Integrand size = 125, antiderivative size = 27 \begin {dmath*} \int \frac {18 x+\left (-2160+876 x-90 x^2\right ) \log \left (\frac {24-5 x}{-5+x}\right )+\left (1440-588 x+60 x^2\right ) \log ^2\left (\frac {24-5 x}{-5+x}\right )+\left (-240+98 x-10 x^2+120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )}{\left (120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )} \, dx=\frac {19}{4}+x+\frac {\left (1-\frac {3}{\log \left (-5+\frac {1}{5-x}\right )}\right )^2}{x^2} \end {dmath*}
Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \begin {dmath*} \int \frac {18 x+\left (-2160+876 x-90 x^2\right ) \log \left (\frac {24-5 x}{-5+x}\right )+\left (1440-588 x+60 x^2\right ) \log ^2\left (\frac {24-5 x}{-5+x}\right )+\left (-240+98 x-10 x^2+120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )}{\left (120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )} \, dx=\frac {1}{x^2}+x+\frac {9}{x^2 \log ^2\left (\frac {24-5 x}{-5+x}\right )}-\frac {6}{x^2 \log \left (\frac {24-5 x}{-5+x}\right )} \end {dmath*}
Integrate[(18*x + (-2160 + 876*x - 90*x^2)*Log[(24 - 5*x)/(-5 + x)] + (144 0 - 588*x + 60*x^2)*Log[(24 - 5*x)/(-5 + x)]^2 + (-240 + 98*x - 10*x^2 + 1 20*x^3 - 49*x^4 + 5*x^5)*Log[(24 - 5*x)/(-5 + x)]^3)/((120*x^3 - 49*x^4 + 5*x^5)*Log[(24 - 5*x)/(-5 + x)]^3),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.
\(\displaystyle \int \frac {\left (60 x^2-588 x+1440\right ) \log ^2\left (\frac {24-5 x}{x-5}\right )+\left (-90 x^2+876 x-2160\right ) \log \left (\frac {24-5 x}{x-5}\right )+\left (5 x^5-49 x^4+120 x^3-10 x^2+98 x-240\right ) \log ^3\left (\frac {24-5 x}{x-5}\right )+18 x}{\left (5 x^5-49 x^4+120 x^3\right ) \log ^3\left (\frac {24-5 x}{x-5}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (60 x^2-588 x+1440\right ) \log ^2\left (\frac {24-5 x}{x-5}\right )+\left (-90 x^2+876 x-2160\right ) \log \left (\frac {24-5 x}{x-5}\right )+\left (5 x^5-49 x^4+120 x^3-10 x^2+98 x-240\right ) \log ^3\left (\frac {24-5 x}{x-5}\right )+18 x}{x^3 \left (5 x^2-49 x+120\right ) \log ^3\left (\frac {24-5 x}{x-5}\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {x^3-2}{x^3}+\frac {12}{x^3 \log \left (\frac {24-5 x}{x-5}\right )}+\frac {18}{(x-5) x^2 (5 x-24) \log ^3\left (\frac {24-5 x}{x-5}\right )}-\frac {6 \left (15 x^2-146 x+360\right )}{(x-5) x^3 (5 x-24) \log ^2\left (\frac {24-5 x}{x-5}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 12 \int \frac {1}{x^3 \log \left (\frac {24-5 x}{x-5}\right )}dx+18 \int \frac {1}{(x-5) x^2 (5 x-24) \log ^3\left (\frac {24-5 x}{x-5}\right )}dx-6 \int \frac {15 x^2-146 x+360}{(x-5) x^3 (5 x-24) \log ^2\left (\frac {24-5 x}{x-5}\right )}dx+\frac {1}{x^2}+x\) |
Int[(18*x + (-2160 + 876*x - 90*x^2)*Log[(24 - 5*x)/(-5 + x)] + (1440 - 58 8*x + 60*x^2)*Log[(24 - 5*x)/(-5 + x)]^2 + (-240 + 98*x - 10*x^2 + 120*x^3 - 49*x^4 + 5*x^5)*Log[(24 - 5*x)/(-5 + x)]^3)/((120*x^3 - 49*x^4 + 5*x^5) *Log[(24 - 5*x)/(-5 + x)]^3),x]
3.11.10.3.1 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])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 61.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70
method | result | size |
risch | \(\frac {x^{3}+1}{x^{2}}-\frac {3 \left (2 \ln \left (\frac {-5 x +24}{-5+x}\right )-3\right )}{x^{2} \ln \left (\frac {-5 x +24}{-5+x}\right )^{2}}\) | \(46\) |
norman | \(\frac {9+\ln \left (\frac {-5 x +24}{-5+x}\right )^{2}+x^{3} \ln \left (\frac {-5 x +24}{-5+x}\right )^{2}-6 \ln \left (\frac {-5 x +24}{-5+x}\right )}{x^{2} \ln \left (\frac {-5 x +24}{-5+x}\right )^{2}}\) | \(67\) |
