Integrand size = 74, antiderivative size = 23 \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=\log \left (-2+x-10 \left (3-\log \left (\frac {4}{3}\right )-\log \left (\log \left (3+x^2\right )\right )\right )\right ) \] Output:
ln(x-32-10*ln(3/4)+10*ln(ln(x^2+3)))
Time = 0.98 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=\log \left (32-x-10 \log \left (\frac {4}{3}\right )-10 \log \left (\log \left (3+x^2\right )\right )\right ) \] Input:
Integrate[(20*x + (3 + x^2)*Log[3 + x^2])/((-96 + 3*x - 32*x^2 + x^3 - (-3 0 - 10*x^2)*Log[4/3])*Log[3 + x^2] + (30 + 10*x^2)*Log[3 + x^2]*Log[Log[3 + x^2]]),x]
Output:
Log[32 - x - 10*Log[4/3] - 10*Log[Log[3 + x^2]]]
Time = 0.76 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {7292, 7235}
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.
\(\displaystyle \int \frac {\left (x^2+3\right ) \log \left (x^2+3\right )+20 x}{\left (10 x^2+30\right ) \log \left (\log \left (x^2+3\right )\right ) \log \left (x^2+3\right )+\left (x^3-32 x^2-\left (-10 x^2-30\right ) \log \left (\frac {4}{3}\right )+3 x-96\right ) \log \left (x^2+3\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (x^2+3\right ) \log \left (x^2+3\right )+20 x}{\left (x^2+3\right ) \log \left (x^2+3\right ) \left (10 \log \left (\log \left (x^2+3\right )\right )+x-32 \left (1-\frac {5}{16} \log \left (\frac {4}{3}\right )\right )\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (10 \log \left (\log \left (x^2+3\right )\right )+x-2 \left (16-5 \log \left (\frac {4}{3}\right )\right )\right )\) |
Input:
Int[(20*x + (3 + x^2)*Log[3 + x^2])/((-96 + 3*x - 32*x^2 + x^3 - (-30 - 10 *x^2)*Log[4/3])*Log[3 + x^2] + (30 + 10*x^2)*Log[3 + x^2]*Log[Log[3 + x^2] ]),x]
Output:
Log[x - 2*(16 - 5*Log[4/3]) + 10*Log[Log[3 + x^2]]]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 1.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78
method | result | size |
parallelrisch | \(\ln \left (x -32-10 \ln \left (\frac {3}{4}\right )+10 \ln \left (\ln \left (x^{2}+3\right )\right )\right )\) | \(18\) |
risch | \(\ln \left (-\ln \left (3\right )+2 \ln \left (2\right )+\frac {x}{10}+\ln \left (\ln \left (x^{2}+3\right )\right )-\frac {16}{5}\right )\) | \(22\) |
Input:
int(((x^2+3)*ln(x^2+3)+20*x)/((10*x^2+30)*ln(x^2+3)*ln(ln(x^2+3))+((-10*x^ 2-30)*ln(3/4)+x^3-32*x^2+3*x-96)*ln(x^2+3)),x,method=_RETURNVERBOSE)
Output:
ln(x-32-10*ln(3/4)+10*ln(ln(x^2+3)))
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=\log \left (x - 10 \, \log \left (\frac {3}{4}\right ) + 10 \, \log \left (\log \left (x^{2} + 3\right )\right ) - 32\right ) \] Input:
integrate(((x^2+3)*log(x^2+3)+20*x)/((10*x^2+30)*log(x^2+3)*log(log(x^2+3) )+((-10*x^2-30)*log(3/4)+x^3-32*x^2+3*x-96)*log(x^2+3)),x, algorithm="fric as")
Output:
log(x - 10*log(3/4) + 10*log(log(x^2 + 3)) - 32)
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=\log {\left (\frac {x}{10} + \log {\left (\log {\left (x^{2} + 3 \right )} \right )} - \frac {16}{5} - \log {\left (3 \right )} + 2 \log {\left (2 \right )} \right )} \] Input:
integrate(((x**2+3)*ln(x**2+3)+20*x)/((10*x**2+30)*ln(x**2+3)*ln(ln(x**2+3 ))+((-10*x**2-30)*ln(3/4)+x**3-32*x**2+3*x-96)*ln(x**2+3)),x)
Output:
log(x/10 + log(log(x**2 + 3)) - 16/5 - log(3) + 2*log(2))
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=\log \left (\frac {1}{10} \, x - \log \left (3\right ) + 2 \, \log \left (2\right ) + \log \left (\log \left (x^{2} + 3\right )\right ) - \frac {16}{5}\right ) \] Input:
integrate(((x^2+3)*log(x^2+3)+20*x)/((10*x^2+30)*log(x^2+3)*log(log(x^2+3) )+((-10*x^2-30)*log(3/4)+x^3-32*x^2+3*x-96)*log(x^2+3)),x, algorithm="maxi ma")
Output:
log(1/10*x - log(3) + 2*log(2) + log(log(x^2 + 3)) - 16/5)
Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=\log \left (x - 10 \, \log \left (3\right ) + 20 \, \log \left (2\right ) + 10 \, \log \left (\log \left (x^{2} + 3\right )\right ) - 32\right ) \] Input:
integrate(((x^2+3)*log(x^2+3)+20*x)/((10*x^2+30)*log(x^2+3)*log(log(x^2+3) )+((-10*x^2-30)*log(3/4)+x^3-32*x^2+3*x-96)*log(x^2+3)),x, algorithm="giac ")
Output:
log(x - 10*log(3) + 20*log(2) + 10*log(log(x^2 + 3)) - 32)
Timed out. \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=-\int \frac {20\,x+\ln \left (x^2+3\right )\,\left (x^2+3\right )}{\ln \left (x^2+3\right )\,\left (\ln \left (\frac {3}{4}\right )\,\left (10\,x^2+30\right )-3\,x+32\,x^2-x^3+96\right )-\ln \left (\ln \left (x^2+3\right )\right )\,\ln \left (x^2+3\right )\,\left (10\,x^2+30\right )} \,d x \] Input:
int(-(20*x + log(x^2 + 3)*(x^2 + 3))/(log(x^2 + 3)*(log(3/4)*(10*x^2 + 30) - 3*x + 32*x^2 - x^3 + 96) - log(log(x^2 + 3))*log(x^2 + 3)*(10*x^2 + 30) ),x)
Output:
-int((20*x + log(x^2 + 3)*(x^2 + 3))/(log(x^2 + 3)*(log(3/4)*(10*x^2 + 30) - 3*x + 32*x^2 - x^3 + 96) - log(log(x^2 + 3))*log(x^2 + 3)*(10*x^2 + 30) ), x)
Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {20 x+\left (3+x^2\right ) \log \left (3+x^2\right )}{\left (-96+3 x-32 x^2+x^3-\left (-30-10 x^2\right ) \log \left (\frac {4}{3}\right )\right ) \log \left (3+x^2\right )+\left (30+10 x^2\right ) \log \left (3+x^2\right ) \log \left (\log \left (3+x^2\right )\right )} \, dx=\mathrm {log}\left (10 \,\mathrm {log}\left (\mathrm {log}\left (x^{2}+3\right )\right )-10 \,\mathrm {log}\left (\frac {3}{4}\right )+x -32\right ) \] Input:
int(((x^2+3)*log(x^2+3)+20*x)/((10*x^2+30)*log(x^2+3)*log(log(x^2+3))+((-1 0*x^2-30)*log(3/4)+x^3-32*x^2+3*x-96)*log(x^2+3)),x)
Output:
log(10*log(log(x**2 + 3)) - 10*log(3/4) + x - 32)