Integrand size = 63, antiderivative size = 28 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=\frac {1}{4}+\frac {1}{3} \left (x+\log (x) \left (-x+\log \left (-\frac {24}{5}-e^{16}+x\right )\right )\right ) \] Output:
1/3*(ln(-exp(16)+x-24/5)-x)*ln(x)+1/3*x+1/4
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.64 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=\frac {1}{3} \left (x+\log \left (\frac {24}{5}+e^{16}\right ) \log \left (24+5 e^{16}-5 x\right )-x \log (x)+\log \left (\frac {x}{\frac {24}{5}+e^{16}}\right ) \log \left (-\frac {24}{5}-e^{16}+x\right )-\operatorname {PolyLog}\left (2,\frac {24+5 e^{16}-5 x}{24+5 e^{16}}\right )+\operatorname {PolyLog}\left (2,-\frac {5 \left (-\frac {24}{5}-e^{16}+x\right )}{24+5 e^{16}}\right )\right ) \] Input:
Integrate[((-29*x - 5*E^16*x + 5*x^2)*Log[x] + (24 + 5*E^16 - 5*x)*Log[(-2 4 - 5*E^16 + 5*x)/5])/(72*x + 15*E^16*x - 15*x^2),x]
Output:
(x + Log[24/5 + E^16]*Log[24 + 5*E^16 - 5*x] - x*Log[x] + Log[x/(24/5 + E^ 16)]*Log[-24/5 - E^16 + x] - PolyLog[2, (24 + 5*E^16 - 5*x)/(24 + 5*E^16)] + PolyLog[2, (-5*(-24/5 - E^16 + x))/(24 + 5*E^16)])/3
Leaf count is larger than twice the leaf count of optimal. \(67\) vs. \(2(28)=56\).
Time = 0.64 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 2026, 7293, 2009}
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.
\(\displaystyle \int \frac {\left (5 x^2-5 e^{16} x-29 x\right ) \log (x)+\left (-5 x+5 e^{16}+24\right ) \log \left (\frac {1}{5} \left (5 x-5 e^{16}-24\right )\right )}{-15 x^2+15 e^{16} x+72 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (5 x^2-5 e^{16} x-29 x\right ) \log (x)+\left (-5 x+5 e^{16}+24\right ) \log \left (\frac {1}{5} \left (5 x-5 e^{16}-24\right )\right )}{\left (72+15 e^{16}\right ) x-15 x^2}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (5 x^2-5 e^{16} x-29 x\right ) \log (x)+\left (-5 x+5 e^{16}+24\right ) \log \left (\frac {1}{5} \left (5 x-5 e^{16}-24\right )\right )}{\left (3 \left (24+5 e^{16}\right )-15 x\right ) x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\log \left (x-e^{16}-\frac {24}{5}\right )}{3 x}-\frac {\left (-5 x+5 e^{16}+29\right ) \log (x)}{3 \left (-5 x+5 e^{16}+24\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x}{3}-\frac {1}{3} x \log (x)+\frac {1}{3} \log \left (\frac {24}{5}+e^{16}\right ) \log \left (-5 x+5 e^{16}+24\right )+\frac {1}{3} \log \left (\frac {5 x}{24+5 e^{16}}\right ) \log \left (x+\frac {1}{5} \left (-24-5 e^{16}\right )\right )\) |
Input:
Int[((-29*x - 5*E^16*x + 5*x^2)*Log[x] + (24 + 5*E^16 - 5*x)*Log[(-24 - 5* E^16 + 5*x)/5])/(72*x + 15*E^16*x - 15*x^2),x]
Output:
x/3 + (Log[24/5 + E^16]*Log[24 + 5*E^16 - 5*x])/3 - (x*Log[x])/3 + (Log[(5 *x)/(24 + 5*E^16)]*Log[(-24 - 5*E^16)/5 + x])/3
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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])
Time = 1.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79
method | result | size |
norman | \(\frac {x}{3}-\frac {x \ln \left (x \right )}{3}+\frac {\ln \left (x \right ) \ln \left (-{\mathrm e}^{16}+x -\frac {24}{5}\right )}{3}\) | \(22\) |
risch | \(\frac {x}{3}-\frac {x \ln \left (x \right )}{3}+\frac {\ln \left (x \right ) \ln \left (-{\mathrm e}^{16}+x -\frac {24}{5}\right )}{3}\) | \(22\) |
parallelrisch | \(-\frac {x \ln \left (x \right )}{3}+\frac {\ln \left (x \right ) \ln \left (-{\mathrm e}^{16}+x -\frac {24}{5}\right )}{3}+\frac {2 \,{\mathrm e}^{16}}{3}+\frac {x}{3}+\frac {16}{5}\) | \(27\) |
