Integrand size = 106, antiderivative size = 33 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx=x \left (2 x+x^4\right ) \left (x+\frac {-1+x}{5 x \left (x^2+\log (1-x)\right )}\right ) \] Output:
x*(x^4+2*x)*(1/5*(-1+x)/(x^2+ln(1-x))/x+x)
Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx=\frac {1}{5} \left (10 x^3+5 x^6+\frac {(-1+x) \left (2 x+x^4\right )}{x^2+\log (1-x)}\right ) \] Input:
Integrate[(-2*x + 2*x^2 - x^4 - 2*x^5 + 33*x^6 + 30*x^9 + (-2 + 4*x - 4*x^ 3 + 65*x^4 + 60*x^7)*Log[1 - x] + (30*x^2 + 30*x^5)*Log[1 - x]^2)/(5*x^4 + 10*x^2*Log[1 - x] + 5*Log[1 - x]^2),x]
Output:
(10*x^3 + 5*x^6 + ((-1 + x)*(2*x + x^4))/(x^2 + Log[1 - x]))/5
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.
\(\displaystyle \int \frac {30 x^9+33 x^6-2 x^5-x^4+2 x^2+\left (30 x^5+30 x^2\right ) \log ^2(1-x)+\left (60 x^7+65 x^4-4 x^3+4 x-2\right ) \log (1-x)-2 x}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {30 x^9+33 x^6-2 x^5-x^4+2 x^2+\left (30 x^5+30 x^2\right ) \log ^2(1-x)+\left (60 x^7+65 x^4-4 x^3+4 x-2\right ) \log (1-x)-2 x}{5 \left (x^2+\log (1-x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {-30 x^9-33 x^6+2 x^5+x^4-2 x^2+2 x-30 \left (x^5+x^2\right ) \log ^2(1-x)+\left (-60 x^7-65 x^4+4 x^3-4 x+2\right ) \log (1-x)}{\left (x^2+\log (1-x)\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {-30 x^9-33 x^6+2 x^5+x^4-2 x^2+2 x-30 \left (x^5+x^2\right ) \log ^2(1-x)+\left (-60 x^7-65 x^4+4 x^3-4 x+2\right ) \log (1-x)}{\left (x^2+\log (1-x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (-30 \left (x^3+1\right ) x^2+\frac {\left (2 x^5-2 x^4+x^3+4 x^2-4 x+2\right ) x}{\left (x^2+\log (1-x)\right )^2}+\frac {-5 x^4+4 x^3-4 x+2}{x^2+\log (1-x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-2 \int \frac {x}{\left (x^2+\log (1-x)\right )^2}dx+4 \int \frac {x^2}{\left (x^2+\log (1-x)\right )^2}dx-2 \int \frac {1}{x^2+\log (1-x)}dx+4 \int \frac {x}{x^2+\log (1-x)}dx-2 \int \frac {x^6}{\left (x^2+\log (1-x)\right )^2}dx+2 \int \frac {x^5}{\left (x^2+\log (1-x)\right )^2}dx-\int \frac {x^4}{\left (x^2+\log (1-x)\right )^2}dx+5 \int \frac {x^4}{x^2+\log (1-x)}dx-4 \int \frac {x^3}{\left (x^2+\log (1-x)\right )^2}dx-4 \int \frac {x^3}{x^2+\log (1-x)}dx+5 x^6+10 x^3\right )\) |
Input:
Int[(-2*x + 2*x^2 - x^4 - 2*x^5 + 33*x^6 + 30*x^9 + (-2 + 4*x - 4*x^3 + 65 *x^4 + 60*x^7)*Log[1 - x] + (30*x^2 + 30*x^5)*Log[1 - x]^2)/(5*x^4 + 10*x^ 2*Log[1 - x] + 5*Log[1 - x]^2),x]
Output:
$Aborted
Time = 3.80 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\left (x^{3}+1\right )^{2}+\frac {\left (x^{3}+2\right ) x \left (-1+x \right )}{5 x^{2}+5 \ln \left (1-x \right )}\) | \(32\) |
