Integrand size = 53, antiderivative size = 23 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=\frac {1}{5} (-1-x-x (2+x \log (3))) \log (\log (6+x)) \] Output:
1/5*(-1-x-x*(x*ln(3)+2))*ln(ln(6+x))
Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=-\frac {1}{5} \left (1+3 x+x^2 \log (3)\right ) \log (\log (6+x)) \] Input:
Integrate[(-1 - 3*x - x^2*Log[3] + (-18 - 3*x + (-12*x - 2*x^2)*Log[3])*Lo g[6 + x]*Log[Log[6 + x]])/((30 + 5*x)*Log[6 + x]),x]
Output:
-1/5*((1 + 3*x + x^2*Log[3])*Log[Log[6 + 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.
\(\displaystyle \int \frac {x^2 (-\log (3))+\left (\left (-2 x^2-12 x\right ) \log (3)-3 x-18\right ) \log (x+6) \log (\log (x+6))-3 x-1}{(5 x+30) \log (x+6)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2 (-\log (3))-3 x-1}{5 (x+6) \log (x+6)}-\frac {1}{5} (x \log (9)+3) \log (\log (x+6))\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{5} \log (9) \int x \log (\log (x+6))dx-\frac {1}{5} \log (3) \operatorname {ExpIntegralEi}(2 \log (x+6))+\frac {3 \operatorname {LogIntegral}(x+6)}{5}-\frac {3}{5} (1-\log (9)) \operatorname {LogIntegral}(x+6)+\frac {6}{5} \log (3) \operatorname {LogIntegral}(x+6)-\frac {3}{5} (x+6) \log (\log (x+6))+\frac {1}{5} (17-36 \log (3)) \log (\log (x+6))\) |
Input:
Int[(-1 - 3*x - x^2*Log[3] + (-18 - 3*x + (-12*x - 2*x^2)*Log[3])*Log[6 + x]*Log[Log[6 + x]])/((30 + 5*x)*Log[6 + x]),x]
Output:
$Aborted
Time = 0.52 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\left (-\frac {x^{2} \ln \left (3\right )}{5}-\frac {3 x}{5}\right ) \ln \left (\ln \left (6+x \right )\right )-\frac {\ln \left (\ln \left (6+x \right )\right )}{5}\) | \(26\) |
default | \(-\frac {\ln \left (3\right ) \ln \left (\ln \left (6+x \right )\right ) x^{2}}{5}-\frac {\ln \left (\ln \left (6+x \right )\right )}{5}-\frac {3 x \ln \left (\ln \left (6+x \right )\right )}{5}\) | \(29\) |
norman | \(-\frac {\ln \left (3\right ) \ln \left (\ln \left (6+x \right )\right ) x^{2}}{5}-\frac {\ln \left (\ln \left (6+x \right )\right )}{5}-\frac {3 x \ln \left (\ln \left (6+x \right )\right )}{5}\) | \(29\) |
parallelrisch | \(-\frac {\ln \left (3\right ) \ln \left (\ln \left (6+x \right )\right ) x^{2}}{5}-\frac {\ln \left (\ln \left (6+x \right )\right )}{5}-\frac {3 x \ln \left (\ln \left (6+x \right )\right )}{5}\) | \(29\) |
parts | \(-\frac {2 \ln \left (3\right ) \left (\left (\frac {\left (6+x \right )^{2}}{2}-36-6 x \right ) \ln \left (\ln \left (6+x \right )\right )+\frac {\operatorname {expIntegral}_{1}\left (-2 \ln \left (6+x \right )\right )}{2}-6 \,\operatorname {expIntegral}_{1}\left (-\ln \left (6+x \right )\right )\right )}{5}-\frac {3 x \ln \left (\ln \left (6+x \right )\right )}{5}-\frac {\ln \left (\ln \left (6+x \right )\right )}{5}+\frac {\ln \left (3\right ) \operatorname {expIntegral}_{1}\left (-2 \ln \left (6+x \right )\right )}{5}-\frac {12 \ln \left (3\right ) \operatorname {expIntegral}_{1}\left (-\ln \left (6+x \right )\right )}{5}-\frac {36 \ln \left (3\right ) \ln \left (\ln \left (6+x \right )\right )}{5}\) | \(93\) |
