Integrand size = 95, antiderivative size = 23 \[ \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx=\log \left (\frac {-5+\frac {x}{\frac {1}{2}+\log (2+2 x)}}{-3+x}\right ) \]
\[ \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx=\int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx \]
Integrate[(1 - 11*x + 4*x^2 + (-8 - 8*x)*Log[2 + 2*x] + (-20 - 20*x)*Log[2 + 2*x]^2)/(-15 - 4*x + 9*x^2 - 2*x^3 + (-60 - 28*x + 28*x^2 - 4*x^3)*Log[ 2 + 2*x] + (-60 - 40*x + 20*x^2)*Log[2 + 2*x]^2),x]
Integrate[(1 - 11*x + 4*x^2 + (-8 - 8*x)*Log[2 + 2*x] + (-20 - 20*x)*Log[2 + 2*x]^2)/(-15 - 4*x + 9*x^2 - 2*x^3 + (-60 - 28*x + 28*x^2 - 4*x^3)*Log[ 2 + 2*x] + (-60 - 40*x + 20*x^2)*Log[2 + 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.
| \(\displaystyle \int \frac {4 x^2-11 x+(-20 x-20) \log ^2(2 x+2)+(-8 x-8) \log (2 x+2)+1}{-2 x^3+9 x^2+\left (20 x^2-40 x-60\right ) \log ^2(2 x+2)+\left (-4 x^3+28 x^2-28 x-60\right ) \log (2 x+2)-4 x-15} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-4 x^2+11 x-(-20 x-20) \log ^2(2 x+2)-(-8 x-8) \log (2 x+2)-1}{\left (-x^2+2 x+3\right ) (2 \log (2 (x+1))+1) (-2 x+10 \log (2 (x+1))+5)}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (-\frac {4 x^2}{(x-3) (x+1) (2 \log (x+1)+1+\log (4)) (2 x-10 \log (2 (x+1))-5)}+\frac {20 \log ^2(2 x+2)}{(3-x) (2 \log (x+1)+1+\log (4)) (-2 x+10 \log (2 (x+1))+5)}+\frac {11 x}{(x-3) (x+1) (2 \log (x+1)+1+\log (4)) (2 x-10 \log (2 (x+1))-5)}+\frac {8 \log (2 x+2)}{(3-x) (2 \log (x+1)+1+\log (4)) (-2 x+10 \log (2 (x+1))+5)}-\frac {1}{(x-3) (x+1) (2 \log (x+1)+1+\log (4)) (2 x-10 \log (2 (x+1))-5)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 20 \int \frac {\log ^2(2 x+2)}{(3-x) (2 \log (x+1)+\log (4)+1) (-2 x+10 \log (2 (x+1))+5)}dx-4 \int \frac {1}{(2 \log (x+1)+\log (4)+1) (2 x-10 \log (2 (x+1))-5)}dx-\int \frac {1}{(x-3) (2 \log (x+1)+\log (4)+1) (2 x-10 \log (2 (x+1))-5)}dx+4 \int \frac {1}{(x+1) (2 \log (x+1)+\log (4)+1) (2 x-10 \log (2 (x+1))-5)}dx+8 \int \frac {\log (2 x+2)}{(3-x) (2 \log (x+1)+\log (4)+1) (-2 x+10 \log (2 (x+1))+5)}dx\) |
Int[(1 - 11*x + 4*x^2 + (-8 - 8*x)*Log[2 + 2*x] + (-20 - 20*x)*Log[2 + 2*x ]^2)/(-15 - 4*x + 9*x^2 - 2*x^3 + (-60 - 28*x + 28*x^2 - 4*x^3)*Log[2 + 2* x] + (-60 - 40*x + 20*x^2)*Log[2 + 2*x]^2),x]
3.7.18.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35
| method | result | size |
| risch | \(-\ln \left (-3+x \right )+\ln \left (\ln \left (2+2 x \right )+\frac {1}{2}-\frac {x}{5}\right )-\ln \left (\frac {1}{2}+\ln \left (2+2 x \right )\right )\) | \(31\) |
| parallelrisch | \(-\ln \left (-3+x \right )-\ln \left (\frac {1}{2}+\ln \left (2+2 x \right )\right )+\ln \left (x -5 \ln \left (2+2 x \right )-\frac {5}{2}\right )\) | \(31\) |
| norman | \(-\ln \left (-3+x \right )-\ln \left (2 \ln \left (2+2 x \right )+1\right )+\ln \left (2 x -10 \ln \left (2+2 x \right )-5\right )\) | \(35\) |