parallelrisch | \(-\frac {-225-25 \ln \left (-\frac {5 x -24}{-5+x}\right )^{2} x^{3}-490 \ln \left (-\frac {5 x -24}{-5+x}\right )^{2} x^{2}-25 \ln \left (-\frac {5 x -24}{-5+x}\right )^{2}+150 \ln \left (-\frac {5 x -24}{-5+x}\right )}{25 \ln \left (-\frac {5 x -24}{-5+x}\right )^{2} x^{2}}\) | \(95\) |
int(((5*x^5-49*x^4+120*x^3-10*x^2+98*x-240)*ln((-5*x+24)/(-5+x))^3+(60*x^2 -588*x+1440)*ln((-5*x+24)/(-5+x))^2+(-90*x^2+876*x-2160)*ln((-5*x+24)/(-5+ x))+18*x)/(5*x^5-49*x^4+120*x^3)/ln((-5*x+24)/(-5+x))^3,x,method=_RETURNVE RBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \begin {dmath*} \int \frac {18 x+\left (-2160+876 x-90 x^2\right ) \log \left (\frac {24-5 x}{-5+x}\right )+\left (1440-588 x+60 x^2\right ) \log ^2\left (\frac {24-5 x}{-5+x}\right )+\left (-240+98 x-10 x^2+120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )}{\left (120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )} \, dx=\frac {{\left (x^{3} + 1\right )} \log \left (-\frac {5 \, x - 24}{x - 5}\right )^{2} - 6 \, \log \left (-\frac {5 \, x - 24}{x - 5}\right ) + 9}{x^{2} \log \left (-\frac {5 \, x - 24}{x - 5}\right )^{2}} \end {dmath*}
integrate(((5*x^5-49*x^4+120*x^3-10*x^2+98*x-240)*log((-5*x+24)/(-5+x))^3+ (60*x^2-588*x+1440)*log((-5*x+24)/(-5+x))^2+(-90*x^2+876*x-2160)*log((-5*x +24)/(-5+x))+18*x)/(5*x^5-49*x^4+120*x^3)/log((-5*x+24)/(-5+x))^3,x, algor ithm=\
((x^3 + 1)*log(-(5*x - 24)/(x - 5))^2 - 6*log(-(5*x - 24)/(x - 5)) + 9)/(x ^2*log(-(5*x - 24)/(x - 5))^2)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {18 x+\left (-2160+876 x-90 x^2\right ) \log \left (\frac {24-5 x}{-5+x}\right )+\left (1440-588 x+60 x^2\right ) \log ^2\left (\frac {24-5 x}{-5+x}\right )+\left (-240+98 x-10 x^2+120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )}{\left (120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )} \, dx=x + \frac {9 - 6 \log {\left (\frac {24 - 5 x}{x - 5} \right )}}{x^{2} \log {\left (\frac {24 - 5 x}{x - 5} \right )}^{2}} + \frac {1}{x^{2}} \end {dmath*}
integrate(((5*x**5-49*x**4+120*x**3-10*x**2+98*x-240)*ln((-5*x+24)/(-5+x)) **3+(60*x**2-588*x+1440)*ln((-5*x+24)/(-5+x))**2+(-90*x**2+876*x-2160)*ln( (-5*x+24)/(-5+x))+18*x)/(5*x**5-49*x**4+120*x**3)/ln((-5*x+24)/(-5+x))**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (25) = 50\).
Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.52 \begin {dmath*} \int \frac {18 x+\left (-2160+876 x-90 x^2\right ) \log \left (\frac {24-5 x}{-5+x}\right )+\left (1440-588 x+60 x^2\right ) \log ^2\left (\frac {24-5 x}{-5+x}\right )+\left (-240+98 x-10 x^2+120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )}{\left (120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )} \, dx=\frac {{\left (x^{3} + 1\right )} \log \left (x - 5\right )^{2} + {\left (x^{3} + 1\right )} \log \left (-5 \, x + 24\right )^{2} - 2 \, {\left ({\left (x^{3} + 1\right )} \log \left (x - 5\right ) + 3\right )} \log \left (-5 \, x + 24\right ) + 6 \, \log \left (x - 5\right ) + 9}{x^{2} \log \left (x - 5\right )^{2} - 2 \, x^{2} \log \left (x - 5\right ) \log \left (-5 \, x + 24\right ) + x^{2} \log \left (-5 \, x + 24\right )^{2}} \end {dmath*}
integrate(((5*x^5-49*x^4+120*x^3-10*x^2+98*x-240)*log((-5*x+24)/(-5+x))^3+ (60*x^2-588*x+1440)*log((-5*x+24)/(-5+x))^2+(-90*x^2+876*x-2160)*log((-5*x +24)/(-5+x))+18*x)/(5*x^5-49*x^4+120*x^3)/log((-5*x+24)/(-5+x))^3,x, algor ithm=\
((x^3 + 1)*log(x - 5)^2 + (x^3 + 1)*log(-5*x + 24)^2 - 2*((x^3 + 1)*log(x - 5) + 3)*log(-5*x + 24) + 6*log(x - 5) + 9)/(x^2*log(x - 5)^2 - 2*x^2*log (x - 5)*log(-5*x + 24) + x^2*log(-5*x + 24)^2)
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (25) = 50\).
