Integrand size = 68, antiderivative size = 20 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=(1-3 x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right ) \] Output:
ln(3/5/x^3/ln(4+x))*(1-3*x)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.10 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=-3 \operatorname {ExpIntegralEi}(\log (4+x))-3 \log (x)-3 x \log \left (\frac {3}{5 x^3 \log (4+x)}\right )-\log (\log (4+x))+3 \operatorname {LogIntegral}(4+x) \] Input:
Integrate[(-x + 3*x^2 + (-12 + 33*x + 9*x^2)*Log[4 + x] + (-12*x - 3*x^2)* Log[4 + x]*Log[3/(5*x^3*Log[4 + x])])/((4*x + x^2)*Log[4 + x]),x]
Output:
-3*ExpIntegralEi[Log[4 + x]] - 3*Log[x] - 3*x*Log[3/(5*x^3*Log[4 + x])] - Log[Log[4 + x]] + 3*LogIntegral[4 + x]
Time = 0.89 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {2026, 7239, 2009}
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.
\(\displaystyle \int \frac {3 x^2+\left (9 x^2+33 x-12\right ) \log (x+4)+\left (-3 x^2-12 x\right ) \log (x+4) \log \left (\frac {3}{5 x^3 \log (x+4)}\right )-x}{\left (x^2+4 x\right ) \log (x+4)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {3 x^2+\left (9 x^2+33 x-12\right ) \log (x+4)+\left (-3 x^2-12 x\right ) \log (x+4) \log \left (\frac {3}{5 x^3 \log (x+4)}\right )-x}{x (x+4) \log (x+4)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-3 \log \left (\frac {3}{5 x^3 \log (x+4)}\right )-\frac {3}{x}+\frac {3 x-1}{(x+4) \log (x+4)}+9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 x \log \left (\frac {3}{5 x^3 \log (x+4)}\right )-3 \log (x)-\log (\log (x+4))\) |
Input:
Int[(-x + 3*x^2 + (-12 + 33*x + 9*x^2)*Log[4 + x] + (-12*x - 3*x^2)*Log[4 + x]*Log[3/(5*x^3*Log[4 + x])])/((4*x + x^2)*Log[4 + x]),x]
Output:
-3*Log[x] - 3*x*Log[3/(5*x^3*Log[4 + x])] - Log[Log[4 + x]]
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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.74 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45
method | result | size |
norman | \(\ln \left (\frac {3}{5 x^{3} \ln \left (4+x \right )}\right )-3 x \ln \left (\frac {3}{5 x^{3} \ln \left (4+x \right )}\right )\) | \(29\) |
default | \(-\ln \left (\ln \left (4+x \right )\right )-3 \ln \left (x \right )-3 x \ln \left (3\right )+3 x \ln \left (5\right )-3 \ln \left (\frac {1}{x^{3} \ln \left (4+x \right )}\right ) x -36\) | \(38\) |
parts | \(-\ln \left (\ln \left (4+x \right )\right )-3 \ln \left (x \right )-3 x \ln \left (3\right )+3 x \ln \left (5\right )-3 \ln \left (\frac {1}{x^{3} \ln \left (4+x \right )}\right ) x -36\) | \(38\) |
parallelrisch | \(-3 x \ln \left (\frac {3}{5 x^{3} \ln \left (4+x \right )}\right )+15 \ln \left (x \right )+5 \ln \left (\ln \left (4+x \right )\right )+6 \ln \left (\frac {3}{5 x^{3} \ln \left (4+x \right )}\right )\) | \(42\) |
risch | \(3 x \ln \left (\ln \left (4+x \right )\right )+9 x \ln \left (x \right )+\frac {3 i \pi x \operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}-\frac {3 i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{3}}\right ) \operatorname {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )^{2}}{2}-\frac {3 i \pi x \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )}{2}-\frac {3 i \pi x \,\operatorname {csgn}\left (\frac {i}{\ln \left (4+x \right )}\right ) \operatorname {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )^{2}}{2}+\frac {3 i \pi x \operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x \right )}{2}-\frac {3 i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+3 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\frac {3 i \pi x \operatorname {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )^{3}}{2}-\frac {3 i \pi x \operatorname {csgn}\left (i x^{3}\right )^{3}}{2}-\frac {3 i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}+\frac {3 i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{3}}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (4+x \right )}\right ) \operatorname {csgn}\left (\frac {i}{x^{3} \ln \left (4+x \right )}\right )}{2}+3 x \ln \left (5\right )-3 x \ln \left (3\right )-3 \ln \left (x \right )-\ln \left (\ln \left (4+x \right )\right )\) | \(278\) |
Input:
int(((-3*x^2-12*x)*ln(4+x)*ln(3/5/x^3/ln(4+x))+(9*x^2+33*x-12)*ln(4+x)+3*x ^2-x)/(x^2+4*x)/ln(4+x),x,method=_RETURNVERBOSE)
Output:
ln(3/5/x^3/ln(4+x))-3*x*ln(3/5/x^3/ln(4+x))
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=-{\left (3 \, x - 1\right )} \log \left (\frac {3}{5 \, x^{3} \log \left (x + 4\right )}\right ) \] Input:
integrate(((-3*x^2-12*x)*log(4+x)*log(3/5/x^3/log(4+x))+(9*x^2+33*x-12)*lo g(4+x)+3*x^2-x)/(x^2+4*x)/log(4+x),x, algorithm="fricas")
Output:
-(3*x - 1)*log(3/5/(x^3*log(x + 4)))
Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=\left (- 3 x - 2\right ) \log {\left (\frac {3}{5 x^{3} \log {\left (x + 4 \right )}} \right )} - 9 \log {\left (x \right )} - 3 \log {\left (\log {\left (x + 4 \right )} \right )} \] Input:
integrate(((-3*x**2-12*x)*ln(4+x)*ln(3/5/x**3/ln(4+x))+(9*x**2+33*x-12)*ln (4+x)+3*x**2-x)/(x**2+4*x)/ln(4+x),x)
Output:
(-3*x - 2)*log(3/(5*x**3*log(x + 4))) - 9*log(x) - 3*log(log(x + 4))
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=3 \, x {\left (\log \left (5\right ) - \log \left (3\right )\right )} + 3 \, {\left (3 \, x - 1\right )} \log \left (x\right ) + 3 \, x \log \left (\log \left (x + 4\right )\right ) - \log \left (\log \left (x + 4\right )\right ) \] Input:
integrate(((-3*x^2-12*x)*log(4+x)*log(3/5/x^3/log(4+x))+(9*x^2+33*x-12)*lo g(4+x)+3*x^2-x)/(x^2+4*x)/log(4+x),x, algorithm="maxima")
Output:
3*x*(log(5) - log(3)) + 3*(3*x - 1)*log(x) + 3*x*log(log(x + 4)) - log(log (x + 4))
Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=-3 \, x \log \left (3\right ) + 3 \, x \log \left (5 \, x^{3} \log \left (x + 4\right )\right ) - 3 \, \log \left (x\right ) - \log \left (\log \left (x + 4\right )\right ) \] Input:
integrate(((-3*x^2-12*x)*log(4+x)*log(3/5/x^3/log(4+x))+(9*x^2+33*x-12)*lo g(4+x)+3*x^2-x)/(x^2+4*x)/log(4+x),x, algorithm="giac")
Output:
-3*x*log(3) + 3*x*log(5*x^3*log(x + 4)) - 3*log(x) - log(log(x + 4))
Time = 0.86 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=-\ln \left (\ln \left (x+4\right )\right )-3\,\ln \left (x\right )-3\,x\,\ln \left (\frac {3}{5\,x^3\,\ln \left (x+4\right )}\right ) \] Input:
int(-(x - log(x + 4)*(33*x + 9*x^2 - 12) - 3*x^2 + log(x + 4)*log(3/(5*x^3 *log(x + 4)))*(12*x + 3*x^2))/(log(x + 4)*(4*x + x^2)),x)
Output:
- log(log(x + 4)) - 3*log(x) - 3*x*log(3/(5*x^3*log(x + 4)))
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-x+3 x^2+\left (-12+33 x+9 x^2\right ) \log (4+x)+\left (-12 x-3 x^2\right ) \log (4+x) \log \left (\frac {3}{5 x^3 \log (4+x)}\right )}{\left (4 x+x^2\right ) \log (4+x)} \, dx=-\mathrm {log}\left (\mathrm {log}\left (x +4\right )\right )-3 \,\mathrm {log}\left (\frac {3}{5 \,\mathrm {log}\left (x +4\right ) x^{3}}\right ) x -3 \,\mathrm {log}\left (x \right ) \] Input:
int(((-3*x^2-12*x)*log(4+x)*log(3/5/x^3/log(4+x))+(9*x^2+33*x-12)*log(4+x) +3*x^2-x)/(x^2+4*x)/log(4+x),x)
Output:
- log(log(x + 4)) - 3*log(3/(5*log(x + 4)*x**3))*x - 3*log(x)