parallelrisch | \(\frac {5 x^{8}+5 \ln \left (1-x \right ) x^{6}+11 x^{5}-x^{4}+10 \ln \left (1-x \right ) x^{3}+2 x^{2}-2 x}{5 x^{2}+5 \ln \left (1-x \right )}\) | \(61\) |
derivativedivides | \(\left (1-x \right )^{6}-6 \left (1-x \right )^{5}+15 \left (1-x \right )^{4}-\frac {111 \left (1-x \right )^{3}}{5}+\frac {107 \left (1-x \right )^{2}}{5}+\frac {\left (-61+\ln \left (1-x \right )\right ) \left (1-x \right )}{5}-\frac {\ln \left (1-x \right )^{2} \left (1-x \right )+2 \ln \left (1-x \right )+2 x}{5 \left (\left (1-x \right )^{2}-1+2 x +\ln \left (1-x \right )\right )}\) | \(108\) |
default | \(\left (1-x \right )^{6}-6 \left (1-x \right )^{5}+15 \left (1-x \right )^{4}-\frac {111 \left (1-x \right )^{3}}{5}+\frac {107 \left (1-x \right )^{2}}{5}+\frac {\left (-61+\ln \left (1-x \right )\right ) \left (1-x \right )}{5}-\frac {\ln \left (1-x \right )^{2} \left (1-x \right )+2 \ln \left (1-x \right )+2 x}{5 \left (\left (1-x \right )^{2}-1+2 x +\ln \left (1-x \right )\right )}\) | \(108\) |
Input:
int(((30*x^5+30*x^2)*ln(1-x)^2+(60*x^7+65*x^4-4*x^3+4*x-2)*ln(1-x)+30*x^9+ 33*x^6-2*x^5-x^4+2*x^2-2*x)/(5*ln(1-x)^2+10*x^2*ln(1-x)+5*x^4),x,method=_R ETURNVERBOSE)
Output:
(x^3+1)^2+1/5*(x^3+2)*x*(-1+x)/(x^2+ln(1-x))
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx=\frac {5 \, x^{8} + 11 \, x^{5} - x^{4} + 2 \, x^{2} + 5 \, {\left (x^{6} + 2 \, x^{3}\right )} \log \left (-x + 1\right ) - 2 \, x}{5 \, {\left (x^{2} + \log \left (-x + 1\right )\right )}} \] Input:
integrate(((30*x^5+30*x^2)*log(1-x)^2+(60*x^7+65*x^4-4*x^3+4*x-2)*log(1-x) +30*x^9+33*x^6-2*x^5-x^4+2*x^2-2*x)/(5*log(1-x)^2+10*x^2*log(1-x)+5*x^4),x , algorithm="fricas")
Output:
1/5*(5*x^8 + 11*x^5 - x^4 + 2*x^2 + 5*(x^6 + 2*x^3)*log(-x + 1) - 2*x)/(x^ 2 + log(-x + 1))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx=x^{6} + 2 x^{3} + \frac {x^{5} - x^{4} + 2 x^{2} - 2 x}{5 x^{2} + 5 \log {\left (1 - x \right )}} \] Input:
integrate(((30*x**5+30*x**2)*ln(1-x)**2+(60*x**7+65*x**4-4*x**3+4*x-2)*ln( 1-x)+30*x**9+33*x**6-2*x**5-x**4+2*x**2-2*x)/(5*ln(1-x)**2+10*x**2*ln(1-x) +5*x**4),x)
Output:
x**6 + 2*x**3 + (x**5 - x**4 + 2*x**2 - 2*x)/(5*x**2 + 5*log(1 - x))
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx=\frac {5 \, x^{8} + 11 \, x^{5} - x^{4} + 2 \, x^{2} + 5 \, {\left (x^{6} + 2 \, x^{3}\right )} \log \left (-x + 1\right ) - 2 \, x}{5 \, {\left (x^{2} + \log \left (-x + 1\right )\right )}} \] Input:
integrate(((30*x^5+30*x^2)*log(1-x)^2+(60*x^7+65*x^4-4*x^3+4*x-2)*log(1-x) +30*x^9+33*x^6-2*x^5-x^4+2*x^2-2*x)/(5*log(1-x)^2+10*x^2*log(1-x)+5*x^4),x , algorithm="maxima")
Output:
1/5*(5*x^8 + 11*x^5 - x^4 + 2*x^2 + 5*(x^6 + 2*x^3)*log(-x + 1) - 2*x)/(x^ 2 + log(-x + 1))
Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (33) = 66\).