Time = 0.39 (sec) , antiderivative size = 235, normalized size of antiderivative = 8.70 \begin {dmath*} \int \frac {18 x+\left (-2160+876 x-90 x^2\right ) \log \left (\frac {24-5 x}{-5+x}\right )+\left (1440-588 x+60 x^2\right ) \log ^2\left (\frac {24-5 x}{-5+x}\right )+\left (-240+98 x-10 x^2+120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )}{\left (120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )} \, dx=-\frac {3 \, {\left (\frac {2 \, {\left (5 \, x - 24\right )}^{2} \log \left (-\frac {5 \, x - 24}{x - 5}\right )}{{\left (x - 5\right )}^{2}} - \frac {20 \, {\left (5 \, x - 24\right )} \log \left (-\frac {5 \, x - 24}{x - 5}\right )}{x - 5} - \frac {3 \, {\left (5 \, x - 24\right )}^{2}}{{\left (x - 5\right )}^{2}} + \frac {30 \, {\left (5 \, x - 24\right )}}{x - 5} + 50 \, \log \left (-\frac {5 \, x - 24}{x - 5}\right ) - 75\right )}}{\frac {25 \, {\left (5 \, x - 24\right )}^{2} \log \left (-\frac {5 \, x - 24}{x - 5}\right )^{2}}{{\left (x - 5\right )}^{2}} - \frac {240 \, {\left (5 \, x - 24\right )} \log \left (-\frac {5 \, x - 24}{x - 5}\right )^{2}}{x - 5} + 576 \, \log \left (-\frac {5 \, x - 24}{x - 5}\right )^{2}} - \frac {\frac {10 \, {\left (5 \, x - 24\right )}}{x - 5} - 49}{25 \, {\left (\frac {25 \, {\left (5 \, x - 24\right )}^{2}}{{\left (x - 5\right )}^{2}} - \frac {240 \, {\left (5 \, x - 24\right )}}{x - 5} + 576\right )}} + \frac {1}{\frac {5 \, x - 24}{x - 5} - 5} \end {dmath*}
integrate(((5*x^5-49*x^4+120*x^3-10*x^2+98*x-240)*log((-5*x+24)/(-5+x))^3+ (60*x^2-588*x+1440)*log((-5*x+24)/(-5+x))^2+(-90*x^2+876*x-2160)*log((-5*x +24)/(-5+x))+18*x)/(5*x^5-49*x^4+120*x^3)/log((-5*x+24)/(-5+x))^3,x, algor ithm=\
-3*(2*(5*x - 24)^2*log(-(5*x - 24)/(x - 5))/(x - 5)^2 - 20*(5*x - 24)*log( -(5*x - 24)/(x - 5))/(x - 5) - 3*(5*x - 24)^2/(x - 5)^2 + 30*(5*x - 24)/(x - 5) + 50*log(-(5*x - 24)/(x - 5)) - 75)/(25*(5*x - 24)^2*log(-(5*x - 24) /(x - 5))^2/(x - 5)^2 - 240*(5*x - 24)*log(-(5*x - 24)/(x - 5))^2/(x - 5) + 576*log(-(5*x - 24)/(x - 5))^2) - 1/25*(10*(5*x - 24)/(x - 5) - 49)/(25* (5*x - 24)^2/(x - 5)^2 - 240*(5*x - 24)/(x - 5) + 576) + 1/((5*x - 24)/(x - 5) - 5)
Time = 18.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.59 \begin {dmath*} \int \frac {18 x+\left (-2160+876 x-90 x^2\right ) \log \left (\frac {24-5 x}{-5+x}\right )+\left (1440-588 x+60 x^2\right ) \log ^2\left (\frac {24-5 x}{-5+x}\right )+\left (-240+98 x-10 x^2+120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )}{\left (120 x^3-49 x^4+5 x^5\right ) \log ^3\left (\frac {24-5 x}{-5+x}\right )} \, dx=x+150\,\ln \left (\frac {5\,x-24}{x-5}\right )-\frac {6}{x^2\,\ln \left (-\frac {5\,x-24}{x-5}\right )}+\frac {9}{x^2\,{\ln \left (-\frac {5\,x-24}{x-5}\right )}^2}+\frac {1}{x^2}+\mathrm {atan}\left (x\,10{}\mathrm {i}-49{}\mathrm {i}\right )\,300{}\mathrm {i} \end {dmath*}
int((18*x - log(-(5*x - 24)/(x - 5))*(90*x^2 - 876*x + 2160) + log(-(5*x - 24)/(x - 5))^3*(98*x - 10*x^2 + 120*x^3 - 49*x^4 + 5*x^5 - 240) + log(-(5 *x - 24)/(x - 5))^2*(60*x^2 - 588*x + 1440))/(log(-(5*x - 24)/(x - 5))^3*( 120*x^3 - 49*x^4 + 5*x^5)),x)