Integrand size = 66, antiderivative size = 23 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=-x+x^2+\left (-3-16 x^2-\log (\log (3+x))\right )^2 \] Output:
x^2-x+(-16*x^2-ln(ln(3+x))-3)^2
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=-x+97 x^2+256 x^4+294 \log (\log (3+x))+32 (-3+x) (3+x) \log (\log (3+x))+\log ^2(\log (3+x)) \] Input:
Integrate[(6 + 32*x^2 + (-3 + 581*x + 194*x^2 + 3072*x^3 + 1024*x^4)*Log[3 + x] + (2 + (192*x + 64*x^2)*Log[3 + x])*Log[Log[3 + x]])/((3 + x)*Log[3 + x]),x]
Output:
-x + 97*x^2 + 256*x^4 + 294*Log[Log[3 + x]] + 32*(-3 + x)*(3 + x)*Log[Log[ 3 + x]] + Log[Log[3 + x]]^2
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.
\(\displaystyle \int \frac {32 x^2+\left (\left (64 x^2+192 x\right ) \log (x+3)+2\right ) \log (\log (x+3))+\left (1024 x^4+3072 x^3+194 x^2+581 x-3\right ) \log (x+3)+6}{(x+3) \log (x+3)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (32 x^2 \log (x+3)+96 x \log (x+3)+1\right ) \log (\log (x+3))}{(x+3) \log (x+3)}+\frac {1024 x^4 \log (x+3)+3072 x^3 \log (x+3)+32 x^2+194 x^2 \log (x+3)+581 x \log (x+3)-3 \log (x+3)+6}{(x+3) \log (x+3)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 64 \int x \log (\log (x+3))dx+32 \operatorname {ExpIntegralEi}(2 \log (x+3))-192 \operatorname {LogIntegral}(x+3)+256 x^4+97 x^2-x+\log ^2(\log (x+3))+294 \log (\log (x+3))\) |
Input:
Int[(6 + 32*x^2 + (-3 + 581*x + 194*x^2 + 3072*x^3 + 1024*x^4)*Log[3 + x] + (2 + (192*x + 64*x^2)*Log[3 + x])*Log[Log[3 + x]])/((3 + x)*Log[3 + x]), x]
Output:
$Aborted
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70
method | result | size |
risch | \(\ln \left (\ln \left (3+x \right )\right )^{2}+32 x^{2} \ln \left (\ln \left (3+x \right )\right )+256 x^{4}+97 x^{2}-x +6 \ln \left (\ln \left (3+x \right )\right )\) | \(39\) |
parallelrisch | \(256 x^{4}+32 x^{2} \ln \left (\ln \left (3+x \right )\right )+97 x^{2}+\ln \left (\ln \left (3+x \right )\right )^{2}+6 \ln \left (\ln \left (3+x \right )\right )-x +\frac {3}{2}\) | \(40\) |
default | \(-x +\ln \left (\ln \left (3+x \right )\right )^{2}+294 \ln \left (\ln \left (3+x \right )\right )+97 x^{2}+256 x^{4}+64 \left (\frac {\left (3+x \right )^{2}}{2}-9-3 x \right ) \ln \left (\ln \left (3+x \right )\right )\) | \(48\) |
parts | \(-x +\ln \left (\ln \left (3+x \right )\right )^{2}+294 \ln \left (\ln \left (3+x \right )\right )+97 x^{2}+256 x^{4}+64 \left (\frac {\left (3+x \right )^{2}}{2}-9-3 x \right ) \ln \left (\ln \left (3+x \right )\right )\) | \(48\) |
Input:
int((((64*x^2+192*x)*ln(3+x)+2)*ln(ln(3+x))+(1024*x^4+3072*x^3+194*x^2+581 *x-3)*ln(3+x)+32*x^2+6)/(3+x)/ln(3+x),x,method=_RETURNVERBOSE)
Output:
ln(ln(3+x))^2+32*x^2*ln(ln(3+x))+256*x^4+97*x^2-x+6*ln(ln(3+x))
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=256 \, x^{4} + 97 \, x^{2} + 2 \, {\left (16 \, x^{2} + 3\right )} \log \left (\log \left (x + 3\right )\right ) + \log \left (\log \left (x + 3\right )\right )^{2} - x \] Input:
