)), x)
Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{\left (15 x-3 x^2+3 x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )} \, dx=\frac {-e^{e}+5 \,\mathrm {log}\left (\mathrm {log}\left (x \right )-x +5\right ) \mathrm {log}\left (\frac {x^{2}}{9}\right )+\mathrm {log}\left (\frac {x^{2}}{9}\right ) x}{3 \,\mathrm {log}\left (\frac {x^{2}}{9}\right )} \] Input:
int(((x*log(x)-x^2+5)*log(1/9*x^2)^2+2*exp(exp(1))*log(x)+(-2*x+10)*exp(ex p(1)))/(3*x*log(x)-3*x^2+15*x)/log(1/9*x^2)^2,x)
Output:
( - e**e + 5*log(log(x) - x + 5)*log(x**2/9) + log(x**2/9)*x)/(3*log(x**2/ 9))