parts | \(\frac {\ln \left (-{\mathrm e}^{16}+x -\frac {24}{5}\right ) \ln \left (\frac {5 x}{5 \,{\mathrm e}^{16}+24}\right )}{3}-\frac {x \ln \left (x \right )}{3}+\frac {x}{3}+\frac {\left (\ln \left (x \right )-\ln \left (\frac {5 x}{5 \,{\mathrm e}^{16}+24}\right )\right ) \ln \left (\frac {5 \,{\mathrm e}^{16}-5 x +24}{5 \,{\mathrm e}^{16}+24}\right )}{3}\) | \(70\) |
default | \(\frac {\ln \left (-5 \,{\mathrm e}^{16}+5 x -24\right ) \ln \left (\frac {5 x}{5 \,{\mathrm e}^{16}+24}\right )}{3}-\frac {\ln \left (5\right ) \ln \left (x \right )}{3}-\frac {x \ln \left (x \right )}{3}+\frac {x}{3}+\frac {\left (\ln \left (x \right )-\ln \left (\frac {5 x}{5 \,{\mathrm e}^{16}+24}\right )\right ) \ln \left (\frac {5 \,{\mathrm e}^{16}-5 x +24}{5 \,{\mathrm e}^{16}+24}\right )}{3}\) | \(78\) |
Input:
int(((-5*x*exp(16)+5*x^2-29*x)*ln(x)+(5*exp(16)-5*x+24)*ln(-exp(16)+x-24/5 ))/(15*x*exp(16)-15*x^2+72*x),x,method=_RETURNVERBOSE)
Output:
1/3*x-1/3*x*ln(x)+1/3*ln(x)*ln(-exp(16)+x-24/5)
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=-\frac {1}{3} \, {\left (x - \log \left (x - e^{16} - \frac {24}{5}\right )\right )} \log \left (x\right ) + \frac {1}{3} \, x \] Input:
integrate(((-5*x*exp(16)+5*x^2-29*x)*log(x)+(5*exp(16)-5*x+24)*log(-exp(16 )+x-24/5))/(15*x*exp(16)-15*x^2+72*x),x, algorithm="fricas")
Output:
-1/3*(x - log(x - e^16 - 24/5))*log(x) + 1/3*x
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=- \frac {x \log {\left (x \right )}}{3} + \frac {x}{3} + \frac {\log {\left (x \right )} \log {\left (x - e^{16} - \frac {24}{5} \right )}}{3} \] Input:
integrate(((-5*x*exp(16)+5*x**2-29*x)*ln(x)+(5*exp(16)-5*x+24)*ln(-exp(16) +x-24/5))/(15*x*exp(16)-15*x**2+72*x),x)
Output:
-x*log(x)/3 + x/3 + log(x)*log(x - exp(16) - 24/5)/3
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=-\frac {1}{3} \, {\left (x + \log \left (5\right )\right )} \log \left (x\right ) + \frac {1}{3} \, \log \left (5 \, x - 5 \, e^{16} - 24\right ) \log \left (x\right ) + \frac {1}{3} \, x \] Input:
integrate(((-5*x*exp(16)+5*x^2-29*x)*log(x)+(5*exp(16)-5*x+24)*log(-exp(16 )+x-24/5))/(15*x*exp(16)-15*x^2+72*x),x, algorithm="maxima")
Output:
-1/3*(x + log(5))*log(x) + 1/3*log(5*x - 5*e^16 - 24)*log(x) + 1/3*x
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=-\frac {1}{3} \, x \log \left (x\right ) - \frac {1}{3} \, \log \left (5\right ) \log \left (x\right ) + \frac {1}{3} \, \log \left (5 \, x - 5 \, e^{16} - 24\right ) \log \left (x\right ) + \frac {1}{3} \, x \] Input:
integrate(((-5*x*exp(16)+5*x^2-29*x)*log(x)+(5*exp(16)-5*x+24)*log(-exp(16 )+x-24/5))/(15*x*exp(16)-15*x^2+72*x),x, algorithm="giac")
Output:
-1/3*x*log(x) - 1/3*log(5)*log(x) + 1/3*log(5*x - 5*e^16 - 24)*log(x) + 1/ 3*x
Time = 2.79 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=\frac {x}{3}+\frac {\ln \left (x-{\mathrm {e}}^{16}-\frac {24}{5}\right )\,\ln \left (x\right )}{3}-\frac {x\,\ln \left (x\right )}{3} \] Input:
int((log(x - exp(16) - 24/5)*(5*exp(16) - 5*x + 24) - log(x)*(29*x + 5*x*e xp(16) - 5*x^2))/(72*x + 15*x*exp(16) - 15*x^2),x)
Output:
x/3 + (log(x - exp(16) - 24/5)*log(x))/3 - (x*log(x))/3
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{72 x+15 e^{16} x-15 x^2} \, dx=\frac {\mathrm {log}\left (-e^{16}+x -\frac {24}{5}\right ) \mathrm {log}\left (x \right )}{3}-\frac {\mathrm {log}\left (x \right ) x}{3}+\frac {x}{3} \] Input:
int(((-5*x*exp(16)+5*x^2-29*x)*log(x)+(5*exp(16)-5*x+24)*log(-exp(16)+x-24 /5))/(15*x*exp(16)-15*x^2+72*x),x)
Output:
(log(( - 5*e**16 + 5*x - 24)/5)*log(x) - log(x)*x + x)/3