Integrand size = 141, antiderivative size = 29 \[ \int \frac {-9 x^5+3 x^6+\left (60 x^2-280 x^3+80 x^4\right ) \log \left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (36-168 x+48 x^2\right ) \log ^3\left (\frac {e^{4 x}}{-3 x+x^2}\right )+(36-12 x) \log ^4\left (\frac {e^{4 x}}{-3 x+x^2}\right )}{-9 x^5+3 x^6} \, dx=x+\left (\frac {5}{3}+\frac {\log ^2\left (\frac {e^{4 x}}{(-3+x) x}\right )}{x^2}\right )^2 \]
Time = 1.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {-9 x^5+3 x^6+\left (60 x^2-280 x^3+80 x^4\right ) \log \left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (36-168 x+48 x^2\right ) \log ^3\left (\frac {e^{4 x}}{-3 x+x^2}\right )+(36-12 x) \log ^4\left (\frac {e^{4 x}}{-3 x+x^2}\right )}{-9 x^5+3 x^6} \, dx=-\frac {928}{3}+x+\frac {10 \log ^2\left (\frac {e^{4 x}}{(-3+x) x}\right )}{3 x^2}+\frac {\log ^4\left (\frac {e^{4 x}}{(-3+x) x}\right )}{x^4} \]
Integrate[(-9*x^5 + 3*x^6 + (60*x^2 - 280*x^3 + 80*x^4)*Log[E^(4*x)/(-3*x + x^2)] + (60*x^2 - 20*x^3)*Log[E^(4*x)/(-3*x + x^2)]^2 + (36 - 168*x + 48 *x^2)*Log[E^(4*x)/(-3*x + x^2)]^3 + (36 - 12*x)*Log[E^(4*x)/(-3*x + x^2)]^ 4)/(-9*x^5 + 3*x^6),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.
\(\displaystyle \int \frac {3 x^6-9 x^5+(36-12 x) \log ^4\left (\frac {e^{4 x}}{x^2-3 x}\right )+\left (48 x^2-168 x+36\right ) \log ^3\left (\frac {e^{4 x}}{x^2-3 x}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{x^2-3 x}\right )+\left (80 x^4-280 x^3+60 x^2\right ) \log \left (\frac {e^{4 x}}{x^2-3 x}\right )}{3 x^6-9 x^5} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {3 x^6-9 x^5+(36-12 x) \log ^4\left (\frac {e^{4 x}}{x^2-3 x}\right )+\left (48 x^2-168 x+36\right ) \log ^3\left (\frac {e^{4 x}}{x^2-3 x}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{x^2-3 x}\right )+\left (80 x^4-280 x^3+60 x^2\right ) \log \left (\frac {e^{4 x}}{x^2-3 x}\right )}{x^5 (3 x-9)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {4 \log ^4\left (\frac {e^{4 x}}{(x-3) x}\right )}{x^5}-\frac {20 \log ^2\left (\frac {e^{4 x}}{(x-3) x}\right )}{3 x^3}+\frac {4 \left (4 x^2-14 x+3\right ) \log ^3\left (\frac {e^{4 x}}{(x-3) x}\right )}{(x-3) x^5}+\frac {20 \left (4 x^2-14 x+3\right ) \log \left (\frac {e^{4 x}}{(x-3) x}\right )}{3 (x-3) x^3}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {\log ^4\left (\frac {e^{4 x}}{(x-3) x}\right )}{x^5}dx-4 \int \frac {\log ^3\left (\frac {e^{4 x}}{(x-3) x}\right )}{x^5}dx+\frac {52}{3} \int \frac {\log ^3\left (\frac {e^{4 x}}{(x-3) x}\right )}{x^4}dx+\frac {4}{9} \int \frac {\log ^3\left (\frac {e^{4 x}}{(x-3) x}\right )}{x^3}dx-\frac {20}{3} \int \frac {\log ^2\left (\frac {e^{4 x}}{(x-3) x}\right )}{x^3}dx+\frac {4}{27} \int \frac {\log ^3\left (\frac {e^{4 x}}{(x-3) x}\right )}{x^2}dx-\frac {4}{81} \int \frac {\log ^3\left (\frac {e^{4 x}}{(x-3) x}\right )}{x-3}dx+\frac {4}{81} \int \frac {\log ^3\left (\frac {e^{4 x}}{(x-3) x}\right )}{x}dx-\frac {20}{27} \int \frac {\log \left (\frac {e^{4 x}}{(x-3) x}\right )}{x-3}dx+\frac {20}{27} \int \frac {\log \left (\frac {e^{4 x}}{(x-3) x}\right )}{x}dx-\frac {5}{3 x^2}+\frac {10 \log \left (-\frac {e^{4 x}}{(3-x) x}\right )}{3 x^2}+x+\frac {130}{3 x}-\frac {250}{27} \log (3-x)+\frac {3370 \log (x)}{27}-\frac {260 \log \left (-\frac {e^{4 x}}{(3-x) x}\right )}{9 x}\) |
Int[(-9*x^5 + 3*x^6 + (60*x^2 - 280*x^3 + 80*x^4)*Log[E^(4*x)/(-3*x + x^2) ] + (60*x^2 - 20*x^3)*Log[E^(4*x)/(-3*x + x^2)]^2 + (36 - 168*x + 48*x^2)* Log[E^(4*x)/(-3*x + x^2)]^3 + (36 - 12*x)*Log[E^(4*x)/(-3*x + x^2)]^4)/(-9 *x^5 + 3*x^6),x]
3.2.44.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(26)=52\).
