Integrand size = 126, antiderivative size = 30 \[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=\log \left (\frac {\frac {e^5}{x}+\frac {x}{-e^3+5 e^8 x}}{\log (x)}\right ) \] Output:
ln((x/(5*x*exp(4)^2-exp(3))+exp(5)/x)/ln(x))
Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=-\log (x)-\log \left (1-5 e^5 x\right )+\log \left (-e^8+5 e^{13} x+x^2\right )-\log (\log (x)) \] Input:
Integrate[(E^3*x^2 - 5*E^8*x^3 + E^5*(-E^6 + 10*E^11*x - 25*E^16*x^2) + (- (E^3*x^2) + E^5*(-E^6 + 10*E^11*x - 25*E^16*x^2))*Log[x])/((-(E^3*x^3) + 5 *E^8*x^4 + E^5*(E^6*x - 10*E^11*x^2 + 25*E^16*x^3))*Log[x]),x]
Output:
-Log[x] - Log[1 - 5*E^5*x] + Log[-E^8 + 5*E^13*x + x^2] - Log[Log[x]]
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.
| \(\displaystyle \int \frac {-5 e^8 x^3+e^3 x^2+e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )+\left (e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )-e^3 x^2\right ) \log (x)}{\left (5 e^8 x^4-e^3 x^3+e^5 \left (25 e^{16} x^3-10 e^{11} x^2+e^6 x\right )\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {-5 e^8 x^3+e^3 x^2+e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )+\left (e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )-e^3 x^2\right ) \log (x)}{x \left (5 e^8 x^3-e^3 \left (1-25 e^{18}\right ) x^2-10 e^{16} x+e^{11}\right ) \log (x)}dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {25 e^7 \left (-5 e^8 x^3+e^3 x^2+e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )+\left (e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )-e^3 x^2\right ) \log (x)\right )}{x \left (5 e^5 x-1\right ) \log (x)}+\frac {\left (-5 e^5 x-25 e^{18}-1\right ) \left (-5 e^8 x^3+e^3 x^2+e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )+\left (e^5 \left (-25 e^{16} x^2+10 e^{11} x-e^6\right )-e^3 x^2\right ) \log (x)\right )}{e^3 x \left (x^2+5 e^{13} x-e^8\right ) \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 25 e^{10} \int \frac {-x^2-5 e^{13} x+e^8}{x \log (x)}dx+\int \frac {25 e^{10} x^2+125 e^{23} x-25 e^{18}-1}{x \log (x)}dx+\log \left (-x^2-5 e^{13} x+e^8\right )-\left (\left (1+25 e^{18}\right ) \log (x)\right )+25 e^{18} \log (x)-\log \left (1-5 e^5 x\right )\) |
Input:
Int[(E^3*x^2 - 5*E^8*x^3 + E^5*(-E^6 + 10*E^11*x - 25*E^16*x^2) + (-(E^3*x ^2) + E^5*(-E^6 + 10*E^11*x - 25*E^16*x^2))*Log[x])/((-(E^3*x^3) + 5*E^8*x ^4 + E^5*(E^6*x - 10*E^11*x^2 + 25*E^16*x^3))*Log[x]),x]
Output:
$Aborted
Time = 0.92 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10
| method | result | size |
| risch | \(\ln \left (-5 \,{\mathrm e}^{13} x +{\mathrm e}^{8}-x^{2}\right )-\ln \left (-5 x^{2} {\mathrm e}^{5}+x \right )-\ln \left (\ln \left (x \right )\right )\) | \(33\) |
