Integrand size = 106, antiderivative size = 20 \[ \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx=-x+x^3 \log ^2(1-x+\log (4+x)) \]
\[ \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx=\int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx \]
Integrate[(-4 + 3*x + x^2 + (-4 - x)*Log[4 + x] + (-6*x^3 - 2*x^4)*Log[1 - x + Log[4 + x]] + (12*x^2 - 9*x^3 - 3*x^4 + (12*x^2 + 3*x^3)*Log[4 + x])* Log[1 - x + Log[4 + x]]^2)/(4 - 3*x - x^2 + (4 + x)*Log[4 + x]),x]
Integrate[(-4 + 3*x + x^2 + (-4 - x)*Log[4 + x] + (-6*x^3 - 2*x^4)*Log[1 - x + Log[4 + x]] + (12*x^2 - 9*x^3 - 3*x^4 + (12*x^2 + 3*x^3)*Log[4 + x])* Log[1 - x + Log[4 + x]]^2)/(4 - 3*x - x^2 + (4 + x)*Log[4 + x]), 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.
\(\displaystyle \int \frac {x^2+\left (-2 x^4-6 x^3\right ) \log (-x+\log (x+4)+1)+\left (-3 x^4-9 x^3+12 x^2+\left (3 x^3+12 x^2\right ) \log (x+4)\right ) \log ^2(-x+\log (x+4)+1)+3 x+(-x-4) \log (x+4)-4}{-x^2-3 x+(x+4) \log (x+4)+4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^2+\left (-2 x^4-6 x^3\right ) \log (-x+\log (x+4)+1)+\left (-3 x^4-9 x^3+12 x^2+\left (3 x^3+12 x^2\right ) \log (x+4)\right ) \log ^2(-x+\log (x+4)+1)+3 x+(-x-4) \log (x+4)-4}{(x+4) (-x+\log (x+4)+1)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 (x+3) x^3 \log (-x+\log (x+4)+1)}{(x+4) (x-\log (x+4)-1)}+3 x^2 \log ^2(-x+\log (x+4)+1)-\frac {x^2}{(x+4) (x-\log (x+4)-1)}-\frac {3 x}{(x+4) (x-\log (x+4)-1)}+\frac {\log (x+4)}{x-\log (x+4)-1}+\frac {4}{(x+4) (x-\log (x+4)-1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {x^3 \log (-x+\log (x+4)+1)}{x-\log (x+4)-1}dx+3 \int x^2 \log ^2(-x+\log (x+4)+1)dx-2 \int \frac {x^2 \log (-x+\log (x+4)+1)}{x-\log (x+4)-1}dx+\int \frac {1}{x-\log (x+4)-1}dx+\int \frac {1}{-x+\log (x+4)+1}dx-32 \int \frac {\log (-x+\log (x+4)+1)}{x-\log (x+4)-1}dx+8 \int \frac {x \log (-x+\log (x+4)+1)}{x-\log (x+4)-1}dx+128 \int \frac {\log (-x+\log (x+4)+1)}{(x+4) (x-\log (x+4)-1)}dx-x\) |
Int[(-4 + 3*x + x^2 + (-4 - x)*Log[4 + x] + (-6*x^3 - 2*x^4)*Log[1 - x + L og[4 + x]] + (12*x^2 - 9*x^3 - 3*x^4 + (12*x^2 + 3*x^3)*Log[4 + x])*Log[1 - x + Log[4 + x]]^2)/(4 - 3*x - x^2 + (4 + x)*Log[4 + x]),x]
3.19.14.3.1 Defintions of rubi rules used
Time = 0.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05
method | result | size |
risch | \(x^{3} \ln \left (\ln \left (4+x \right )-x +1\right )^{2}-x\) | \(21\) |
parallelrisch | \(x^{3} \ln \left (\ln \left (4+x \right )-x +1\right )^{2}+1-x\) | \(22\) |
int((((3*x^3+12*x^2)*ln(4+x)-3*x^4-9*x^3+12*x^2)*ln(ln(4+x)-x+1)^2+(-2*x^4 -6*x^3)*ln(ln(4+x)-x+1)+(-4-x)*ln(4+x)+x^2+3*x-4)/((4+x)*ln(4+x)-x^2-3*x+4 ),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx=x^{3} \log \left (-x + \log \left (x + 4\right ) + 1\right )^{2} - x \]
integrate((((3*x^3+12*x^2)*log(4+x)-3*x^4-9*x^3+12*x^2)*log(log(4+x)-x+1)^ 2+(-2*x^4-6*x^3)*log(log(4+x)-x+1)+(-4-x)*log(4+x)+x^2+3*x-4)/((4+x)*log(4 +x)-x^2-3*x+4),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx=x^{3} \log {\left (- x + \log {\left (x + 4 \right )} + 1 \right )}^{2} - x \]
integrate((((3*x**3+12*x**2)*ln(4+x)-3*x**4-9*x**3+12*x**2)*ln(ln(4+x)-x+1 )**2+(-2*x**4-6*x**3)*ln(ln(4+x)-x+1)+(-4-x)*ln(4+x)+x**2+3*x-4)/((4+x)*ln (4+x)-x**2-3*x+4),x)
Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx=x^{3} \log \left (-x + \log \left (x + 4\right ) + 1\right )^{2} - x \]
integrate((((3*x^3+12*x^2)*log(4+x)-3*x^4-9*x^3+12*x^2)*log(log(4+x)-x+1)^ 2+(-2*x^4-6*x^3)*log(log(4+x)-x+1)+(-4-x)*log(4+x)+x^2+3*x-4)/((4+x)*log(4 +x)-x^2-3*x+4),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx=x^{3} \log \left (-x + \log \left (x + 4\right ) + 1\right )^{2} - x \]
integrate((((3*x^3+12*x^2)*log(4+x)-3*x^4-9*x^3+12*x^2)*log(log(4+x)-x+1)^ 2+(-2*x^4-6*x^3)*log(log(4+x)-x+1)+(-4-x)*log(4+x)+x^2+3*x-4)/((4+x)*log(4 +x)-x^2-3*x+4),x, algorithm=\
Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4+3 x+x^2+(-4-x) \log (4+x)+\left (-6 x^3-2 x^4\right ) \log (1-x+\log (4+x))+\left (12 x^2-9 x^3-3 x^4+\left (12 x^2+3 x^3\right ) \log (4+x)\right ) \log ^2(1-x+\log (4+x))}{4-3 x-x^2+(4+x) \log (4+x)} \, dx=x^3\,{\ln \left (\ln \left (x+4\right )-x+1\right )}^2-x \]