Integrand size = 183, antiderivative size = 27 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \left (1-x+\log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )\right )^2 \]
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \left ((-2+x) x-2 (-1+x) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\log ^4\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )\right ) \]
Integrate[((-8*x + 8*x^2 + (3*x - 4*x^2 + x^3)*Log[x])*Log[(8 + (-3 + x)*L og[x])/3] + (-6 + 8*x - 2*x^2 + (2*x - 2*x^2)*Log[x])*Log[Log[(8 + (-3 + x )*Log[x])/3]] + (-8*x + (3*x - x^2)*Log[x])*Log[(8 + (-3 + x)*Log[x])/3]*L og[Log[(8 + (-3 + x)*Log[x])/3]]^2 + (-6 + 2*x + 2*x*Log[x])*Log[Log[(8 + (-3 + x)*Log[x])/3]]^3)/((16*x + (-6*x + 2*x^2)*Log[x])*Log[(8 + (-3 + x)* Log[x])/3]),x]
((-2 + x)*x - 2*(-1 + x)*Log[Log[(8 + (-3 + x)*Log[x])/3]]^2 + Log[Log[(8 + (-3 + x)*Log[x])/3]]^4)/4
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.
| \(\displaystyle \int \frac {\left (\left (3 x-x^2\right ) \log (x)-8 x\right ) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right ) \log ^2\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )+\left (-2 x^2+\left (2 x-2 x^2\right ) \log (x)+8 x-6\right ) \log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )+\left (8 x^2+\left (x^3-4 x^2+3 x\right ) \log (x)-8 x\right ) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )+(2 x+2 x \log (x)-6) \log ^3\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{\left (\left (2 x^2-6 x\right ) \log (x)+16 x\right ) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (x ((x-3) \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )-2 (x+x \log (x)-3) \log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )\right ) \left (x-\log ^2\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )-1\right )}{2 x ((x-3) \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -\frac {\left (x (8-(3-x) \log (x)) \log \left (\frac {1}{3} (8-(3-x) \log (x))\right )+2 (-\log (x) x-x+3) \log \left (\log \left (\frac {1}{3} (8-(3-x) \log (x))\right )\right )\right ) \left (\log ^2\left (\log \left (\frac {1}{3} (8-(3-x) \log (x))\right )\right )-x+1\right )}{x (8-(3-x) \log (x)) \log \left (\frac {1}{3} (8-(3-x) \log (x))\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {\left (x (8-(3-x) \log (x)) \log \left (\frac {1}{3} (8-(3-x) \log (x))\right )+2 (-\log (x) x-x+3) \log \left (\log \left (\frac {1}{3} (8-(3-x) \log (x))\right )\right )\right ) \left (\log ^2\left (\log \left (\frac {1}{3} (8-(3-x) \log (x))\right )\right )-x+1\right )}{x (8-(3-x) \log (x)) \log \left (\frac {1}{3} (8-(3-x) \log (x))\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{2} \int \left (-\frac {2 (\log (x) x+x-3) \log ^3\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{x (x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}+\log ^2\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )+\frac {2 (x-1) (\log (x) x+x-3) \log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{x (x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}-x+1\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (2 \int \frac {\log ^3\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{(x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx-6 \int \frac {\log ^3\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{x (x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx+2 \int \frac {\log (x) \log ^3\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{(x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx-\int \log ^2\left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )dx+8 \int \frac {\log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{(x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx-6 \int \frac {\log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{x (x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx-2 \int \frac {x \log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{(x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx+2 \int \frac {\log (x) \log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{(x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx-2 \int \frac {x \log (x) \log \left (\log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )\right )}{(x \log (x)-3 \log (x)+8) \log \left (\frac {1}{3} ((x-3) \log (x)+8)\right )}dx+\frac {x^2}{2}-x\right )\) |
Int[((-8*x + 8*x^2 + (3*x - 4*x^2 + x^3)*Log[x])*Log[(8 + (-3 + x)*Log[x]) /3] + (-6 + 8*x - 2*x^2 + (2*x - 2*x^2)*Log[x])*Log[Log[(8 + (-3 + x)*Log[ x])/3]] + (-8*x + (3*x - x^2)*Log[x])*Log[(8 + (-3 + x)*Log[x])/3]*Log[Log [(8 + (-3 + x)*Log[x])/3]]^2 + (-6 + 2*x + 2*x*Log[x])*Log[Log[(8 + (-3 + x)*Log[x])/3]]^3)/((16*x + (-6*x + 2*x^2)*Log[x])*Log[(8 + (-3 + x)*Log[x] )/3]),x]
3.24.88.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 3.86 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63
| method | result | size |
| risch | \(\frac {\ln \left (\ln \left (\frac {\ln \left (x \right ) \left (-3+x \right )}{3}+\frac {8}{3}\right )\right )^{4}}{4}+\left (-\frac {x}{2}+\frac {1}{2}\right ) \ln \left (\ln \left (\frac {\ln \left (x \right ) \left (-3+x \right )}{3}+\frac {8}{3}\right )\right )^{2}+\frac {x^{2}}{4}-\frac {x}{2}\) | \(44\) |
int(((2*x*ln(x)+2*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))^3+((-x^2+3*x)*ln(x)-8* x)*ln(1/3*ln(x)*(-3+x)+8/3)*ln(ln(1/3*ln(x)*(-3+x)+8/3))^2+((-2*x^2+2*x)*l n(x)-2*x^2+8*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))+((x^3-4*x^2+3*x)*ln(x)+8*x^ 2-8*x)*ln(1/3*ln(x)*(-3+x)+8/3))/((2*x^2-6*x)*ln(x)+16*x)/ln(1/3*ln(x)*(-3 +x)+8/3),x,method=_RETURNVERBOSE)
1/4*ln(ln(1/3*ln(x)*(-3+x)+8/3))^4+(-1/2*x+1/2)*ln(ln(1/3*ln(x)*(-3+x)+8/3 ))^2+1/4*x^2-1/2*x
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \, \log \left (\log \left (\frac {1}{3} \, {\left (x - 3\right )} \log \left (x\right ) + \frac {8}{3}\right )\right )^{4} - \frac {1}{2} \, {\left (x - 1\right )} \log \left (\log \left (\frac {1}{3} \, {\left (x - 3\right )} \log \left (x\right ) + \frac {8}{3}\right )\right )^{2} + \frac {1}{4} \, x^{2} - \frac {1}{2} \, x \]
integrate(((2*x*log(x)+2*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))^3+((-x^2+3*x )*log(x)-8*x)*log(1/3*log(x)*(-3+x)+8/3)*log(log(1/3*log(x)*(-3+x)+8/3))^2 +((-2*x^2+2*x)*log(x)-2*x^2+8*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))+((x^3-4 *x^2+3*x)*log(x)+8*x^2-8*x)*log(1/3*log(x)*(-3+x)+8/3))/((2*x^2-6*x)*log(x )+16*x)/log(1/3*log(x)*(-3+x)+8/3),x, algorithm=\
1/4*log(log(1/3*(x - 3)*log(x) + 8/3))^4 - 1/2*(x - 1)*log(log(1/3*(x - 3) *log(x) + 8/3))^2 + 1/4*x^2 - 1/2*x
Exception generated. \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\text {Exception raised: TypeError} \]
integrate(((2*x*ln(x)+2*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))**3+((-x**2+3*x)* ln(x)-8*x)*ln(1/3*ln(x)*(-3+x)+8/3)*ln(ln(1/3*ln(x)*(-3+x)+8/3))**2+((-2*x **2+2*x)*ln(x)-2*x**2+8*x-6)*ln(ln(1/3*ln(x)*(-3+x)+8/3))+((x**3-4*x**2+3* x)*ln(x)+8*x**2-8*x)*ln(1/3*ln(x)*(-3+x)+8/3))/((2*x**2-6*x)*ln(x)+16*x)/l n(1/3*ln(x)*(-3+x)+8/3),x)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (22) = 44\).
Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \, \log \left (-\log \left (3\right ) + \log \left ({\left (x - 3\right )} \log \left (x\right ) + 8\right )\right )^{4} - \frac {1}{2} \, {\left (x - 1\right )} \log \left (-\log \left (3\right ) + \log \left ({\left (x - 3\right )} \log \left (x\right ) + 8\right )\right )^{2} + \frac {1}{4} \, x^{2} - \frac {1}{2} \, x \]
integrate(((2*x*log(x)+2*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))^3+((-x^2+3*x )*log(x)-8*x)*log(1/3*log(x)*(-3+x)+8/3)*log(log(1/3*log(x)*(-3+x)+8/3))^2 +((-2*x^2+2*x)*log(x)-2*x^2+8*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))+((x^3-4 *x^2+3*x)*log(x)+8*x^2-8*x)*log(1/3*log(x)*(-3+x)+8/3))/((2*x^2-6*x)*log(x )+16*x)/log(1/3*log(x)*(-3+x)+8/3),x, algorithm=\
1/4*log(-log(3) + log((x - 3)*log(x) + 8))^4 - 1/2*(x - 1)*log(-log(3) + l og((x - 3)*log(x) + 8))^2 + 1/4*x^2 - 1/2*x
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 1.67 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx=\frac {1}{4} \, \log \left (-\log \left (3\right ) + \log \left (x \log \left (x\right ) - 3 \, \log \left (x\right ) + 8\right )\right )^{4} - \frac {1}{2} \, {\left (x - 1\right )} \log \left (-\log \left (3\right ) + \log \left (x \log \left (x\right ) - 3 \, \log \left (x\right ) + 8\right )\right )^{2} + \frac {1}{4} \, x^{2} - \frac {1}{2} \, x \]
integrate(((2*x*log(x)+2*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))^3+((-x^2+3*x )*log(x)-8*x)*log(1/3*log(x)*(-3+x)+8/3)*log(log(1/3*log(x)*(-3+x)+8/3))^2 +((-2*x^2+2*x)*log(x)-2*x^2+8*x-6)*log(log(1/3*log(x)*(-3+x)+8/3))+((x^3-4 *x^2+3*x)*log(x)+8*x^2-8*x)*log(1/3*log(x)*(-3+x)+8/3))/((2*x^2-6*x)*log(x )+16*x)/log(1/3*log(x)*(-3+x)+8/3),x, algorithm=\
1/4*log(-log(3) + log(x*log(x) - 3*log(x) + 8))^4 - 1/2*(x - 1)*log(-log(3 ) + log(x*log(x) - 3*log(x) + 8))^2 + 1/4*x^2 - 1/2*x
Time = 13.60 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {\left (-8 x+8 x^2+\left (3 x-4 x^2+x^3\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )+\left (-6+8 x-2 x^2+\left (2 x-2 x^2\right ) \log (x)\right ) \log \left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+\left (-8 x+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right ) \log ^2\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )+(-6+2 x+2 x \log (x)) \log ^3\left (\log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )\right )}{\left (16 x+\left (-6 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {1}{3} (8+(-3+x) \log (x))\right )} \, dx={\ln \left (\ln \left (\frac {\ln \left (x\right )\,\left (x-3\right )}{3}+\frac {8}{3}\right )\right )}^2\,\left (\frac {\frac {3\,x^2}{2}-\frac {x^3}{2}}{x\,\left (x-3\right )}+\frac {1}{2}\right )-\frac {x}{2}+\frac {{\ln \left (\ln \left (\frac {\ln \left (x\right )\,\left (x-3\right )}{3}+\frac {8}{3}\right )\right )}^4}{4}+\frac {x^2}{4} \]
int((log(log((log(x)*(x - 3))/3 + 8/3))^3*(2*x + 2*x*log(x) - 6) + log((lo g(x)*(x - 3))/3 + 8/3)*(log(x)*(3*x - 4*x^2 + x^3) - 8*x + 8*x^2) + log(lo g((log(x)*(x - 3))/3 + 8/3))*(8*x + log(x)*(2*x - 2*x^2) - 2*x^2 - 6) - lo g(log((log(x)*(x - 3))/3 + 8/3))^2*log((log(x)*(x - 3))/3 + 8/3)*(8*x - lo g(x)*(3*x - x^2)))/(log((log(x)*(x - 3))/3 + 8/3)*(16*x - log(x)*(6*x - 2* x^2))),x)