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.70 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx={\left (x - 1\right )}^{6} + 6 \, {\left (x - 1\right )}^{5} + 15 \, {\left (x - 1\right )}^{4} + 22 \, {\left (x - 1\right )}^{3} + 21 \, {\left (x - 1\right )}^{2} + 12 \, x + \frac {{\left (x - 1\right )}^{5} + 4 \, {\left (x - 1\right )}^{4} + 6 \, {\left (x - 1\right )}^{3} + 6 \, {\left (x - 1\right )}^{2} + 3 \, x - 3}{5 \, {\left ({\left (x - 1\right )}^{2} + 2 \, x + \log \left (-x + 1\right ) - 1\right )}} - 12 \] Input:
integrate(((30*x^5+30*x^2)*log(1-x)^2+(60*x^7+65*x^4-4*x^3+4*x-2)*log(1-x) +30*x^9+33*x^6-2*x^5-x^4+2*x^2-2*x)/(5*log(1-x)^2+10*x^2*log(1-x)+5*x^4),x , algorithm="giac")
Output:
(x - 1)^6 + 6*(x - 1)^5 + 15*(x - 1)^4 + 22*(x - 1)^3 + 21*(x - 1)^2 + 12* x + 1/5*((x - 1)^5 + 4*(x - 1)^4 + 6*(x - 1)^3 + 6*(x - 1)^2 + 3*x - 3)/(( x - 1)^2 + 2*x + log(-x + 1) - 1) - 12
Time = 2.91 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.76 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx=\frac {5\,x^3}{2}-\frac {\frac {x}{8}-\frac {1}{40}}{x^2-x+\frac {1}{2}}-\frac {2\,x^2}{5}-\frac {x}{4}+x^6+\frac {\frac {\left (x-1\right )\,\left (-3\,x^6+2\,x^5+x^4-2\,x^2+2\,x\right )}{5\,\left (2\,x^2-2\,x+1\right )}-\frac {\ln \left (1-x\right )\,\left (x-1\right )\,\left (5\,x^4-4\,x^3+4\,x-2\right )}{5\,\left (2\,x^2-2\,x+1\right )}}{\ln \left (1-x\right )+x^2} \] Input:
int((log(1 - x)^2*(30*x^2 + 30*x^5) - 2*x + 2*x^2 - x^4 - 2*x^5 + 33*x^6 + 30*x^9 + log(1 - x)*(4*x - 4*x^3 + 65*x^4 + 60*x^7 - 2))/(5*log(1 - x)^2 + 5*x^4 + 10*x^2*log(1 - x)),x)
Output:
(5*x^3)/2 - (x/8 - 1/40)/(x^2 - x + 1/2) - (2*x^2)/5 - x/4 + x^6 + (((x - 1)*(2*x - 2*x^2 + x^4 + 2*x^5 - 3*x^6))/(5*(2*x^2 - 2*x + 1)) - (log(1 - x )*(x - 1)*(4*x - 4*x^3 + 5*x^4 - 2))/(5*(2*x^2 - 2*x + 1)))/(log(1 - x) + x^2)
Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int \frac {-2 x+2 x^2-x^4-2 x^5+33 x^6+30 x^9+\left (-2+4 x-4 x^3+65 x^4+60 x^7\right ) \log (1-x)+\left (30 x^2+30 x^5\right ) \log ^2(1-x)}{5 x^4+10 x^2 \log (1-x)+5 \log ^2(1-x)} \, dx=\frac {5 \,\mathrm {log}\left (1-x \right ) x^{6}+10 \,\mathrm {log}\left (1-x \right ) x^{3}-2 \,\mathrm {log}\left (1-x \right )+5 x^{8}+11 x^{5}-x^{4}-2 x}{5 \,\mathrm {log}\left (1-x \right )+5 x^{2}} \] Input:
int(((30*x^5+30*x^2)*log(1-x)^2+(60*x^7+65*x^4-4*x^3+4*x-2)*log(1-x)+30*x^ 9+33*x^6-2*x^5-x^4+2*x^2-2*x)/(5*log(1-x)^2+10*x^2*log(1-x)+5*x^4),x)
Output:
(5*log( - x + 1)*x**6 + 10*log( - x + 1)*x**3 - 2*log( - x + 1) + 5*x**8 + 11*x**5 - x**4 - 2*x)/(5*(log( - x + 1) + x**2))