Integrand size = 83, antiderivative size = 23 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 x^2}{3-x \log (x)}+4 \log ^2(5+x) \] Output:
4*ln(5+x)^2+4*x^2/(3-x*ln(x))
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=-\frac {4 x^2}{-3+x \log (x)}+4 \log ^2(5+x) \] Input:
Integrate[(120*x + 44*x^2 + 4*x^3 + (-20*x^2 - 4*x^3)*Log[x] + (72 - 48*x* Log[x] + 8*x^2*Log[x]^2)*Log[5 + x])/(45 + 9*x + (-30*x - 6*x^2)*Log[x] + (5*x^2 + x^3)*Log[x]^2),x]
Output:
(-4*x^2)/(-3 + x*Log[x]) + 4*Log[5 + 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.
\(\displaystyle \int \frac {4 x^3+44 x^2+\left (8 x^2 \log ^2(x)-48 x \log (x)+72\right ) \log (x+5)+\left (-4 x^3-20 x^2\right ) \log (x)+120 x}{\left (-6 x^2-30 x\right ) \log (x)+\left (x^3+5 x^2\right ) \log ^2(x)+9 x+45} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 x^3+44 x^2+\left (8 x^2 \log ^2(x)-48 x \log (x)+72\right ) \log (x+5)+\left (-4 x^3-20 x^2\right ) \log (x)+120 x}{(x+5) (3-x \log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 x^3}{(x+5) (x \log (x)-3)^2}-\frac {4 x^2 \log (x)}{(x \log (x)-3)^2}+\frac {44 x^2}{(x+5) (x \log (x)-3)^2}+\frac {120 x}{(x+5) (x \log (x)-3)^2}+\frac {8 \log (x+5)}{x+5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {x^2}{(x \log (x)-3)^2}dx+12 \int \frac {x}{(x \log (x)-3)^2}dx-4 \int \frac {x}{x \log (x)-3}dx+4 \log ^2(x+5)\) |
Input:
Int[(120*x + 44*x^2 + 4*x^3 + (-20*x^2 - 4*x^3)*Log[x] + (72 - 48*x*Log[x] + 8*x^2*Log[x]^2)*Log[5 + x])/(45 + 9*x + (-30*x - 6*x^2)*Log[x] + (5*x^2 + x^3)*Log[x]^2),x]
Output:
$Aborted
Time = 4.72 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
default | \(4 \ln \left (5+x \right )^{2}-\frac {4 x^{2}}{x \ln \left (x \right )-3}\) | \(23\) |
risch | \(4 \ln \left (5+x \right )^{2}-\frac {4 x^{2}}{x \ln \left (x \right )-3}\) | \(23\) |
parallelrisch | \(-\frac {-400 \ln \left (5+x \right )^{2} \ln \left (x \right ) x +400 x^{2}+1200 \ln \left (5+x \right )^{2}}{100 \left (x \ln \left (x \right )-3\right )}\) | \(36\) |
Input:
int(((8*x^2*ln(x)^2-48*x*ln(x)+72)*ln(5+x)+(-4*x^3-20*x^2)*ln(x)+4*x^3+44* x^2+120*x)/((x^3+5*x^2)*ln(x)^2+(-6*x^2-30*x)*ln(x)+9*x+45),x,method=_RETU RNVERBOSE)
Output:
4*ln(5+x)^2-4*x^2/(x*ln(x)-3)
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \, {\left ({\left (x \log \left (x\right ) - 3\right )} \log \left (x + 5\right )^{2} - x^{2}\right )}}{x \log \left (x\right ) - 3} \] Input:
integrate(((8*x^2*log(x)^2-48*x*log(x)+72)*log(5+x)+(-4*x^3-20*x^2)*log(x) +4*x^3+44*x^2+120*x)/((x^3+5*x^2)*log(x)^2+(-6*x^2-30*x)*log(x)+9*x+45),x, algorithm="fricas")
Output:
4*((x*log(x) - 3)*log(x + 5)^2 - x^2)/(x*log(x) - 3)
Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=- \frac {4 x^{2}}{x \log {\left (x \right )} - 3} + 4 \log {\left (x + 5 \right )}^{2} \] Input:
integrate(((8*x**2*ln(x)**2-48*x*ln(x)+72)*ln(5+x)+(-4*x**3-20*x**2)*ln(x) +4*x**3+44*x**2+120*x)/((x**3+5*x**2)*ln(x)**2+(-6*x**2-30*x)*ln(x)+9*x+45 ),x)
Output:
-4*x**2/(x*log(x) - 3) + 4*log(x + 5)**2
Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \, {\left ({\left (x \log \left (x\right ) - 3\right )} \log \left (x + 5\right )^{2} - x^{2}\right )}}{x \log \left (x\right ) - 3} \] Input:
integrate(((8*x^2*log(x)^2-48*x*log(x)+72)*log(5+x)+(-4*x^3-20*x^2)*log(x) +4*x^3+44*x^2+120*x)/((x^3+5*x^2)*log(x)^2+(-6*x^2-30*x)*log(x)+9*x+45),x, algorithm="maxima")
Output:
4*((x*log(x) - 3)*log(x + 5)^2 - x^2)/(x*log(x) - 3)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=4 \, \log \left (x + 5\right )^{2} - \frac {4 \, x^{2}}{x \log \left (x\right ) - 3} \] Input:
integrate(((8*x^2*log(x)^2-48*x*log(x)+72)*log(5+x)+(-4*x^3-20*x^2)*log(x) +4*x^3+44*x^2+120*x)/((x^3+5*x^2)*log(x)^2+(-6*x^2-30*x)*log(x)+9*x+45),x, algorithm="giac")
Output:
4*log(x + 5)^2 - 4*x^2/(x*log(x) - 3)
Time = 4.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=4\,{\ln \left (x+5\right )}^2-\frac {4\,x^2}{x\,\ln \left (x\right )-3} \] Input:
int((120*x - log(x)*(20*x^2 + 4*x^3) + log(x + 5)*(8*x^2*log(x)^2 - 48*x*l og(x) + 72) + 44*x^2 + 4*x^3)/(9*x + log(x)^2*(5*x^2 + x^3) - log(x)*(30*x + 6*x^2) + 45),x)
Output:
4*log(x + 5)^2 - (4*x^2)/(x*log(x) - 3)
Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {120 x+44 x^2+4 x^3+\left (-20 x^2-4 x^3\right ) \log (x)+\left (72-48 x \log (x)+8 x^2 \log ^2(x)\right ) \log (5+x)}{45+9 x+\left (-30 x-6 x^2\right ) \log (x)+\left (5 x^2+x^3\right ) \log ^2(x)} \, dx=\frac {4 \mathrm {log}\left (x +5\right )^{2} \mathrm {log}\left (x \right ) x -12 \mathrm {log}\left (x +5\right )^{2}-4 x^{2}}{\mathrm {log}\left (x \right ) x -3} \] Input:
int(((8*x^2*log(x)^2-48*x*log(x)+72)*log(5+x)+(-4*x^3-20*x^2)*log(x)+4*x^3 +44*x^2+120*x)/((x^3+5*x^2)*log(x)^2+(-6*x^2-30*x)*log(x)+9*x+45),x)
Output:
(4*(log(x + 5)**2*log(x)*x - 3*log(x + 5)**2 - x**2))/(log(x)*x - 3)