| norman | \(-\ln \left (x \right )-\ln \left (\ln \left (x \right )\right )-\ln \left (5 x \,{\mathrm e}^{8}-{\mathrm e}^{3}\right )+\ln \left (5 \,{\mathrm e}^{5} {\mathrm e}^{8} x -{\mathrm e}^{5} {\mathrm e}^{3}+x^{2}\right )\) | \(46\) |
| parallelrisch | \(-\ln \left (\ln \left (x \right )\right )-\ln \left (\frac {\left (5 x \,{\mathrm e}^{8}-{\mathrm e}^{3}\right ) {\mathrm e}^{-8}}{5}\right )+\ln \left (5 \,{\mathrm e}^{5} {\mathrm e}^{8} x -{\mathrm e}^{5} {\mathrm e}^{3}+x^{2}\right )-\ln \left (x \right )\) | \(52\) |
| default | \(-\ln \left (\ln \left (x \right )\right )+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (5 \textit {\_Z}^{3} {\mathrm e}^{8}+\left (25 \,{\mathrm e}^{21}-{\mathrm e}^{3}\right ) \textit {\_Z}^{2}-10 \textit {\_Z} \,{\mathrm e}^{16}+{\mathrm e}^{11}\right )}{\sum }\frac {\textit {\_R} \left (5 \textit {\_R} \,{\mathrm e}^{8}-2 \,{\mathrm e}^{3}\right ) \ln \left (x -\textit {\_R} \right )}{50 \textit {\_R} \,{\mathrm e}^{21}-10 \,{\mathrm e}^{16}+15 \textit {\_R}^{2} {\mathrm e}^{8}-2 \textit {\_R} \,{\mathrm e}^{3}}\right )-\ln \left (x \right )\) | \(85\) |
| parts | \(-\ln \left (\ln \left (x \right )\right )+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (5 \textit {\_Z}^{3} {\mathrm e}^{8}+\left (25 \,{\mathrm e}^{21}-{\mathrm e}^{3}\right ) \textit {\_Z}^{2}-10 \textit {\_Z} \,{\mathrm e}^{16}+{\mathrm e}^{11}\right )}{\sum }\frac {\textit {\_R} \left (5 \textit {\_R} \,{\mathrm e}^{8}-2 \,{\mathrm e}^{3}\right ) \ln \left (x -\textit {\_R} \right )}{50 \textit {\_R} \,{\mathrm e}^{21}-10 \,{\mathrm e}^{16}+15 \textit {\_R}^{2} {\mathrm e}^{8}-2 \textit {\_R} \,{\mathrm e}^{3}}\right )-\ln \left (x \right )\) | \(85\) |
Input:
int((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*exp(3))* ln(x)+(-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3*exp(4) ^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^2)*exp(5) +5*x^4*exp(4)^2-x^3*exp(3))/ln(x),x,method=_RETURNVERBOSE)
Output:
ln(-5*exp(13)*x+exp(8)-x^2)-ln(-5*x^2*exp(5)+x)-ln(ln(x))
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=-\log \left (5 \, x^{2} e^{5} - x\right ) + \log \left (x^{2} + 5 \, x e^{13} - e^{8}\right ) - \log \left (\log \left (x\right )\right ) \] Input:
integrate((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*ex p(3))*log(x)+(-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3 *exp(4)^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^2) *exp(5)+5*x^4*exp(4)^2-x^3*exp(3))/log(x),x, algorithm="fricas")
Output:
-log(5*x^2*e^5 - x) + log(x^2 + 5*x*e^13 - e^8) - log(log(x))
