Integrand size = 131, antiderivative size = 28 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (-2-x) \log (1-x)+\frac {e^{-x} \log (-2-x) \left (\left (-2 x-x^2\right ) \log (-2-x)+\left (-x+x^2+\left (2+x-2 x^2-x^3\right ) \log (-2-x)\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (-2 x+x^2+x^3\right ) \log (-2-x) \log (1-x)} \, dx=3-x+\frac {e^{-x} \log (-2-x)}{x \log (1-x)} \]
Time = 5.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (-2-x) \log (1-x)+\frac {e^{-x} \log (-2-x) \left (\left (-2 x-x^2\right ) \log (-2-x)+\left (-x+x^2+\left (2+x-2 x^2-x^3\right ) \log (-2-x)\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (-2 x+x^2+x^3\right ) \log (-2-x) \log (1-x)} \, dx=-x+\frac {e^{-x} \log (-2-x)}{x \log (1-x)} \]
Integrate[((2*x - x^2 - x^3)*Log[-2 - x]*Log[1 - x] + (Log[-2 - x]*((-2*x - x^2)*Log[-2 - x] + (-x + x^2 + (2 + x - 2*x^2 - x^3)*Log[-2 - x])*Log[1 - x]))/(E^x*x*Log[1 - x]))/((-2*x + x^2 + x^3)*Log[-2 - x]*Log[1 - 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.
\(\displaystyle \int \frac {\left (-x^3-x^2+2 x\right ) \log (-x-2) \log (1-x)+\frac {e^{-x} \log (-x-2) \left (\left (-x^2-2 x\right ) \log (-x-2)+\left (x^2+\left (-x^3-2 x^2+x+2\right ) \log (-x-2)-x\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (x^3+x^2-2 x\right ) \log (-x-2) \log (1-x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-x^3-x^2+2 x\right ) \log (-x-2) \log (1-x)+\frac {e^{-x} \log (-x-2) \left (\left (-x^2-2 x\right ) \log (-x-2)+\left (x^2+\left (-x^3-2 x^2+x+2\right ) \log (-x-2)-x\right ) \log (1-x)\right )}{x \log (1-x)}}{x \left (x^2+x-2\right ) \log (-x-2) \log (1-x)}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (-\frac {e^{-x} \left (x^3 \log (-x-2) \log (1-x)+x^2 \log (-x-2)+2 x^2 \log (-x-2) \log (1-x)-x^2 \log (1-x)+2 x \log (-x-2)-x \log (-x-2) \log (1-x)+x \log (1-x)-2 \log (-x-2) \log (1-x)\right )}{x^2 \left (x^2+x-2\right ) \log ^2(1-x)}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{-x} \log (-x-2)}{x^2 \log (1-x)}dx-\int \frac {e^{-x} \log (-x-2)}{(x-1) \log ^2(1-x)}dx+\int \frac {e^{-x} \log (-x-2)}{x \log ^2(1-x)}dx+\frac {1}{2} \int \frac {e^{-x}}{x \log (1-x)}dx-\frac {1}{2} \int \frac {e^{-x}}{(x+2) \log (1-x)}dx-\int \frac {e^{-x} \log (-x-2)}{x \log (1-x)}dx-x\) |
Int[((2*x - x^2 - x^3)*Log[-2 - x]*Log[1 - x] + (Log[-2 - x]*((-2*x - x^2) *Log[-2 - x] + (-x + x^2 + (2 + x - 2*x^2 - x^3)*Log[-2 - x])*Log[1 - x])) /(E^x*x*Log[1 - x]))/((-2*x + x^2 + x^3)*Log[-2 - x]*Log[1 - x]),x]
3.3.61.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
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 = 165.61 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\frac {3}{2}-x +{\mathrm e}^{\ln \left (\frac {\ln \left (-2-x \right )}{x \ln \left (1-x \right )}\right )-x}\) | \(30\) |
int(((((-x^3-2*x^2+x+2)*ln(-2-x)+x^2-x)*ln(1-x)+(-x^2-2*x)*ln(-2-x))*exp(l n(ln(-2-x)/x/ln(1-x))-x)+(-x^3-x^2+2*x)*ln(-2-x)*ln(1-x))/(x^3+x^2-2*x)/ln (-2-x)/ln(1-x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (-2-x) \log (1-x)+\frac {e^{-x} \log (-2-x) \left (\left (-2 x-x^2\right ) \log (-2-x)+\left (-x+x^2+\left (2+x-2 x^2-x^3\right ) \log (-2-x)\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (-2 x+x^2+x^3\right ) \log (-2-x) \log (1-x)} \, dx=-x + e^{\left (-x + \log \left (\frac {\log \left (-x - 2\right )}{x \log \left (-x + 1\right )}\right )\right )} \]