integrate((((64*x^2+192*x)*log(3+x)+2)*log(log(3+x))+(1024*x^4+3072*x^3+19 4*x^2+581*x-3)*log(3+x)+32*x^2+6)/(3+x)/log(3+x),x, algorithm="fricas")
Output:
256*x^4 + 97*x^2 + 2*(16*x^2 + 3)*log(log(x + 3)) + log(log(x + 3))^2 - x
Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=256 x^{4} + 32 x^{2} \log {\left (\log {\left (x + 3 \right )} \right )} + 97 x^{2} - x + \log {\left (\log {\left (x + 3 \right )} \right )}^{2} + 6 \log {\left (\log {\left (x + 3 \right )} \right )} \] Input:
integrate((((64*x**2+192*x)*ln(3+x)+2)*ln(ln(3+x))+(1024*x**4+3072*x**3+19 4*x**2+581*x-3)*ln(3+x)+32*x**2+6)/(3+x)/ln(3+x),x)
Output:
256*x**4 + 32*x**2*log(log(x + 3)) + 97*x**2 - x + log(log(x + 3))**2 + 6* log(log(x + 3))
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=256 \, x^{4} + 32 \, x^{2} \log \left (\log \left (x + 3\right )\right ) + 97 \, x^{2} + \log \left (\log \left (x + 3\right )\right )^{2} - x + 6 \, \log \left (\log \left (x + 3\right )\right ) \] Input:
integrate((((64*x^2+192*x)*log(3+x)+2)*log(log(3+x))+(1024*x^4+3072*x^3+19 4*x^2+581*x-3)*log(3+x)+32*x^2+6)/(3+x)/log(3+x),x, algorithm="maxima")
Output:
256*x^4 + 32*x^2*log(log(x + 3)) + 97*x^2 + log(log(x + 3))^2 - x + 6*log( log(x + 3))
Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=256 \, x^{4} + 32 \, x^{2} \log \left (\log \left (x + 3\right )\right ) + 97 \, x^{2} + \log \left (\log \left (x + 3\right )\right )^{2} - x + 6 \, \log \left (\log \left (x + 3\right )\right ) \] Input:
integrate((((64*x^2+192*x)*log(3+x)+2)*log(log(3+x))+(1024*x^4+3072*x^3+19 4*x^2+581*x-3)*log(3+x)+32*x^2+6)/(3+x)/log(3+x),x, algorithm="giac")
Output:
256*x^4 + 32*x^2*log(log(x + 3)) + 97*x^2 + log(log(x + 3))^2 - x + 6*log( log(x + 3))
Time = 4.60 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=256\,x^4+32\,x^2\,\ln \left (\ln \left (x+3\right )\right )+97\,x^2-x+{\ln \left (\ln \left (x+3\right )\right )}^2+6\,\ln \left (\ln \left (x+3\right )\right ) \] Input:
int((log(x + 3)*(581*x + 194*x^2 + 3072*x^3 + 1024*x^4 - 3) + log(log(x + 3))*(log(x + 3)*(192*x + 64*x^2) + 2) + 32*x^2 + 6)/(log(x + 3)*(x + 3)),x )
Output:
6*log(log(x + 3)) - x + log(log(x + 3))^2 + 97*x^2 + 256*x^4 + 32*x^2*log( log(x + 3))
Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {6+32 x^2+\left (-3+581 x+194 x^2+3072 x^3+1024 x^4\right ) \log (3+x)+\left (2+\left (192 x+64 x^2\right ) \log (3+x)\right ) \log (\log (3+x))}{(3+x) \log (3+x)} \, dx=\mathrm {log}\left (\mathrm {log}\left (x +3\right )\right )^{2}+32 \,\mathrm {log}\left (\mathrm {log}\left (x +3\right )\right ) x^{2}+6 \,\mathrm {log}\left (\mathrm {log}\left (x +3\right )\right )+256 x^{4}+97 x^{2}-x \] Input:
int((((64*x^2+192*x)*log(3+x)+2)*log(log(3+x))+(1024*x^4+3072*x^3+194*x^2+ 581*x-3)*log(3+x)+32*x^2+6)/(3+x)/log(3+x),x)
Output:
log(log(x + 3))**2 + 32*log(log(x + 3))*x**2 + 6*log(log(x + 3)) + 256*x** 4 + 97*x**2 - x