Integrand size = 176, antiderivative size = 28 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx=\frac {3}{x}+2 x+\left (4+\sqrt [5]{e}+x+\log (\log (-5-x+\log (x)))\right )^2 \] Output:
2*x+3/x+(exp(1/5)+ln(ln(ln(x)-5-x))+4+x)^2
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx=\frac {3}{x}+2 \left (5+\sqrt [5]{e}\right ) x+x^2+2 \left (4+\sqrt [5]{e}+x\right ) \log (\log (-5-x+\log (x)))+\log ^2(\log (-5-x+\log (x))) \] Input:
Integrate[(8*x - 6*x^2 - 2*x^3 + E^(1/5)*(2*x - 2*x^2) + (15 + 3*x - 50*x^ 2 - 20*x^3 - 2*x^4 + E^(1/5)*(-10*x^2 - 2*x^3) + (-3 + 10*x^2 + 2*E^(1/5)* x^2 + 2*x^3)*Log[x])*Log[-5 - x + Log[x]] + (2*x - 2*x^2 + (-10*x^2 - 2*x^ 3 + 2*x^2*Log[x])*Log[-5 - x + Log[x]])*Log[Log[-5 - x + Log[x]]])/((-5*x^ 2 - x^3 + x^2*Log[x])*Log[-5 - x + Log[x]]),x]
Output:
3/x + 2*(5 + E^(1/5))*x + x^2 + 2*(4 + E^(1/5) + x)*Log[Log[-5 - x + Log[x ]]] + Log[Log[-5 - x + Log[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.
\(\displaystyle \int \frac {-2 x^3-6 x^2+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (-2 x^2+\left (-2 x^3-10 x^2+2 x^2 \log (x)\right ) \log (-x+\log (x)-5)+2 x\right ) \log (\log (-x+\log (x)-5))+\left (-2 x^4-20 x^3-50 x^2+\sqrt [5]{e} \left (-2 x^3-10 x^2\right )+\left (2 x^3+2 \sqrt [5]{e} x^2+10 x^2-3\right ) \log (x)+3 x+15\right ) \log (-x+\log (x)-5)+8 x}{\left (-x^3-5 x^2+x^2 \log (x)\right ) \log (-x+\log (x)-5)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\frac {3}{x^2}+2 x+2 \log (\log (-x+\log (x)-5))+\frac {2 (x-1) \left (x+\log (\log (-x+\log (x)-5))+4 \left (1+\frac {\sqrt [5]{e}}{4}\right )\right )}{x (x-\log (x)+5) \log (-x+\log (x)-5)}+2 \left (5+\sqrt [5]{e}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (3+\sqrt [5]{e}\right ) \int \frac {1}{(x-\log (x)+5) \log (-x+\log (x)-5)}dx-2 \left (4+\sqrt [5]{e}\right ) \int \frac {1}{x (x-\log (x)+5) \log (-x+\log (x)-5)}dx+2 \int \frac {x}{(x-\log (x)+5) \log (-x+\log (x)-5)}dx+2 \int \log (\log (-x+\log (x)-5))dx+x^2+2 \left (5+\sqrt [5]{e}\right ) x+\frac {3}{x}+\log ^2(\log (-x+\log (x)-5))\) |
Input:
Int[(8*x - 6*x^2 - 2*x^3 + E^(1/5)*(2*x - 2*x^2) + (15 + 3*x - 50*x^2 - 20 *x^3 - 2*x^4 + E^(1/5)*(-10*x^2 - 2*x^3) + (-3 + 10*x^2 + 2*E^(1/5)*x^2 + 2*x^3)*Log[x])*Log[-5 - x + Log[x]] + (2*x - 2*x^2 + (-10*x^2 - 2*x^3 + 2* x^2*Log[x])*Log[-5 - x + Log[x]])*Log[Log[-5 - x + Log[x]]])/((-5*x^2 - x^ 3 + x^2*Log[x])*Log[-5 - x + Log[x]]),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(71\) vs. \(2(25)=50\).