Time = 0.90 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=- \log {\left (x^{2} - \frac {x}{5 e^{5}} \right )} + \log {\left (x^{2} + 5 x e^{13} - e^{8} \right )} - \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate((((-25*x**2*exp(4)**4+10*x*exp(3)*exp(4)**2-exp(3)**2)*exp(5)-x* *2*exp(3))*ln(x)+(-25*x**2*exp(4)**4+10*x*exp(3)*exp(4)**2-exp(3)**2)*exp( 5)-5*x**3*exp(4)**2+x**2*exp(3))/((25*x**3*exp(4)**4-10*x**2*exp(3)*exp(4) **2+x*exp(3)**2)*exp(5)+5*x**4*exp(4)**2-x**3*exp(3))/ln(x),x)
Output:
-log(x**2 - x*exp(-5)/5) + log(x**2 + 5*x*exp(13) - exp(8)) - log(log(x))
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=\log \left (x^{2} + 5 \, x e^{13} - e^{8}\right ) - \log \left (5 \, x e^{5} - 1\right ) - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \] Input:
integrate((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*ex p(3))*log(x)+(-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3 *exp(4)^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^2) *exp(5)+5*x^4*exp(4)^2-x^3*exp(3))/log(x),x, algorithm="maxima")
Output:
log(x^2 + 5*x*e^13 - e^8) - log(5*x*e^5 - 1) - log(x) - log(log(x))
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=\log \left (x^{2} + 5 \, x e^{13} - e^{8}\right ) - \log \left (5 \, x e^{5} - 1\right ) - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \] Input:
integrate((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*ex p(3))*log(x)+(-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3 *exp(4)^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^2) *exp(5)+5*x^4*exp(4)^2-x^3*exp(3))/log(x),x, algorithm="giac")
Output:
log(x^2 + 5*x*e^13 - e^8) - log(5*x*e^5 - 1) - log(x) - log(log(x))
Time = 2.80 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.83 \[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=-\ln \left (\ln \left (x\right )\right )-\mathrm {atan}\left (\frac {{\mathrm {e}}^8\,1{}\mathrm {i}-x\,{\mathrm {e}}^{13}\,25{}\mathrm {i}-x\,{\mathrm {e}}^{31}\,125{}\mathrm {i}+x^2\,{\mathrm {e}}^{18}\,100{}\mathrm {i}+x^2\,{\mathrm {e}}^{36}\,625{}\mathrm {i}-x^2\,1{}\mathrm {i}}{{\mathrm {e}}^8+15\,x\,{\mathrm {e}}^{13}+125\,x\,{\mathrm {e}}^{31}-100\,x^2\,{\mathrm {e}}^{18}-625\,x^2\,{\mathrm {e}}^{36}-x^2}\right )\,2{}\mathrm {i} \] Input:
int(-(exp(5)*(exp(6) - 10*x*exp(11) + 25*x^2*exp(16)) - x^2*exp(3) + 5*x^3 *exp(8) + log(x)*(exp(5)*(exp(6) - 10*x*exp(11) + 25*x^2*exp(16)) + x^2*ex p(3)))/(log(x)*(exp(5)*(x*exp(6) - 10*x^2*exp(11) + 25*x^3*exp(16)) - x^3* exp(3) + 5*x^4*exp(8))),x)
Output:
- log(log(x)) - atan((exp(8)*1i - x*exp(13)*25i - x*exp(31)*125i + x^2*exp (18)*100i + x^2*exp(36)*625i - x^2*1i)/(exp(8) + 15*x*exp(13) + 125*x*exp( 31) - 100*x^2*exp(18) - 625*x^2*exp(36) - x^2))*2i
