Integrand size = 98, antiderivative size = 19 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {x+\log (x)+\log \left (x \left (3+\log \left (\frac {1}{3+x}\right )\right )\right )}{x} \] Output:
(ln((ln(1/(3+x))+3)*x)+x+ln(x))/x
Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\log (x)}{x}+\frac {\log \left (x \left (3+\log \left (\frac {1}{3+x}\right )\right )\right )}{x} \] Input:
Integrate[(18 + 5*x + (6 + 2*x)*Log[(3 + x)^(-1)] + Log[x]*(-9 - 3*x + (-3 - x)*Log[(3 + x)^(-1)]) + (-9 - 3*x + (-3 - x)*Log[(3 + x)^(-1)])*Log[3*x + x*Log[(3 + x)^(-1)]])/(9*x^2 + 3*x^3 + (3*x^2 + x^3)*Log[(3 + x)^(-1)]) ,x]
Output:
Log[x]/x + Log[x*(3 + Log[(3 + x)^(-1)])]/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.
\(\displaystyle \int \frac {5 x+(2 x+6) \log \left (\frac {1}{x+3}\right )+\log (x) \left (-3 x+(-x-3) \log \left (\frac {1}{x+3}\right )-9\right )+\left (-3 x+(-x-3) \log \left (\frac {1}{x+3}\right )-9\right ) \log \left (3 x+x \log \left (\frac {1}{x+3}\right )\right )+18}{3 x^3+9 x^2+\left (x^3+3 x^2\right ) \log \left (\frac {1}{x+3}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 x+(2 x+6) \log \left (\frac {1}{x+3}\right )+\log (x) \left (-3 x+(-x-3) \log \left (\frac {1}{x+3}\right )-9\right )+\left (-3 x+(-x-3) \log \left (\frac {1}{x+3}\right )-9\right ) \log \left (3 x+x \log \left (\frac {1}{x+3}\right )\right )+18}{x^2 (x+3) \left (\log \left (\frac {1}{x+3}\right )+3\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 x-3 x \log (x)-x \log (x) \log \left (\frac {1}{x+3}\right )+2 x \log \left (\frac {1}{x+3}\right )-9 \log (x)-3 \log (x) \log \left (\frac {1}{x+3}\right )+6 \log \left (\frac {1}{x+3}\right )+18}{x^2 (x+3) \left (\log \left (\frac {1}{x+3}\right )+3\right )}-\frac {\log \left (3 x+x \log \left (\frac {1}{x+3}\right )\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\log \left (\log \left (\frac {1}{x+3}\right ) x+3 x\right )}{x^2}dx-\frac {1}{3} \int \frac {1}{x \left (\log \left (\frac {1}{x+3}\right )+3\right )}dx+\frac {1}{x}-\frac {2-\log (x)}{x}-\frac {1}{3} \log \left (\log \left (\frac {1}{x+3}\right )+3\right )\) |
Input:
Int[(18 + 5*x + (6 + 2*x)*Log[(3 + x)^(-1)] + Log[x]*(-9 - 3*x + (-3 - x)* Log[(3 + x)^(-1)]) + (-9 - 3*x + (-3 - x)*Log[(3 + x)^(-1)])*Log[3*x + x*L og[(3 + x)^(-1)]])/(9*x^2 + 3*x^3 + (3*x^2 + x^3)*Log[(3 + x)^(-1)]),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(54\) vs. \(2(19)=38\).
