Integrand size = 105, antiderivative size = 24 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (\log \left (\frac {-2+\frac {\log (15)}{x}}{e^5 \left (-5+e^5-x\right )^2}\right )\right ) \] Output:
ln(ln((ln(15)/x-2)/(exp(5)-x-5)^2/exp(5)))
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (\log \left (\frac {-2 x+\log (15)}{e^5 x \left (5-e^5+x\right )^2}\right )\right ) \] Input:
Integrate[(-4*x^2 + (5 - E^5 + 3*x)*Log[15])/((10*x^2 - 2*E^5*x^2 + 2*x^3 + (-5*x + E^5*x - x^2)*Log[15])*Log[(-2*x + Log[15])/(E^15*x + E^10*(-10*x - 2*x^2) + E^5*(25*x + 10*x^2 + x^3))]),x]
Output:
Log[Log[(-2*x + Log[15])/(E^5*x*(5 - E^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.
\(\displaystyle \int \frac {\left (3 x-e^5+5\right ) \log (15)-4 x^2}{\left (2 x^3-2 e^5 x^2+10 x^2+\left (-x^2+e^5 x-5 x\right ) \log (15)\right ) \log \left (\frac {\log (15)-2 x}{e^{10} \left (-2 x^2-10 x\right )+e^5 \left (x^3+10 x^2+25 x\right )+e^{15} x}\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (3 x-e^5+5\right ) \log (15)-4 x^2}{\left (2 x^3+\left (10-2 e^5\right ) x^2+\left (-x^2+e^5 x-5 x\right ) \log (15)\right ) \log \left (\frac {\log (15)-2 x}{e^{10} \left (-2 x^2-10 x\right )+e^5 \left (x^3+10 x^2+25 x\right )+e^{15} x}\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (3 x-e^5+5\right ) \log (15)-4 x^2}{x \left (2 x^2+x \left (10-2 e^5-\log (15)\right )-\left (5-e^5\right ) \log (15)\right ) \log \left (\frac {\log (15)-2 x}{e^{10} \left (-2 x^2-10 x\right )+e^5 \left (x^3+10 x^2+25 x\right )+e^{15} x}\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {2}{\left (-x+e^5-5\right ) \log \left (\frac {\log (15)-2 x}{e^5 x \left (x-e^5+5\right )^2}\right )}-\frac {1}{x \log \left (\frac {\log (15)-2 x}{e^5 x \left (x-e^5+5\right )^2}\right )}+\frac {2}{(2 x-\log (15)) \log \left (\frac {\log (15)-2 x}{e^5 x \left (x-e^5+5\right )^2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {1}{\left (-x+e^5-5\right ) \log \left (\frac {\log (15)-2 x}{e^5 x \left (x-e^5+5\right )^2}\right )}dx-\int \frac {1}{x \log \left (\frac {\log (15)-2 x}{e^5 x \left (x-e^5+5\right )^2}\right )}dx+2 \int \frac {1}{(2 x-\log (15)) \log \left (\frac {\log (15)-2 x}{e^5 x \left (x-e^5+5\right )^2}\right )}dx\) |
Input:
Int[(-4*x^2 + (5 - E^5 + 3*x)*Log[15])/((10*x^2 - 2*E^5*x^2 + 2*x^3 + (-5* x + E^5*x - x^2)*Log[15])*Log[(-2*x + Log[15])/(E^15*x + E^10*(-10*x - 2*x ^2) + E^5*(25*x + 10*x^2 + x^3))]),x]
Output:
$Aborted
Time = 1.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58
method | result | size |
default | \(\ln \left (-5+\ln \left (\frac {\ln \left (15\right )-2 x}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}-10 \,{\mathrm e}^{5}+10 x +25\right )}\right )\right )\) | \(38\) |
parallelrisch | \(\ln \left (\ln \left (\frac {\left (\ln \left (15\right )-2 x \right ) {\mathrm e}^{-5}}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}-10 \,{\mathrm e}^{5}+10 x +25\right )}\right )\right )\) | \(40\) |
risch | \(\ln \left (\ln \left (\frac {\ln \left (3\right )+\ln \left (5\right )-2 x}{x \,{\mathrm e}^{15}+\left (-2 x^{2}-10 x \right ) {\mathrm e}^{10}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{5}}\right )\right )\) | \(46\) |
norman | \(\ln \left (\ln \left (\frac {\ln \left (15\right )-2 x}{x \,{\mathrm e}^{15}+\left (-2 x^{2}-10 x \right ) {\mathrm e}^{10}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{5}}\right )\right )\) | \(48\) |
Input:
