\(\int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+(100 x^3+25 x^4) \log (-4-x)+(80 x^3+20 x^4+(40 x^3+10 x^4) \log (-4-x)) \log (2+\log (-4-x))+(8 x^3+2 x^4+(4 x^3+x^4) \log (-4-x)) \log ^2(2+\log (-4-x))} \, dx\) [2074]

Optimal result

Integrand size = 154, antiderivative size = 18 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^2 (5+\log (2+\log (-4-x)))} \]

[Out]

5/(5+ln(ln(-4-x)+2))/x^2

Rubi [F]

\[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx \]

[In]

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 +
Log[-4 - x]] + (8*x^3 + 2*x^4 + (4*x^3 + x^4)*Log[-4 - x])*Log[2 + Log[-4 - x]]^2),x]

[Out]

5/(16*(5 + Log[2 + Log[-4 - x]])) - (5*Defer[Int][1/(x^2*(2 + Log[-4 - x])*(5 + Log[2 + Log[-4 - x]])^2), x])/
4 + (5*Defer[Int][1/(x*(2 + Log[-4 - x])*(5 + Log[2 + Log[-4 - x]])^2), x])/16 - 10*Defer[Int][1/(x^3*(5 + Log
[2 + Log[-4 - x]])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 (-80-21 x-4 (4+x) \log (2+\log (-4-x))-2 (4+x) \log (-4-x) (5+\log (2+\log (-4-x))))}{x^3 (4+x) (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx \\ & = 5 \int \frac {-80-21 x-4 (4+x) \log (2+\log (-4-x))-2 (4+x) \log (-4-x) (5+\log (2+\log (-4-x)))}{x^3 (4+x) (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx \\ & = 5 \int \left (-\frac {1}{x^2 (4+x) (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2}-\frac {2}{x^3 (5+\log (2+\log (-4-x)))}\right ) \, dx \\ & = -\left (5 \int \frac {1}{x^2 (4+x) (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx\right )-10 \int \frac {1}{x^3 (5+\log (2+\log (-4-x)))} \, dx \\ & = -\left (5 \int \left (\frac {1}{4 x^2 (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2}-\frac {1}{16 x (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2}+\frac {1}{16 (4+x) (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2}\right ) \, dx\right )-10 \int \frac {1}{x^3 (5+\log (2+\log (-4-x)))} \, dx \\ & = \frac {5}{16} \int \frac {1}{x (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx-\frac {5}{16} \int \frac {1}{(4+x) (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx-\frac {5}{4} \int \frac {1}{x^2 (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx-10 \int \frac {1}{x^3 (5+\log (2+\log (-4-x)))} \, dx \\ & = \frac {5}{16 (5+\log (2+\log (-4-x)))}+\frac {5}{16} \int \frac {1}{x (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx-\frac {5}{4} \int \frac {1}{x^2 (2+\log (-4-x)) (5+\log (2+\log (-4-x)))^2} \, dx-10 \int \frac {1}{x^3 (5+\log (2+\log (-4-x)))} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^2 (5+\log (2+\log (-4-x)))} \]

[In]

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 + 25*x^4)*Log[-4 - x] + (80*x^3 + 20*x^4 + (40*x^3 + 10*x^4)*Log[-4 - x])*L
og[2 + Log[-4 - x]] + (8*x^3 + 2*x^4 + (4*x^3 + x^4)*Log[-4 - x])*Log[2 + Log[-4 - x]]^2),x]

[Out]

5/(x^2*(5 + Log[2 + Log[-4 - x]]))

Maple [A] (verified)

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	\(\frac {5}{\left (5+\ln \left (\ln \left (-4-x \right )+2\right )\right ) x^{2}}\)	\(19\)

[In]

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)

[Out]

5/(5+ln(ln(-4-x)+2))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log \left (\log \left (-x - 4\right ) + 2\right ) + 5 \, x^{2}} \]

[In]

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="fricas")

[Out]

5/(x^2*log(log(-x - 4) + 2) + 5*x^2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log {\left (\log {\left (- x - 4 \right )} + 2 \right )} + 5 x^{2}} \]

[In]

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)

[Out]

5/(x**2*log(log(-x - 4) + 2) + 5*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log \left (\log \left (-x - 4\right ) + 2\right ) + 5 \, x^{2}} \]

[In]

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="maxima")

[Out]

5/(x^2*log(log(-x - 4) + 2) + 5*x^2)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^{2} \log \left (\log \left (-x - 4\right ) + 2\right ) + 5 \, x^{2}} \]

[In]

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="giac")

[Out]

5/(x^2*log(log(-x - 4) + 2) + 5*x^2)

Mupad [B] (verification not implemented)

Time = 10.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-400-105 x+(-200-50 x) \log (-4-x)+(-80-20 x+(-40-10 x) \log (-4-x)) \log (2+\log (-4-x))}{200 x^3+50 x^4+\left (100 x^3+25 x^4\right ) \log (-4-x)+\left (80 x^3+20 x^4+\left (40 x^3+10 x^4\right ) \log (-4-x)\right ) \log (2+\log (-4-x))+\left (8 x^3+2 x^4+\left (4 x^3+x^4\right ) \log (-4-x)\right ) \log ^2(2+\log (-4-x))} \, dx=\frac {5}{x^2\,\left (\ln \left (\ln \left (-x-4\right )+2\right )+5\right )} \]

[In]

int(-(105*x + log(- x - 4)*(50*x + 200) + log(log(- x - 4) + 2)*(20*x + log(- 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)

[Out]

5/(x^2*(log(log(- x - 4) + 2) + 5))