Integrand size = 96, antiderivative size = 29 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=x+(2-x) \left (5+\frac {2}{\left (3-\frac {x}{5}\right ) \log (\log (\log (2 x)))}\right ) \]
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-2 \left (2 x-\frac {5 (-2+x)}{(-15+x) \log (\log (\log (2 x)))}\right ) \]
Integrate[(-300 + 170*x - 10*x^2 - 130*x*Log[2*x]*Log[Log[2*x]]*Log[Log[Lo g[2*x]]] + (-900*x + 120*x^2 - 4*x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2 *x]]]^2)/((225*x - 30*x^2 + x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x]]] ^2),x]
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.
\(\displaystyle \int \frac {-10 x^2+\left (-4 x^3+120 x^2-900 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+170 x-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))-300}{\left (x^3-30 x^2+225 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-10 x^2+\left (-4 x^3+120 x^2-900 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+170 x-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))-300}{x \left (x^2-30 x+225\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {5 x^2+65 \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x))) x-85 x+2 \left (x^3-30 x^2+225 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+150}{2 (15-x)^2 x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -2 \int \frac {5 x^2+65 \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x))) x-85 x+2 \left (x^3-30 x^2+225 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+150}{(15-x)^2 x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {5 (x-2)}{(x-15) x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}+\frac {65}{(x-15)^2 \log (\log (\log (2 x)))}+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\frac {13}{3} \int \frac {1}{(x-15) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx+65 \int \frac {1}{(x-15)^2 \log (\log (\log (2 x)))}dx+2 x-\frac {2}{3 \log (\log (\log (2 x)))}\right )\) |
Int[(-300 + 170*x - 10*x^2 - 130*x*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x] ]] + (-900*x + 120*x^2 - 4*x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x]]]^ 2)/((225*x - 30*x^2 + x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x]]]^2),x]
3.14.57.3.1 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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 4.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-4 x +\frac {10 x -20}{\left (x -15\right ) \ln \left (\ln \left (\ln \left (2 x \right )\right )\right )}\) | \(23\) |
parallelrisch | \(\frac {-300-60 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right ) x^{2}+150 x +13500 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right )}{15 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right ) \left (x -15\right )}\) | \(40\) |
int(((-4*x^3+120*x^2-900*x)*ln(2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))^2-130*x*ln (2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))-10*x^2+170*x-300)/(x^3-30*x^2+225*x)/ln( 2*x)/ln(ln(2*x))/ln(ln(ln(2*x)))^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-\frac {2 \, {\left (2 \, {\left (x^{2} - 15 \, x\right )} \log \left (\log \left (\log \left (2 \, x\right )\right )\right ) - 5 \, x + 10\right )}}{{\left (x - 15\right )} \log \left (\log \left (\log \left (2 \, x\right )\right )\right )} \]
integrate(((-4*x^3+120*x^2-900*x)*log(2*x)*log(log(2*x))*log(log(log(2*x)) )^2-130*x*log(2*x)*log(log(2*x))*log(log(log(2*x)))-10*x^2+170*x-300)/(x^3 -30*x^2+225*x)/log(2*x)/log(log(2*x))/log(log(log(2*x)))^2,x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=- 4 x + \frac {10 x - 20}{\left (x - 15\right ) \log {\left (\log {\left (\log {\left (2 x \right )} \right )} \right )}} \]
integrate(((-4*x**3+120*x**2-900*x)*ln(2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))**2 -130*x*ln(2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))-10*x**2+170*x-300)/(x**3-30*x** 2+225*x)/ln(2*x)/ln(ln(2*x))/ln(ln(ln(2*x)))**2,x)
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-\frac {2 \, {\left (2 \, {\left (x^{2} - 15 \, x\right )} \log \left (\log \left (\log \left (2\right ) + \log \left (x\right )\right )\right ) - 5 \, x + 10\right )}}{{\left (x - 15\right )} \log \left (\log \left (\log \left (2\right ) + \log \left (x\right )\right )\right )} \]
integrate(((-4*x^3+120*x^2-900*x)*log(2*x)*log(log(2*x))*log(log(log(2*x)) )^2-130*x*log(2*x)*log(log(2*x))*log(log(log(2*x)))-10*x^2+170*x-300)/(x^3 -30*x^2+225*x)/log(2*x)/log(log(2*x))/log(log(log(2*x)))^2,x, algorithm=\
-2*(2*(x^2 - 15*x)*log(log(log(2) + log(x))) - 5*x + 10)/((x - 15)*log(log (log(2) + log(x))))
Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-4 \, x + \frac {10 \, {\left (x - 2\right )}}{x \log \left (\log \left (\log \left (2 \, x\right )\right )\right ) - 15 \, \log \left (\log \left (\log \left (2 \, x\right )\right )\right )} \]
integrate(((-4*x^3+120*x^2-900*x)*log(2*x)*log(log(2*x))*log(log(log(2*x)) )^2-130*x*log(2*x)*log(log(2*x))*log(log(log(2*x)))-10*x^2+170*x-300)/(x^3 -30*x^2+225*x)/log(2*x)/log(log(2*x))/log(log(log(2*x)))^2,x, algorithm=\
Time = 11.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=\frac {2\,\left (5\,x-2\,x^2\,\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )+30\,x\,\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )-10\right )}{\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )\,\left (x-15\right )} \]
int(-(10*x^2 - 170*x + log(log(log(2*x)))^2*log(2*x)*log(log(2*x))*(900*x - 120*x^2 + 4*x^3) + 130*x*log(log(log(2*x)))*log(2*x)*log(log(2*x)) + 300 )/(log(log(log(2*x)))^2*log(2*x)*log(log(2*x))*(225*x - 30*x^2 + x^3)),x)