int(((-exp(5)+3*x+5)*ln(15)-4*x^2)/((x*exp(5)-x^2-5*x)*ln(15)-2*x^2*exp(5) +2*x^3+10*x^2)/ln((ln(15)-2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10* x^2+25*x)*exp(5))),x,method=_RETURNVERBOSE)
Output:
ln(-5+ln((ln(15)-2*x)/x/(exp(5)^2-2*x*exp(5)+x^2-10*exp(5)+10*x+25)))
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (\log \left (-\frac {2 \, x - \log \left (15\right )}{x e^{15} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{10} + {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{5}}\right )\right ) \] Input:
integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^ 2*exp(5)+2*x^3+10*x^2)/log((log(15)-2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^ 2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="fricas")
Output:
log(log(-(2*x - log(15))/(x*e^15 - 2*(x^2 + 5*x)*e^10 + (x^3 + 10*x^2 + 25 *x)*e^5)))
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log {\left (\log {\left (\frac {- 2 x + \log {\left (15 \right )}}{x e^{15} + \left (- 2 x^{2} - 10 x\right ) e^{10} + \left (x^{3} + 10 x^{2} + 25 x\right ) e^{5}} \right )} \right )} \] Input:
integrate(((-exp(5)+3*x+5)*ln(15)-4*x**2)/((x*exp(5)-x**2-5*x)*ln(15)-2*x* *2*exp(5)+2*x**3+10*x**2)/ln((ln(15)-2*x)/(x*exp(5)**3+(-2*x**2-10*x)*exp( 5)**2+(x**3+10*x**2+25*x)*exp(5))),x)
Output:
log(log((-2*x + log(15))/(x*exp(15) + (-2*x**2 - 10*x)*exp(10) + (x**3 + 1 0*x**2 + 25*x)*exp(5))))
Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (-2 \, \log \left (x - e^{5} + 5\right ) - \log \left (x\right ) + \log \left (-2 \, x + \log \left (5\right ) + \log \left (3\right )\right ) - 5\right ) \] Input:
integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^ 2*exp(5)+2*x^3+10*x^2)/log((log(15)-2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^ 2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="maxima")
Output:
log(-2*log(x - e^5 + 5) - log(x) + log(-2*x + log(5) + log(3)) - 5)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (-\log \left (x^{3} e^{5} - 2 \, x^{2} e^{10} + 10 \, x^{2} e^{5} + x e^{15} - 10 \, x e^{10} + 25 \, x e^{5}\right ) + \log \left (-2 \, x + \log \left (15\right )\right )\right ) \] Input:
integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^ 2*exp(5)+2*x^3+10*x^2)/log((log(15)-2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^ 2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="giac")
Output:
log(-log(x^3*e^5 - 2*x^2*e^10 + 10*x^2*e^5 + x*e^15 - 10*x*e^10 + 25*x*e^5 ) + log(-2*x + log(15)))
Time = 4.46 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\ln \left (\ln \left (-\frac {2\,x-\ln \left (15\right )}{{\mathrm {e}}^5\,\left (x^3+10\,x^2+25\,x\right )-{\mathrm {e}}^{10}\,\left (2\,x^2+10\,x\right )+x\,{\mathrm {e}}^{15}}\right )\right ) \] Input:
int(-(log(15)*(3*x - exp(5) + 5) - 4*x^2)/(log(-(2*x - log(15))/(exp(5)*(2 5*x + 10*x^2 + x^3) - exp(10)*(10*x + 2*x^2) + x*exp(15)))*(log(15)*(5*x - x*exp(5) + x^2) + 2*x^2*exp(5) - 10*x^2 - 2*x^3)),x)
Output:
log(log(-(2*x - log(15))/(exp(5)*(25*x + 10*x^2 + x^3) - exp(10)*(10*x + 2 *x^2) + x*exp(15))))
Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (15\right )-2 x}{e^{15} x -2 e^{10} x^{2}-10 e^{10} x +e^{5} x^{3}+10 e^{5} x^{2}+25 e^{5} x}\right )\right ) \] Input:
int(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^2*exp( 5)+2*x^3+10*x^2)/log((log(15)-2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3 +10*x^2+25*x)*exp(5))),x)
Output:
log(log((log(15) - 2*x)/(e**15*x - 2*e**10*x**2 - 10*e**10*x + e**5*x**3 + 10*e**5*x**2 + 25*e**5*x)))