integrate(((((-x^3-2*x^2+x+2)*log(-2-x)+x^2-x)*log(1-x)+(-x^2-2*x)*log(-2- x))*exp(log(log(-2-x)/x/log(1-x))-x)+(-x^3-x^2+2*x)*log(-2-x)*log(1-x))/(x ^3+x^2-2*x)/log(-2-x)/log(1-x),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (-2-x) \log (1-x)+\frac {e^{-x} \log (-2-x) \left (\left (-2 x-x^2\right ) \log (-2-x)+\left (-x+x^2+\left (2+x-2 x^2-x^3\right ) \log (-2-x)\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (-2 x+x^2+x^3\right ) \log (-2-x) \log (1-x)} \, dx=- x + \frac {e^{- x} \log {\left (- x - 2 \right )}}{x \log {\left (1 - x \right )}} \]
integrate(((((-x**3-2*x**2+x+2)*ln(-2-x)+x**2-x)*ln(1-x)+(-x**2-2*x)*ln(-2 -x))*exp(ln(ln(-2-x)/x/ln(1-x))-x)+(-x**3-x**2+2*x)*ln(-2-x)*ln(1-x))/(x** 3+x**2-2*x)/ln(-2-x)/ln(1-x),x)
Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (-2-x) \log (1-x)+\frac {e^{-x} \log (-2-x) \left (\left (-2 x-x^2\right ) \log (-2-x)+\left (-x+x^2+\left (2+x-2 x^2-x^3\right ) \log (-2-x)\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (-2 x+x^2+x^3\right ) \log (-2-x) \log (1-x)} \, dx=-x + \frac {e^{\left (-x\right )} \log \left (-x - 2\right )}{x \log \left (-x + 1\right )} \]
integrate(((((-x^3-2*x^2+x+2)*log(-2-x)+x^2-x)*log(1-x)+(-x^2-2*x)*log(-2- x))*exp(log(log(-2-x)/x/log(1-x))-x)+(-x^3-x^2+2*x)*log(-2-x)*log(1-x))/(x ^3+x^2-2*x)/log(-2-x)/log(1-x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (29) = 58\).
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (-2-x) \log (1-x)+\frac {e^{-x} \log (-2-x) \left (\left (-2 x-x^2\right ) \log (-2-x)+\left (-x+x^2+\left (2+x-2 x^2-x^3\right ) \log (-2-x)\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (-2 x+x^2+x^3\right ) \log (-2-x) \log (1-x)} \, dx=-\frac {{\left (x - 1\right )}^{2} e \log \left (-x + 1\right ) + {\left (x - 1\right )} e \log \left (-x + 1\right ) - e^{\left (-x + 1\right )} \log \left (-x - 2\right )}{{\left (x - 1\right )} e \log \left (-x + 1\right ) + e \log \left (-x + 1\right )} \]
integrate(((((-x^3-2*x^2+x+2)*log(-2-x)+x^2-x)*log(1-x)+(-x^2-2*x)*log(-2- x))*exp(log(log(-2-x)/x/log(1-x))-x)+(-x^3-x^2+2*x)*log(-2-x)*log(1-x))/(x ^3+x^2-2*x)/log(-2-x)/log(1-x),x, algorithm=\
-((x - 1)^2*e*log(-x + 1) + (x - 1)*e*log(-x + 1) - e^(-x + 1)*log(-x - 2) )/((x - 1)*e*log(-x + 1) + e*log(-x + 1))
Timed out. \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (-2-x) \log (1-x)+\frac {e^{-x} \log (-2-x) \left (\left (-2 x-x^2\right ) \log (-2-x)+\left (-x+x^2+\left (2+x-2 x^2-x^3\right ) \log (-2-x)\right ) \log (1-x)\right )}{x \log (1-x)}}{\left (-2 x+x^2+x^3\right ) \log (-2-x) \log (1-x)} \, dx=\int -\frac {{\mathrm {e}}^{\ln \left (\frac {\ln \left (-x-2\right )}{x\,\ln \left (1-x\right )}\right )-x}\,\left (\ln \left (-x-2\right )\,\left (x^2+2\,x\right )-\ln \left (1-x\right )\,\left (\ln \left (-x-2\right )\,\left (-x^3-2\,x^2+x+2\right )-x+x^2\right )\right )+\ln \left (1-x\right )\,\ln \left (-x-2\right )\,\left (x^3+x^2-2\,x\right )}{\ln \left (1-x\right )\,\ln \left (-x-2\right )\,\left (x^3+x^2-2\,x\right )} \,d x \]
int(-(exp(log(log(- x - 2)/(x*log(1 - x))) - x)*(log(- x - 2)*(2*x + x^2) - log(1 - x)*(log(- x - 2)*(x - 2*x^2 - x^3 + 2) - x + x^2)) + log(1 - x)* log(- x - 2)*(x^2 - 2*x + x^3))/(log(1 - x)*log(- x - 2)*(x^2 - 2*x + x^3) ),x)