Integrand size = 73, antiderivative size = 29 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (\frac {3}{2}+\frac {5}{x}-\frac {\left (12-e^4\right ) (4+x)}{x \log (x)}\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(29)=58\).
Time = 9.91 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=-2 \left (\frac {\log (x)}{2}-\frac {1}{2} \log (10+3 x)+\frac {1}{2} \log (\log (x))\right )-2 \left (\frac {1}{2} \log (10+3 x)-\frac {1}{2} \log \left (96-8 e^4+24 x-2 e^4 x-10 \log (x)-3 x \log (x)\right )\right ) \]
Integrate[(96 + E^4*(-8 - 2*x) + 24*x + (96 - 8*E^4)*Log[x] - 10*Log[x]^2) /((-96*x - 24*x^2 + E^4*(8*x + 2*x^2))*Log[x] + (10*x + 3*x^2)*Log[x]^2),x ]
-2*(Log[x]/2 - Log[10 + 3*x]/2 + Log[Log[x]]/2) - 2*(Log[10 + 3*x]/2 - Log [96 - 8*E^4 + 24*x - 2*E^4*x - 10*Log[x] - 3*x*Log[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.
| \(\displaystyle \int \frac {e^4 (-2 x-8)+24 x-10 \log ^2(x)+\left (96-8 e^4\right ) \log (x)+96}{\left (3 x^2+10 x\right ) \log ^2(x)+\left (-24 x^2+e^4 \left (2 x^2+8 x\right )-96 x\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-e^4 (-2 x-8)-24 x+10 \log ^2(x)-\left (96-8 e^4\right ) \log (x)-96}{x \log (x) \left (24 \left (1-\frac {e^4}{12}\right ) x-3 x \log (x)-10 \log (x)+96 \left (1-\frac {e^4}{12}\right )\right )}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 \left (-12 \left (1-\frac {e^4}{12}\right ) x+5 \log ^2(x)-48 \left (1-\frac {e^4}{12}\right ) \log (x)-48 \left (1-\frac {e^4}{12}\right )\right )}{x \log (x) \left (24 \left (1-\frac {e^4}{12}\right ) x-3 x \log (x)-10 \log (x)+96 \left (1-\frac {e^4}{12}\right )\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int -\frac {-5 \log ^2(x)+4 \left (12-e^4\right ) \log (x)+\left (12-e^4\right ) x+4 \left (12-e^4\right )}{x \log (x) \left (-3 \log (x) x+2 \left (12-e^4\right ) x-10 \log (x)+8 \left (12-e^4\right )\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {-5 \log ^2(x)+4 \left (12-e^4\right ) \log (x)+\left (12-e^4\right ) x+4 \left (12-e^4\right )}{x \log (x) \left (-3 \log (x) x+2 \left (12-e^4\right ) x-10 \log (x)+8 \left (12-e^4\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -2 \int \left (\frac {3 x+10}{2 x \left (-3 \log (x) x+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)+96 \left (1-\frac {e^4}{12}\right )\right )}+\frac {1}{2 x \log (x)}+\frac {5}{x (3 x+10)}+\frac {2 \left (12-e^4\right )}{\left (-3 \log (x) x+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)+96 \left (1-\frac {e^4}{12}\right )\right ) (3 x+10)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \left (\frac {3}{2} \int \frac {1}{-3 \log (x) x+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)+96 \left (1-\frac {e^4}{12}\right )}dx+5 \int \frac {1}{x \left (-3 \log (x) x+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)+96 \left (1-\frac {e^4}{12}\right )\right )}dx+2 \left (12-e^4\right ) \int \frac {1}{(3 x+10) \left (-3 \log (x) x+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)+96 \left (1-\frac {e^4}{12}\right )\right )}dx+\frac {\log (x)}{2}-\frac {1}{2} \log (3 x+10)+\frac {1}{2} \log (\log (x))\right )\) |
Int[(96 + E^4*(-8 - 2*x) + 24*x + (96 - 8*E^4)*Log[x] - 10*Log[x]^2)/((-96 *x - 24*x^2 + E^4*(8*x + 2*x^2))*Log[x] + (10*x + 3*x^2)*Log[x]^2),x]
