Integrand size = 85, antiderivative size = 25 \begin {dmath*} \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-7+\frac {\log (2 x)}{x \log \left (\frac {4+x}{2 (1+x)}\right )} \end {dmath*}
Time = 0.40 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log (2 x)}{x \log \left (\frac {4+x}{2+2 x}\right )} \end {dmath*}
Integrate[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/((4*x^2 + 5*x^3 + x^4)*Log[(4 + x)/(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.
\(\displaystyle \int \frac {\left (x^2+5 x+4\right ) \log \left (\frac {x+4}{2 x+2}\right )+\log (2 x) \left (\left (-x^2-5 x-4\right ) \log \left (\frac {x+4}{2 x+2}\right )+3 x\right )}{\left (x^4+5 x^3+4 x^2\right ) \log ^2\left (\frac {x+4}{2 x+2}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^2+5 x+4\right ) \log \left (\frac {x+4}{2 x+2}\right )+\log (2 x) \left (\left (-x^2-5 x-4\right ) \log \left (\frac {x+4}{2 x+2}\right )+3 x\right )}{x^2 \left (x^2+5 x+4\right ) \log ^2\left (\frac {x+4}{2 x+2}\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {1-\log (2 x)}{x^2 \log \left (\frac {x+4}{2 x+2}\right )}+\frac {3 \log (2 x)}{x (x+1) (x+4) \log ^2\left (\frac {x+4}{2 x+2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x^2 \log \left (\frac {x+4}{2 x+2}\right )}dx-\int \frac {\log (2 x)}{x^2 \log \left (\frac {x+4}{2 x+2}\right )}dx+\frac {3}{4} \int \frac {\log (2 x)}{x \log ^2\left (\frac {x+4}{2 x+2}\right )}dx-\int \frac {\log (2 x)}{(x+1) \log ^2\left (\frac {x+4}{2 x+2}\right )}dx+\frac {1}{4} \int \frac {\log (2 x)}{(x+4) \log ^2\left (\frac {x+4}{2 x+2}\right )}dx\) |
Int[((4 + 5*x + x^2)*Log[(4 + x)/(2 + 2*x)] + Log[2*x]*(3*x + (-4 - 5*x - x^2)*Log[(4 + x)/(2 + 2*x)]))/((4*x^2 + 5*x^3 + x^4)*Log[(4 + x)/(2 + 2*x) ]^2),x]
3.9.79.3.1 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 = 11.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {\ln \left (2 x \right )}{\ln \left (\frac {4+x}{2+2 x}\right ) x}\) | \(22\) |
risch | \(\frac {2 i \ln \left (2 x \right )}{x \left (\pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )-\pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{1+x}\right ) \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (4+x \right )}{1+x}\right )^{3}-2 i \ln \left (2\right )+2 i \ln \left (4+x \right )-2 i \ln \left (1+x \right )\right )}\) | \(129\) |
int((((-x^2-5*x-4)*ln((4+x)/(2+2*x))+3*x)*ln(2*x)+(x^2+5*x+4)*ln((4+x)/(2+ 2*x)))/(x^4+5*x^3+4*x^2)/ln((4+x)/(2+2*x))^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \begin {dmath*} \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log \left (2 \, x\right )}{x \log \left (\frac {x + 4}{2 \, {\left (x + 1\right )}}\right )} \end {dmath*}
integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log( (4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/log((4+x)/(2+2*x))^2,x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60 \begin {dmath*} \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\log {\left (2 x \right )}}{x \log {\left (\frac {x + 4}{2 x + 2} \right )}} \end {dmath*}
integrate((((-x**2-5*x-4)*ln((4+x)/(2+2*x))+3*x)*ln(2*x)+(x**2+5*x+4)*ln(( 4+x)/(2+2*x)))/(x**4+5*x**3+4*x**2)/ln((4+x)/(2+2*x))**2,x)
Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \end {dmath*}
integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log( (4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/log((4+x)/(2+2*x))^2,x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=-\frac {\log \left (2\right ) + \log \left (x\right )}{x \log \left (2\right ) - x \log \left (x + 4\right ) + x \log \left (x + 1\right )} \end {dmath*}
integrate((((-x^2-5*x-4)*log((4+x)/(2+2*x))+3*x)*log(2*x)+(x^2+5*x+4)*log( (4+x)/(2+2*x)))/(x^4+5*x^3+4*x^2)/log((4+x)/(2+2*x))^2,x, algorithm=\
Time = 15.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {\left (4+5 x+x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )+\log (2 x) \left (3 x+\left (-4-5 x-x^2\right ) \log \left (\frac {4+x}{2+2 x}\right )\right )}{\left (4 x^2+5 x^3+x^4\right ) \log ^2\left (\frac {4+x}{2+2 x}\right )} \, dx=\frac {\ln \left (2\,x\right )}{x\,\ln \left (\frac {x+4}{2\,x+2}\right )} \end {dmath*}