| derivativedivides | \(\ln \left (10 \ln \left (2+2 x \right )+5-2 x \right )-\ln \left (2 \ln \left (2+2 x \right )+1\right )-\ln \left (2 x -6\right )\) | \(37\) |
| default | \(\ln \left (10 \ln \left (2+2 x \right )+5-2 x \right )-\ln \left (2 \ln \left (2+2 x \right )+1\right )-\ln \left (2 x -6\right )\) | \(37\) |
int(((-20*x-20)*ln(2+2*x)^2+(-8*x-8)*ln(2+2*x)+4*x^2-11*x+1)/((20*x^2-40*x -60)*ln(2+2*x)^2+(-4*x^3+28*x^2-28*x-60)*ln(2+2*x)-2*x^3+9*x^2-4*x-15),x,m ethod=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx=-\log \left (x - 3\right ) + \log \left (-2 \, x + 10 \, \log \left (2 \, x + 2\right ) + 5\right ) - \log \left (2 \, \log \left (2 \, x + 2\right ) + 1\right ) \]
integrate(((-20*x-20)*log(2+2*x)^2+(-8*x-8)*log(2+2*x)+4*x^2-11*x+1)/((20* x^2-40*x-60)*log(2+2*x)^2+(-4*x^3+28*x^2-28*x-60)*log(2+2*x)-2*x^3+9*x^2-4 *x-15),x, algorithm=\
Exception generated. \[ \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx=\text {Exception raised: PolynomialError} \]
integrate(((-20*x-20)*ln(2+2*x)**2+(-8*x-8)*ln(2+2*x)+4*x**2-11*x+1)/((20* x**2-40*x-60)*ln(2+2*x)**2+(-4*x**3+28*x**2-28*x-60)*ln(2+2*x)-2*x**3+9*x* *2-4*x-15),x)
Exception raised: PolynomialError >> 1/(5*x + 5) contains an element of th e set of generators.
Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx=-\log \left (x - 3\right ) + \log \left (-\frac {1}{5} \, x + \log \left (2\right ) + \log \left (x + 1\right ) + \frac {1}{2}\right ) - \log \left (\log \left (2\right ) + \log \left (x + 1\right ) + \frac {1}{2}\right ) \]
integrate(((-20*x-20)*log(2+2*x)^2+(-8*x-8)*log(2+2*x)+4*x^2-11*x+1)/((20* x^2-40*x-60)*log(2+2*x)^2+(-4*x^3+28*x^2-28*x-60)*log(2+2*x)-2*x^3+9*x^2-4 *x-15),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx=-\log \left (x - 3\right ) + \log \left (-2 \, x + 10 \, \log \left (2 \, x + 2\right ) + 5\right ) - \log \left (2 \, \log \left (2 \, x + 2\right ) + 1\right ) \]
integrate(((-20*x-20)*log(2+2*x)^2+(-8*x-8)*log(2+2*x)+4*x^2-11*x+1)/((20* x^2-40*x-60)*log(2+2*x)^2+(-4*x^3+28*x^2-28*x-60)*log(2+2*x)-2*x^3+9*x^2-4 *x-15),x, algorithm=\
Time = 9.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {1-11 x+4 x^2+(-8-8 x) \log (2+2 x)+(-20-20 x) \log ^2(2+2 x)}{-15-4 x+9 x^2-2 x^3+\left (-60-28 x+28 x^2-4 x^3\right ) \log (2+2 x)+\left (-60-40 x+20 x^2\right ) \log ^2(2+2 x)} \, dx=\ln \left (\frac {10\,\ln \left (2\,x+2\right )-2\,x+5}{x+1}\right )-\ln \left (\frac {\left (2\,\ln \left (2\,x+2\right )+1\right )\,\left (x-4\right )}{x+1}\right )-2\,\mathrm {atanh}\left (2\,x-7\right ) \]
int((11*x + log(2*x + 2)*(8*x + 8) + log(2*x + 2)^2*(20*x + 20) - 4*x^2 - 1)/(4*x + log(2*x + 2)*(28*x - 28*x^2 + 4*x^3 + 60) + log(2*x + 2)^2*(40*x - 20*x^2 + 60) - 9*x^2 + 2*x^3 + 15),x)