Integrand size = 154, antiderivative size = 18 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^2 (5+\log (2+\log (-4-x)))} \]
Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^2 (5+\log (2+\log (-4-x)))} \]
Integrate[(-400 - 105*x + (-200 - 50*x)*Log[-4 - x] + (-80 - 20*x + (-40 - 10*x)*Log[-4 - x])*Log[2 + Log[-4 - x]])/(200*x^3 + 50*x^4 + (100*x^3 + 2 5*x^4)*Log[-4 - x] + (80*x^3 + 20*x^4 + (40*x^3 + 10*x^4)*Log[-4 - x])*Log [2 + Log[-4 - x]] + (8*x^3 + 2*x^4 + (4*x^3 + x^4)*Log[-4 - x])*Log[2 + Lo g[-4 - 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.
\(\displaystyle \int \frac {-105 x+(-50 x-200) \log (-x-4)+(-20 x+(-10 x-40) \log (-x-4)-80) \log (\log (-x-4)+2)-400}{50 x^4+200 x^3+\left (2 x^4+8 x^3+\left (x^4+4 x^3\right ) \log (-x-4)\right ) \log ^2(\log (-x-4)+2)+\left (25 x^4+100 x^3\right ) \log (-x-4)+\left (20 x^4+80 x^3+\left (10 x^4+40 x^3\right ) \log (-x-4)\right ) \log (\log (-x-4)+2)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 (-21 x-4 (x+4) \log (\log (-x-4)+2)-2 (x+4) \log (-x-4) (\log (\log (-x-4)+2)+5)-80)}{x^3 (x+4) (\log (-x-4)+2) (\log (\log (-x-4)+2)+5)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {21 x+4 (x+4) \log (\log (-x-4)+2)+2 (x+4) \log (-x-4) (\log (\log (-x-4)+2)+5)+80}{x^3 (x+4) (\log (-x-4)+2) (\log (\log (-x-4)+2)+5)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {21 x+4 (x+4) \log (\log (-x-4)+2)+2 (x+4) \log (-x-4) (\log (\log (-x-4)+2)+5)+80}{x^3 (x+4) (\log (-x-4)+2) (\log (\log (-x-4)+2)+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (\frac {2}{x^3 (\log (\log (-x-4)+2)+5)}+\frac {1}{x^2 (x+4) (\log (\log (-x-4)+2)+5)^2 (\log (-x-4)+2)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (2 \int \frac {1}{x^3 (\log (\log (-x-4)+2)+5)}dx+\frac {1}{4} \int \frac {1}{x^2 (\log (-x-4)+2) (\log (\log (-x-4)+2)+5)^2}dx-\frac {1}{16} \int \frac {1}{x (\log (-x-4)+2) (\log (\log (-x-4)+2)+5)^2}dx-\frac {1}{16 (\log (\log (-x-4)+2)+5)}\right )\) |
Int[(-400 - 105*x + (-200 - 50*x)*Log[-4 - x] + (-80 - 20*x + (-40 - 10*x) *Log[-4 - x])*Log[2 + Log[-4 - x]])/(200*x^3 + 50*x^4 + (100*x^3 + 25*x^4) *Log[-4 - x] + (80*x^3 + 20*x^4 + (40*x^3 + 10*x^4)*Log[-4 - x])*Log[2 + L og[-4 - x]] + (8*x^3 + 2*x^4 + (4*x^3 + x^4)*Log[-4 - x])*Log[2 + Log[-4 - x]]^2),x]
3.21.74.3.1 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 = 0.78 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06
method | result | size |
default | \(\frac {5}{\left (5+\ln \left (\ln \left (-4-x \right )+2\right )\right ) x^{2}}\) | \(19\) |
risch | \(\frac {5}{\left (5+\ln \left (\ln \left (-4-x \right )+2\right )\right ) x^{2}}\) | \(19\) |
parallelrisch | \(\frac {5}{\left (5+\ln \left (\ln \left (-4-x \right )+2\right )\right ) x^{2}}\) | \(19\) |
int((((-10*x-40)*ln(-4-x)-20*x-80)*ln(ln(-4-x)+2)+(-50*x-200)*ln(-4-x)-105 *x-400)/(((x^4+4*x^3)*ln(-4-x)+2*x^4+8*x^3)*ln(ln(-4-x)+2)^2+((10*x^4+40*x ^3)*ln(-4-x)+20*x^4+80*x^3)*ln(ln(-4-x)+2)+(25*x^4+100*x^3)*ln(-4-x)+50*x^ 4+200*x^3),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log \left (\log \left (-x - 4\right ) + 2\right ) + 5 \, x^{2}} \]
integrate((((-10*x-40)*log(-4-x)-20*x-80)*log(log(-4-x)+2)+(-50*x-200)*log (-4-x)-105*x-400)/(((x^4+4*x^3)*log(-4-x)+2*x^4+8*x^3)*log(log(-4-x)+2)^2+ ((10*x^4+40*x^3)*log(-4-x)+20*x^4+80*x^3)*log(log(-4-x)+2)+(25*x^4+100*x^3 )*log(-4-x)+50*x^4+200*x^3),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log {\left (\log {\left (- x - 4 \right )} + 2 \right )} + 5 x^{2}} \]
integrate((((-10*x-40)*ln(-4-x)-20*x-80)*ln(ln(-4-x)+2)+(-50*x-200)*ln(-4- x)-105*x-400)/(((x**4+4*x**3)*ln(-4-x)+2*x**4+8*x**3)*ln(ln(-4-x)+2)**2+(( 10*x**4+40*x**3)*ln(-4-x)+20*x**4+80*x**3)*ln(ln(-4-x)+2)+(25*x**4+100*x** 3)*ln(-4-x)+50*x**4+200*x**3),x)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log \left (\log \left (-x - 4\right ) + 2\right ) + 5 \, x^{2}} \]
integrate((((-10*x-40)*log(-4-x)-20*x-80)*log(log(-4-x)+2)+(-50*x-200)*log (-4-x)-105*x-400)/(((x^4+4*x^3)*log(-4-x)+2*x^4+8*x^3)*log(log(-4-x)+2)^2+ ((10*x^4+40*x^3)*log(-4-x)+20*x^4+80*x^3)*log(log(-4-x)+2)+(25*x^4+100*x^3 )*log(-4-x)+50*x^4+200*x^3),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log \left (\log \left (-x - 4\right ) + 2\right ) + 5 \, x^{2}} \]
integrate((((-10*x-40)*log(-4-x)-20*x-80)*log(log(-4-x)+2)+(-50*x-200)*log (-4-x)-105*x-400)/(((x^4+4*x^3)*log(-4-x)+2*x^4+8*x^3)*log(log(-4-x)+2)^2+ ((10*x^4+40*x^3)*log(-4-x)+20*x^4+80*x^3)*log(log(-4-x)+2)+(25*x^4+100*x^3 )*log(-4-x)+50*x^4+200*x^3),x, algorithm=\
Time = 10.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^2\,\left (\ln \left (\ln \left (-x-4\right )+2\right )+5\right )} \]
int(-(105*x + log(- x - 4)*(50*x + 200) + log(log(- x - 4) + 2)*(20*x + lo g(- x - 4)*(10*x + 40) + 80) + 400)/(log(log(- x - 4) + 2)*(log(- x - 4)*( 40*x^3 + 10*x^4) + 80*x^3 + 20*x^4) + log(log(- x - 4) + 2)^2*(log(- x - 4 )*(4*x^3 + x^4) + 8*x^3 + 2*x^4) + log(- x - 4)*(100*x^3 + 25*x^4) + 200*x ^3 + 50*x^4),x)