Time = 8.79 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.57
method | result | size |
risch | \(\ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right )^{2}+2 \ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right ) x +\frac {2 \ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right ) x \,{\mathrm e}^{\frac {1}{5}}+2 x^{2} {\mathrm e}^{\frac {1}{5}}+x^{3}+8 \ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right ) x +10 x^{2}+3}{x}\) | \(72\) |
parallelrisch | \(\frac {\ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right )^{2} x +2 \ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right ) x \,{\mathrm e}^{\frac {1}{5}}+2 \ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right ) x^{2}+2 x^{2} {\mathrm e}^{\frac {1}{5}}+x^{3}+8 \ln \left (\ln \left (\ln \left (x \right )-5-x \right )\right ) x +10 x^{2}+3}{x}\) | \(75\) |
Input:
int((((2*x^2*ln(x)-2*x^3-10*x^2)*ln(ln(x)-5-x)-2*x^2+2*x)*ln(ln(ln(x)-5-x) )+((2*x^2*exp(1/5)+2*x^3+10*x^2-3)*ln(x)+(-2*x^3-10*x^2)*exp(1/5)-2*x^4-20 *x^3-50*x^2+3*x+15)*ln(ln(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2*x^3-6*x^2+8*x)/( x^2*ln(x)-x^3-5*x^2)/ln(ln(x)-5-x),x,method=_RETURNVERBOSE)
Output:
ln(ln(ln(x)-5-x))^2+2*ln(ln(ln(x)-5-x))*x+(2*ln(ln(ln(x)-5-x))*x*exp(1/5)+ 2*x^2*exp(1/5)+x^3+8*ln(ln(ln(x)-5-x))*x+10*x^2+3)/x
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx=\frac {x^{3} + 2 \, x^{2} e^{\frac {1}{5}} + x \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right )^{2} + 10 \, x^{2} + 2 \, {\left (x^{2} + x e^{\frac {1}{5}} + 4 \, x\right )} \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right ) + 3}{x} \] Input:
integrate((((2*x^2*log(x)-2*x^3-10*x^2)*log(log(x)-5-x)-2*x^2+2*x)*log(log (log(x)-5-x))+((2*x^2*exp(1/5)+2*x^3+10*x^2-3)*log(x)+(-2*x^3-10*x^2)*exp( 1/5)-2*x^4-20*x^3-50*x^2+3*x+15)*log(log(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2*x ^3-6*x^2+8*x)/(x^2*log(x)-x^3-5*x^2)/log(log(x)-5-x),x, algorithm="fricas" )
Output:
(x^3 + 2*x^2*e^(1/5) + x*log(log(-x + log(x) - 5))^2 + 10*x^2 + 2*(x^2 + x *e^(1/5) + 4*x)*log(log(-x + log(x) - 5)) + 3)/x
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (26) = 52\).
Time = 0.67 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.14 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx=x^{2} + 2 x \log {\left (\log {\left (- x + \log {\left (x \right )} - 5 \right )} \right )} + x \left (2 e^{\frac {1}{5}} + 10\right ) + \log {\left (\log {\left (- x + \log {\left (x \right )} - 5 \right )} \right )}^{2} + 2 \left (e^{\frac {1}{5}} + 4\right ) \log {\left (\log {\left (- x + \log {\left (x \right )} - 5 \right )} \right )} + \frac {3}{x} \] Input:
integrate((((2*x**2*ln(x)-2*x**3-10*x**2)*ln(ln(x)-5-x)-2*x**2+2*x)*ln(ln( ln(x)-5-x))+((2*x**2*exp(1/5)+2*x**3+10*x**2-3)*ln(x)+(-2*x**3-10*x**2)*ex p(1/5)-2*x**4-20*x**3-50*x**2+3*x+15)*ln(ln(x)-5-x)+(-2*x**2+2*x)*exp(1/5) -2*x**3-6*x**2+8*x)/(x**2*ln(x)-x**3-5*x**2)/ln(ln(x)-5-x),x)
Output:
x**2 + 2*x*log(log(-x + log(x) - 5)) + x*(2*exp(1/5) + 10) + log(log(-x + log(x) - 5))**2 + 2*(exp(1/5) + 4)*log(log(-x + log(x) - 5)) + 3/x
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (25) = 50\).
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx=\frac {x^{3} + 2 \, x^{2} {\left (e^{\frac {1}{5}} + 5\right )} + x \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right )^{2} + 2 \, {\left (x^{2} + x {\left (e^{\frac {1}{5}} + 4\right )}\right )} \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right ) + 3}{x} \] Input:
integrate((((2*x^2*log(x)-2*x^3-10*x^2)*log(log(x)-5-x)-2*x^2+2*x)*log(log (log(x)-5-x))+((2*x^2*exp(1/5)+2*x^3+10*x^2-3)*log(x)+(-2*x^3-10*x^2)*exp( 1/5)-2*x^4-20*x^3-50*x^2+3*x+15)*log(log(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2*x ^3-6*x^2+8*x)/(x^2*log(x)-x^3-5*x^2)/log(log(x)-5-x),x, algorithm="maxima" )
Output:
(x^3 + 2*x^2*(e^(1/5) + 5) + x*log(log(-x + log(x) - 5))^2 + 2*(x^2 + x*(e ^(1/5) + 4))*log(log(-x + log(x) - 5)) + 3)/x
Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (25) = 50\).
Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.64 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx=\frac {x^{3} + 2 \, x^{2} e^{\frac {1}{5}} + 2 \, x^{2} \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right ) + 2 \, x e^{\frac {1}{5}} \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right ) + x \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right )^{2} + 10 \, x^{2} + 8 \, x \log \left (\log \left (-x + \log \left (x\right ) - 5\right )\right ) + 3}{x} \] Input:
integrate((((2*x^2*log(x)-2*x^3-10*x^2)*log(log(x)-5-x)-2*x^2+2*x)*log(log (log(x)-5-x))+((2*x^2*exp(1/5)+2*x^3+10*x^2-3)*log(x)+(-2*x^3-10*x^2)*exp( 1/5)-2*x^4-20*x^3-50*x^2+3*x+15)*log(log(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2*x ^3-6*x^2+8*x)/(x^2*log(x)-x^3-5*x^2)/log(log(x)-5-x),x, algorithm="giac")
Output:
(x^3 + 2*x^2*e^(1/5) + 2*x^2*log(log(-x + log(x) - 5)) + 2*x*e^(1/5)*log(l og(-x + log(x) - 5)) + x*log(log(-x + log(x) - 5))^2 + 10*x^2 + 8*x*log(lo g(-x + log(x) - 5)) + 3)/x
Time = 2.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx={\ln \left (\ln \left (\ln \left (x\right )-x-5\right )\right )}^2+\ln \left (\ln \left (\ln \left (x\right )-x-5\right )\right )\,\left (2\,{\mathrm {e}}^{1/5}+8\right )+2\,x\,\ln \left (\ln \left (\ln \left (x\right )-x-5\right )\right )+\frac {3}{x}+x^2+x\,\left (2\,{\mathrm {e}}^{1/5}+10\right ) \] Input:
int((log(log(x) - x - 5)*(exp(1/5)*(10*x^2 + 2*x^3) - 3*x - log(x)*(2*x^2* exp(1/5) + 10*x^2 + 2*x^3 - 3) + 50*x^2 + 20*x^3 + 2*x^4 - 15) - 8*x - exp (1/5)*(2*x - 2*x^2) + log(log(log(x) - x - 5))*(log(log(x) - x - 5)*(10*x^ 2 - 2*x^2*log(x) + 2*x^3) - 2*x + 2*x^2) + 6*x^2 + 2*x^3)/(log(log(x) - x - 5)*(5*x^2 - x^2*log(x) + x^3)),x)
Output:
log(log(log(x) - x - 5))^2 + log(log(log(x) - x - 5))*(2*exp(1/5) + 8) + 2 *x*log(log(log(x) - x - 5)) + 3/x + x^2 + x*(2*exp(1/5) + 10)
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.71 \[ \int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx=\frac {2 e^{\frac {1}{5}} \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )-x -5\right )\right ) x +2 e^{\frac {1}{5}} x^{2}+\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )-x -5\right )\right )^{2} x +2 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )-x -5\right )\right ) x^{2}+8 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )-x -5\right )\right ) x +x^{3}+10 x^{2}+3}{x} \] Input:
int((((2*x^2*log(x)-2*x^3-10*x^2)*log(log(x)-5-x)-2*x^2+2*x)*log(log(log(x )-5-x))+((2*x^2*exp(1/5)+2*x^3+10*x^2-3)*log(x)+(-2*x^3-10*x^2)*exp(1/5)-2 *x^4-20*x^3-50*x^2+3*x+15)*log(log(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2*x^3-6*x ^2+8*x)/(x^2*log(x)-x^3-5*x^2)/log(log(x)-5-x),x)
Output:
(2*e**(1/5)*log(log(log(x) - x - 5))*x + 2*e**(1/5)*x**2 + log(log(log(x) - x - 5))**2*x + 2*log(log(log(x) - x - 5))*x**2 + 8*log(log(log(x) - x - 5))*x + x**3 + 10*x**2 + 3)/x