Input:
int((((-2*x^2-12*x)*ln(3)-3*x-18)*ln(6+x)*ln(ln(6+x))-x^2*ln(3)-3*x-1)/(5* x+30)/ln(6+x),x,method=_RETURNVERBOSE)
Output:
(-1/5*x^2*ln(3)-3/5*x)*ln(ln(6+x))-1/5*ln(ln(6+x))
Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=-\frac {1}{5} \, {\left (x^{2} \log \left (3\right ) + 3 \, x + 1\right )} \log \left (\log \left (x + 6\right )\right ) \] Input:
integrate((((-2*x^2-12*x)*log(3)-3*x-18)*log(6+x)*log(log(6+x))-x^2*log(3) -3*x-1)/(5*x+30)/log(6+x),x, algorithm="fricas")
Output:
-1/5*(x^2*log(3) + 3*x + 1)*log(log(x + 6))
Time = 0.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=\left (- \frac {x^{2} \log {\left (3 \right )}}{5} - \frac {3 x}{5} - \frac {9}{5} + \frac {12 \log {\left (3 \right )}}{5}\right ) \log {\left (\log {\left (x + 6 \right )} \right )} - \frac {4 \left (-2 + 3 \log {\left (3 \right )}\right ) \log {\left (\log {\left (x + 6 \right )} \right )}}{5} \] Input:
integrate((((-2*x**2-12*x)*ln(3)-3*x-18)*ln(6+x)*ln(ln(6+x))-x**2*ln(3)-3* x-1)/(5*x+30)/ln(6+x),x)
Output:
(-x**2*log(3)/5 - 3*x/5 - 9/5 + 12*log(3)/5)*log(log(x + 6)) - 4*(-2 + 3*l og(3))*log(log(x + 6))/5
Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=-\frac {1}{5} \, {\left (x^{2} \log \left (3\right ) + 3 \, x\right )} \log \left (\log \left (x + 6\right )\right ) - \frac {1}{5} \, \log \left (\log \left (x + 6\right )\right ) \] Input:
integrate((((-2*x^2-12*x)*log(3)-3*x-18)*log(6+x)*log(log(6+x))-x^2*log(3) -3*x-1)/(5*x+30)/log(6+x),x, algorithm="maxima")
Output:
-1/5*(x^2*log(3) + 3*x)*log(log(x + 6)) - 1/5*log(log(x + 6))
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=-\frac {1}{5} \, {\left (x^{2} \log \left (3\right ) + 3 \, x\right )} \log \left (\log \left (x + 6\right )\right ) - \frac {1}{5} \, \log \left (\log \left (x + 6\right )\right ) \] Input:
integrate((((-2*x^2-12*x)*log(3)-3*x-18)*log(6+x)*log(log(6+x))-x^2*log(3) -3*x-1)/(5*x+30)/log(6+x),x, algorithm="giac")
Output:
-1/5*(x^2*log(3) + 3*x)*log(log(x + 6)) - 1/5*log(log(x + 6))
Time = 0.72 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=-\frac {\ln \left (\ln \left (x+6\right )\right )\,\left (\ln \left (3\right )\,x^2+3\,x+1\right )}{5} \] Input:
int(-(3*x + x^2*log(3) + log(x + 6)*log(log(x + 6))*(3*x + log(3)*(12*x + 2*x^2) + 18) + 1)/(log(x + 6)*(5*x + 30)),x)
Output:
-(log(log(x + 6))*(3*x + x^2*log(3) + 1))/5
Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-1-3 x-x^2 \log (3)+\left (-18-3 x+\left (-12 x-2 x^2\right ) \log (3)\right ) \log (6+x) \log (\log (6+x))}{(30+5 x) \log (6+x)} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (x +6\right )\right ) \left (-\mathrm {log}\left (3\right ) x^{2}-3 x -1\right )}{5} \] Input:
int((((-2*x^2-12*x)*log(3)-3*x-18)*log(6+x)*log(log(6+x))-x^2*log(3)-3*x-1 )/(5*x+30)/log(6+x),x)
Output:
(log(log(x + 6))*( - log(3)*x**2 - 3*x - 1))/5