Time = 28.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93
method | result | size |
parallelrisch | \(-\frac {-18 x^{5}-27 x^{4}-60 \ln \left (\frac {{\mathrm e}^{4 x}}{x \left (-3+x \right )}\right )^{2} x^{2}-18 \ln \left (\frac {{\mathrm e}^{4 x}}{x \left (-3+x \right )}\right )^{4}}{18 x^{4}}\) | \(56\) |
risch | \(\text {Expression too large to display}\) | \(489742\) |
int(((-12*x+36)*ln(exp(x)^4/(x^2-3*x))^4+(48*x^2-168*x+36)*ln(exp(x)^4/(x^ 2-3*x))^3+(-20*x^3+60*x^2)*ln(exp(x)^4/(x^2-3*x))^2+(80*x^4-280*x^3+60*x^2 )*ln(exp(x)^4/(x^2-3*x))+3*x^6-9*x^5)/(3*x^6-9*x^5),x,method=_RETURNVERBOS E)
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79 \[ \int \frac {-9 x^5+3 x^6+\left (60 x^2-280 x^3+80 x^4\right ) \log \left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (36-168 x+48 x^2\right ) \log ^3\left (\frac {e^{4 x}}{-3 x+x^2}\right )+(36-12 x) \log ^4\left (\frac {e^{4 x}}{-3 x+x^2}\right )}{-9 x^5+3 x^6} \, dx=\frac {3 \, x^{5} + 10 \, x^{2} \log \left (\frac {e^{\left (4 \, x\right )}}{x^{2} - 3 \, x}\right )^{2} + 3 \, \log \left (\frac {e^{\left (4 \, x\right )}}{x^{2} - 3 \, x}\right )^{4}}{3 \, x^{4}} \]
integrate(((-12*x+36)*log(exp(x)^4/(x^2-3*x))^4+(48*x^2-168*x+36)*log(exp( x)^4/(x^2-3*x))^3+(-20*x^3+60*x^2)*log(exp(x)^4/(x^2-3*x))^2+(80*x^4-280*x ^3+60*x^2)*log(exp(x)^4/(x^2-3*x))+3*x^6-9*x^5)/(3*x^6-9*x^5),x, algorithm =\
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {-9 x^5+3 x^6+\left (60 x^2-280 x^3+80 x^4\right ) \log \left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (36-168 x+48 x^2\right ) \log ^3\left (\frac {e^{4 x}}{-3 x+x^2}\right )+(36-12 x) \log ^4\left (\frac {e^{4 x}}{-3 x+x^2}\right )}{-9 x^5+3 x^6} \, dx=x + \frac {10 \log {\left (\frac {e^{4 x}}{x^{2} - 3 x} \right )}^{2}}{3 x^{2}} + \frac {\log {\left (\frac {e^{4 x}}{x^{2} - 3 x} \right )}^{4}}{x^{4}} \]
integrate(((-12*x+36)*ln(exp(x)**4/(x**2-3*x))**4+(48*x**2-168*x+36)*ln(ex p(x)**4/(x**2-3*x))**3+(-20*x**3+60*x**2)*ln(exp(x)**4/(x**2-3*x))**2+(80* x**4-280*x**3+60*x**2)*ln(exp(x)**4/(x**2-3*x))+3*x**6-9*x**5)/(3*x**6-9*x **5),x)
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (29) = 58\).
Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 4.07 \[ \int \frac {-9 x^5+3 x^6+\left (60 x^2-280 x^3+80 x^4\right ) \log \left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (36-168 x+48 x^2\right ) \log ^3\left (\frac {e^{4 x}}{-3 x+x^2}\right )+(36-12 x) \log ^4\left (\frac {e^{4 x}}{-3 x+x^2}\right )}{-9 x^5+3 x^6} \, dx=x - \frac {12 \, {\left (4 \, x - \log \left (x\right )\right )} \log \left (x - 3\right )^{3} - 3 \, \log \left (x - 3\right )^{4} + 848 \, x^{3} \log \left (x\right ) - 298 \, x^{2} \log \left (x\right )^{2} + 48 \, x \log \left (x\right )^{3} - 3 \, \log \left (x\right )^{4} - 2 \, {\left (149 \, x^{2} - 72 \, x \log \left (x\right ) + 9 \, \log \left (x\right )^{2}\right )} \log \left (x - 3\right )^{2} + 4 \, {\left (212 \, x^{3} - 149 \, x^{2} \log \left (x\right ) + 36 \, x \log \left (x\right )^{2} - 3 \, \log \left (x\right )^{3}\right )} \log \left (x - 3\right )}{3 \, x^{4}} \]
integrate(((-12*x+36)*log(exp(x)^4/(x^2-3*x))^4+(48*x^2-168*x+36)*log(exp( x)^4/(x^2-3*x))^3+(-20*x^3+60*x^2)*log(exp(x)^4/(x^2-3*x))^2+(80*x^4-280*x ^3+60*x^2)*log(exp(x)^4/(x^2-3*x))+3*x^6-9*x^5)/(3*x^6-9*x^5),x, algorithm =\
x - 1/3*(12*(4*x - log(x))*log(x - 3)^3 - 3*log(x - 3)^4 + 848*x^3*log(x) - 298*x^2*log(x)^2 + 48*x*log(x)^3 - 3*log(x)^4 - 2*(149*x^2 - 72*x*log(x) + 9*log(x)^2)*log(x - 3)^2 + 4*(212*x^3 - 149*x^2*log(x) + 36*x*log(x)^2 - 3*log(x)^3)*log(x - 3))/x^4
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (29) = 58\).
Time = 1.70 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.03 \[ \int \frac {-9 x^5+3 x^6+\left (60 x^2-280 x^3+80 x^4\right ) \log \left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (36-168 x+48 x^2\right ) \log ^3\left (\frac {e^{4 x}}{-3 x+x^2}\right )+(36-12 x) \log ^4\left (\frac {e^{4 x}}{-3 x+x^2}\right )}{-9 x^5+3 x^6} \, dx=x - \frac {848 \, \log \left (x^{2} - 3 \, x\right )}{3 \, x} + \frac {298 \, \log \left (x^{2} - 3 \, x\right )^{2}}{3 \, x^{2}} - \frac {16 \, \log \left (x^{2} - 3 \, x\right )^{3}}{x^{3}} + \frac {\log \left (x^{2} - 3 \, x\right )^{4}}{x^{4}} \]
integrate(((-12*x+36)*log(exp(x)^4/(x^2-3*x))^4+(48*x^2-168*x+36)*log(exp( x)^4/(x^2-3*x))^3+(-20*x^3+60*x^2)*log(exp(x)^4/(x^2-3*x))^2+(80*x^4-280*x ^3+60*x^2)*log(exp(x)^4/(x^2-3*x))+3*x^6-9*x^5)/(3*x^6-9*x^5),x, algorithm =\
x - 848/3*log(x^2 - 3*x)/x + 298/3*log(x^2 - 3*x)^2/x^2 - 16*log(x^2 - 3*x )^3/x^3 + log(x^2 - 3*x)^4/x^4
Time = 9.86 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {-9 x^5+3 x^6+\left (60 x^2-280 x^3+80 x^4\right ) \log \left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (60 x^2-20 x^3\right ) \log ^2\left (\frac {e^{4 x}}{-3 x+x^2}\right )+\left (36-168 x+48 x^2\right ) \log ^3\left (\frac {e^{4 x}}{-3 x+x^2}\right )+(36-12 x) \log ^4\left (\frac {e^{4 x}}{-3 x+x^2}\right )}{-9 x^5+3 x^6} \, dx=\frac {x^5+\frac {10\,x^2\,{\ln \left (-\frac {{\mathrm {e}}^{4\,x}}{3\,x-x^2}\right )}^2}{3}+{\ln \left (-\frac {{\mathrm {e}}^{4\,x}}{3\,x-x^2}\right )}^4}{x^4} \]