\[ \int \frac {e^3 x^2-5 e^8 x^3+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )+\left (-e^3 x^2+e^5 \left (-e^6+10 e^{11} x-25 e^{16} x^2\right )\right ) \log (x)}{\left (-e^3 x^3+5 e^8 x^4+e^5 \left (e^6 x-10 e^{11} x^2+25 e^{16} x^3\right )\right ) \log (x)} \, dx=\frac {1250 \left (\int \frac {1}{625 e^{36} x^{3}-250 e^{31} x^{2}+25 e^{26} x +125 e^{23} x^{4}-50 e^{18} x^{3}+10 e^{13} x^{2}-e^{8} x -5 e^{5} x^{4}+x^{3}}d x \right ) e^{44}+50 \left (\int \frac {1}{625 e^{36} x^{3}-250 e^{31} x^{2}+25 e^{26} x +125 e^{23} x^{4}-50 e^{18} x^{3}+10 e^{13} x^{2}-e^{8} x -5 e^{5} x^{4}+x^{3}}d x \right ) e^{26}-4 \left (\int \frac {1}{625 e^{36} x^{3}-250 e^{31} x^{2}+25 e^{26} x +125 e^{23} x^{4}-50 e^{18} x^{3}+10 e^{13} x^{2}-e^{8} x -5 e^{5} x^{4}+x^{3}}d x \right ) e^{8}-6250 \left (\int \frac {1}{625 e^{36} x^{2}-250 e^{31} x +25 e^{26}+125 e^{23} x^{3}-50 e^{18} x^{2}+10 e^{13} x -e^{8}-5 e^{5} x^{3}+x^{2}}d x \right ) e^{49}-500 \left (\int \frac {1}{625 e^{36} x^{2}-250 e^{31} x +25 e^{26}+125 e^{23} x^{3}-50 e^{18} x^{2}+10 e^{13} x -e^{8}-5 e^{5} x^{3}+x^{2}}d x \right ) e^{31}+30 \left (\int \frac {1}{625 e^{36} x^{2}-250 e^{31} x +25 e^{26}+125 e^{23} x^{3}-50 e^{18} x^{2}+10 e^{13} x -e^{8}-5 e^{5} x^{3}+x^{2}}d x \right ) e^{13}-25 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) e^{18}+\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+25 \,\mathrm {log}\left (25 e^{18} x^{2}-10 e^{13} x +e^{8}+5 e^{5} x^{3}-x^{2}\right ) e^{18}+\mathrm {log}\left (25 e^{18} x^{2}-10 e^{13} x +e^{8}+5 e^{5} x^{3}-x^{2}\right )-75 \,\mathrm {log}\left (x \right ) e^{18}-3 \,\mathrm {log}\left (x \right )}{25 e^{18}-1} \] Input:
int((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*exp(3))* log(x)+(-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3*exp(4 )^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^2)*exp(5 )+5*x^4*exp(4)^2-x^3*exp(3))/log(x),x)
Output:
(1250*int(1/(625*e**36*x**3 - 250*e**31*x**2 + 25*e**26*x + 125*e**23*x**4 - 50*e**18*x**3 + 10*e**13*x**2 - e**8*x - 5*e**5*x**4 + x**3),x)*e**44 + 50*int(1/(625*e**36*x**3 - 250*e**31*x**2 + 25*e**26*x + 125*e**23*x**4 - 50*e**18*x**3 + 10*e**13*x**2 - e**8*x - 5*e**5*x**4 + x**3),x)*e**26 - 4 *int(1/(625*e**36*x**3 - 250*e**31*x**2 + 25*e**26*x + 125*e**23*x**4 - 50 *e**18*x**3 + 10*e**13*x**2 - e**8*x - 5*e**5*x**4 + x**3),x)*e**8 - 6250* int(1/(625*e**36*x**2 - 250*e**31*x + 25*e**26 + 125*e**23*x**3 - 50*e**18 *x**2 + 10*e**13*x - e**8 - 5*e**5*x**3 + x**2),x)*e**49 - 500*int(1/(625* e**36*x**2 - 250*e**31*x + 25*e**26 + 125*e**23*x**3 - 50*e**18*x**2 + 10* e**13*x - e**8 - 5*e**5*x**3 + x**2),x)*e**31 + 30*int(1/(625*e**36*x**2 - 250*e**31*x + 25*e**26 + 125*e**23*x**3 - 50*e**18*x**2 + 10*e**13*x - e* *8 - 5*e**5*x**3 + x**2),x)*e**13 - 25*log(log(x))*e**18 + log(log(x)) + 2 5*log(25*e**18*x**2 - 10*e**13*x + e**8 + 5*e**5*x**3 - x**2)*e**18 + log( 25*e**18*x**2 - 10*e**13*x + e**8 + 5*e**5*x**3 - x**2) - 75*log(x)*e**18 - 3*log(x))/(25*e**18 - 1)