3.8.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 4.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31
| method | result | size |
| parallelrisch | \(-\ln \left (\ln \left (x \right )\right )+\ln \left (\frac {2 x \,{\mathrm e}^{4}}{3}+\frac {8 \,{\mathrm e}^{4}}{3}+x \ln \left (x \right )+\frac {10 \ln \left (x \right )}{3}-8 x -32\right )-\ln \left (x \right )\) | \(38\) |
| default | \(-\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (3 x \ln \left (x \right )+2 x \,{\mathrm e}^{4}+10 \ln \left (x \right )+8 \,{\mathrm e}^{4}-24 x -96\right )\) | \(39\) |
| norman | \(-\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )+\ln \left (3 x \ln \left (x \right )+2 x \,{\mathrm e}^{4}+10 \ln \left (x \right )+8 \,{\mathrm e}^{4}-24 x -96\right )\) | \(39\) |
| risch | \(\ln \left (3 x +10\right )-\ln \left (x \right )+\ln \left (\ln \left (x \right )+\frac {2 x \,{\mathrm e}^{4}+8 \,{\mathrm e}^{4}-24 x -96}{3 x +10}\right )-\ln \left (\ln \left (x \right )\right )\) | \(43\) |
int((-10*ln(x)^2+(-8*exp(2)^2+96)*ln(x)+(-2*x-8)*exp(2)^2+24*x+96)/((3*x^2 +10*x)*ln(x)^2+((2*x^2+8*x)*exp(2)^2-24*x^2-96*x)*ln(x)),x,method=_RETURNV ERBOSE)
Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (3 \, x + 10\right ) - \log \left (x\right ) + \log \left (\frac {2 \, {\left (x + 4\right )} e^{4} + {\left (3 \, x + 10\right )} \log \left (x\right ) - 24 \, x - 96}{3 \, x + 10}\right ) - \log \left (\log \left (x\right )\right ) \]
integrate((-10*log(x)^2+(-8*exp(2)^2+96)*log(x)+(-2*x-8)*exp(2)^2+24*x+96) /((3*x^2+10*x)*log(x)^2+((2*x^2+8*x)*exp(2)^2-24*x^2-96*x)*log(x)),x, algo rithm=\
log(3*x + 10) - log(x) + log((2*(x + 4)*e^4 + (3*x + 10)*log(x) - 24*x - 9 6)/(3*x + 10)) - log(log(x))
Exception generated. \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\text {Exception raised: PolynomialError} \]
integrate((-10*ln(x)**2+(-8*exp(2)**2+96)*ln(x)+(-2*x-8)*exp(2)**2+24*x+96 )/((3*x**2+10*x)*ln(x)**2+((2*x**2+8*x)*exp(2)**2-24*x**2-96*x)*ln(x)),x)
Exception raised: PolynomialError >> 1/(9*x**3 + 60*x**2 + 100*x) contains an element of the set of generators.
Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (3 \, x + 10\right ) - \log \left (x\right ) + \log \left (\frac {2 \, x {\left (e^{4} - 12\right )} + {\left (3 \, x + 10\right )} \log \left (x\right ) + 8 \, e^{4} - 96}{3 \, x + 10}\right ) - \log \left (\log \left (x\right )\right ) \]
integrate((-10*log(x)^2+(-8*exp(2)^2+96)*log(x)+(-2*x-8)*exp(2)^2+24*x+96) /((3*x^2+10*x)*log(x)^2+((2*x^2+8*x)*exp(2)^2-24*x^2-96*x)*log(x)),x, algo rithm=\
log(3*x + 10) - log(x) + log((2*x*(e^4 - 12) + (3*x + 10)*log(x) + 8*e^4 - 96)/(3*x + 10)) - log(log(x))
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\log \left (2 \, x e^{4} + 3 \, x \log \left (x\right ) - 24 \, x + 8 \, e^{4} + 10 \, \log \left (x\right ) - 96\right ) - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \]
integrate((-10*log(x)^2+(-8*exp(2)^2+96)*log(x)+(-2*x-8)*exp(2)^2+24*x+96) /((3*x^2+10*x)*log(x)^2+((2*x^2+8*x)*exp(2)^2-24*x^2-96*x)*log(x)),x, algo rithm=\
Time = 8.99 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(x)} \, dx=\ln \left (8\,{\mathrm {e}}^4-24\,x+10\,\ln \left (x\right )+2\,x\,{\mathrm {e}}^4+3\,x\,\ln \left (x\right )-96\right )-\ln \left (\ln \left (x\right )\right )-\ln \left (x\right ) \]