Time = 1.57 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.89
method | result | size |
parallelrisch | \(\frac {27 \ln \left (\ln \left (\frac {1}{3+x}\right )+3\right ) x +27 x \ln \left (x \right )-27 \ln \left (\left (\ln \left (\frac {1}{3+x}\right )+3\right ) x \right ) x +162 \ln \left (x \right )+162 \ln \left (\left (\ln \left (\frac {1}{3+x}\right )+3\right ) x \right )}{162 x}\) | \(55\) |
risch | \(\frac {\ln \left (\ln \left (3+x \right )-3\right )}{x}+\frac {-2 i \pi \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i \left (\ln \left (3+x \right )-3\right )\right ) \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (\ln \left (3+x \right )-3\right )\right ) \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right ) \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{2} \operatorname {csgn}\left (i x \right )+2 i \pi +4 \ln \left (x \right )}{2 x}\) | \(140\) |
Input:
int((((-3-x)*ln(1/(3+x))-3*x-9)*ln(x*ln(1/(3+x))+3*x)+((-3-x)*ln(1/(3+x))- 3*x-9)*ln(x)+(2*x+6)*ln(1/(3+x))+5*x+18)/((x^3+3*x^2)*ln(1/(3+x))+3*x^3+9* x^2),x,method=_RETURNVERBOSE)
Output:
1/162/x*(27*ln(ln(1/(3+x))+3)*x+27*x*ln(x)-27*ln((ln(1/(3+x))+3)*x)*x+162* ln(x)+162*ln((ln(1/(3+x))+3)*x))
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\log \left (x \log \left (\frac {1}{x + 3}\right ) + 3 \, x\right ) + \log \left (x\right )}{x} \] Input:
integrate((((-3-x)*log(1/(3+x))-3*x-9)*log(x*log(1/(3+x))+3*x)+((-3-x)*log (1/(3+x))-3*x-9)*log(x)+(2*x+6)*log(1/(3+x))+5*x+18)/((x^3+3*x^2)*log(1/(3 +x))+3*x^3+9*x^2),x, algorithm="fricas")
Output:
(log(x*log(1/(x + 3)) + 3*x) + log(x))/x
Time = 0.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\log {\left (x \right )}}{x} + \frac {\log {\left (x \log {\left (\frac {1}{x + 3} \right )} + 3 x \right )}}{x} \] Input:
integrate((((-3-x)*ln(1/(3+x))-3*x-9)*ln(x*ln(1/(3+x))+3*x)+((-3-x)*ln(1/( 3+x))-3*x-9)*ln(x)+(2*x+6)*ln(1/(3+x))+5*x+18)/((x**3+3*x**2)*ln(1/(3+x))+ 3*x**3+9*x**2),x)
Output:
log(x)/x + log(x*log(1/(x + 3)) + 3*x)/x
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {2 \, \log \left (x\right ) + \log \left (-\log \left (x + 3\right ) + 3\right )}{x} \] Input:
integrate((((-3-x)*log(1/(3+x))-3*x-9)*log(x*log(1/(3+x))+3*x)+((-3-x)*log (1/(3+x))-3*x-9)*log(x)+(2*x+6)*log(1/(3+x))+5*x+18)/((x^3+3*x^2)*log(1/(3 +x))+3*x^3+9*x^2),x, algorithm="maxima")
Output:
(2*log(x) + log(-log(x + 3) + 3))/x
Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {2 \, \log \left (x\right )}{x} + \frac {\log \left (-\log \left (x + 3\right ) + 3\right )}{x} \] Input:
integrate((((-3-x)*log(1/(3+x))-3*x-9)*log(x*log(1/(3+x))+3*x)+((-3-x)*log (1/(3+x))-3*x-9)*log(x)+(2*x+6)*log(1/(3+x))+5*x+18)/((x^3+3*x^2)*log(1/(3 +x))+3*x^3+9*x^2),x, algorithm="giac")
Output:
2*log(x)/x + log(-log(x + 3) + 3)/x
Time = 3.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\ln \left (3\,x+x\,\ln \left (\frac {1}{x+3}\right )\right )+\ln \left (x\right )}{x} \] Input:
int((5*x + log(1/(x + 3))*(2*x + 6) - log(x)*(3*x + log(1/(x + 3))*(x + 3) + 9) - log(3*x + x*log(1/(x + 3)))*(3*x + log(1/(x + 3))*(x + 3) + 9) + 1 8)/(log(1/(x + 3))*(3*x^2 + x^3) + 9*x^2 + 3*x^3),x)
Output:
(log(3*x + x*log(1/(x + 3))) + log(x))/x
Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\mathrm {log}\left (-\mathrm {log}\left (x +3\right ) x +3 x \right )+\mathrm {log}\left (x \right )}{x} \] Input:
int((((-3-x)*log(1/(3+x))-3*x-9)*log(x*log(1/(3+x))+3*x)+((-3-x)*log(1/(3+ x))-3*x-9)*log(x)+(2*x+6)*log(1/(3+x))+5*x+18)/((x^3+3*x^2)*log(1/(3+x))+3 *x^3+9*x^2),x)
Output:
(log( - log(x + 3)*x + 3*x) + log(x))/x