Integrand size = 86, antiderivative size = 28 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 x \log \left (\frac {4 \left (-3+\frac {x^2}{4}\right )}{e^2 \left (2 x+\log \left (\frac {5}{2}\right )\right )}\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(28)=56\).
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 \left (-2 x+x \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )+\frac {1}{2} \log \left (\frac {5}{2}\right ) \log \left (2 x+\log \left (\frac {5}{2}\right )\right )-\frac {1}{2} \log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )\right ) \]
Integrate[(48*x + 4*x^3 + 4*x^2*Log[5/2] + (-48*x + 4*x^3 + (-24 + 2*x^2)* Log[5/2])*Log[(-12 + x^2)/(2*E^2*x + E^2*Log[5/2])])/(-24*x + 2*x^3 + (-12 + x^2)*Log[5/2]),x]
2*(-2*x + x*Log[(-12 + x^2)/(2*x + Log[5/2])] + (Log[5/2]*Log[2*x + Log[5/ 2]])/2 - (Log[5/2]*Log[4*x + Log[25/4]])/2)
Leaf count is larger than twice the leaf count of optimal. \(269\) vs. \(2(28)=56\).
Time = 1.09 (sec) , antiderivative size = 269, normalized size of antiderivative = 9.61, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2463, 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.
\(\displaystyle \int \frac {4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (4 x^3+\left (2 x^2-24\right ) \log \left (\frac {5}{2}\right )-48 x\right ) \log \left (\frac {x^2-12}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )+48 x}{2 x^3+\left (x^2-12\right ) \log \left (\frac {5}{2}\right )-24 x} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (\log \left (\frac {5}{2}\right )-2 x\right ) \left (4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (4 x^3+\left (2 x^2-24\right ) \log \left (\frac {5}{2}\right )-48 x\right ) \log \left (\frac {x^2-12}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )+48 x\right )}{\left (x^2-12\right ) \left (\log ^2\left (\frac {5}{2}\right )-48\right )}+\frac {4 \left (4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (4 x^3+\left (2 x^2-24\right ) \log \left (\frac {5}{2}\right )-48 x\right ) \log \left (\frac {x^2-12}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )+48 x\right )}{\left (\log ^2\left (\frac {5}{2}\right )-48\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \sqrt {3} \text {arctanh}\left (\frac {x}{2 \sqrt {3}}\right )+\frac {384 \sqrt {3} \text {arctanh}\left (\frac {x}{2 \sqrt {3}}\right )}{48-\log ^2\left (\frac {5}{2}\right )}-\frac {8 \sqrt {3} \log ^2\left (\frac {5}{2}\right ) \text {arctanh}\left (\frac {x}{2 \sqrt {3}}\right )}{48-\log ^2\left (\frac {5}{2}\right )}-\frac {2 x \log ^2\left (\frac {5}{2}\right ) \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )}{48-\log ^2\left (\frac {5}{2}\right )}+\frac {96 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )}{48-\log ^2\left (\frac {5}{2}\right )}+2 x+\frac {6 x \log ^2\left (\frac {5}{2}\right )}{48-\log ^2\left (\frac {5}{2}\right )}-\frac {288 x}{48-\log ^2\left (\frac {5}{2}\right )}-\frac {48 \log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )}{48-\log ^2\left (\frac {5}{2}\right )}+\frac {\log ^3\left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )}{48-\log ^2\left (\frac {5}{2}\right )}+\log \left (\frac {5}{2}\right ) \log \left (2 x+\log \left (\frac {5}{2}\right )\right )\) |
Int[(48*x + 4*x^3 + 4*x^2*Log[5/2] + (-48*x + 4*x^3 + (-24 + 2*x^2)*Log[5/ 2])*Log[(-12 + x^2)/(2*E^2*x + E^2*Log[5/2])])/(-24*x + 2*x^3 + (-12 + x^2 )*Log[5/2]),x]
2*x - 8*Sqrt[3]*ArcTanh[x/(2*Sqrt[3])] - (288*x)/(48 - Log[5/2]^2) + (384* Sqrt[3]*ArcTanh[x/(2*Sqrt[3])])/(48 - Log[5/2]^2) + (6*x*Log[5/2]^2)/(48 - Log[5/2]^2) - (8*Sqrt[3]*ArcTanh[x/(2*Sqrt[3])]*Log[5/2]^2)/(48 - Log[5/2 ]^2) + (96*x*Log[-((12 - x^2)/(2*x + Log[5/2]))])/(48 - Log[5/2]^2) - (2*x *Log[5/2]^2*Log[-((12 - x^2)/(2*x + Log[5/2]))])/(48 - Log[5/2]^2) + Log[5 /2]*Log[2*x + Log[5/2]] - (48*Log[5/2]*Log[4*x + Log[25/4]])/(48 - Log[5/2 ]^2) + (Log[5/2]^3*Log[4*x + Log[25/4]])/(48 - Log[5/2]^2)
3.16.92.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
parallelrisch | \(2 \ln \left (\frac {\left (x^{2}-12\right ) {\mathrm e}^{-2}}{\ln \left (\frac {5}{2}\right )+2 x}\right ) x\) | \(23\) |
norman | \(2 x \ln \left (\frac {x^{2}-12}{{\mathrm e}^{2} \ln \left (\frac {5}{2}\right )+2 \,{\mathrm e}^{2} x}\right )\) | \(24\) |
default | \(-4 x +2 x \ln \left (\frac {x^{2}-12}{\ln \left (5\right )-\ln \left (2\right )+2 x}\right )\) | \(27\) |
risch | \(2 x \ln \left (\frac {x^{2}-12}{{\mathrm e}^{2} \left (\ln \left (5\right )-\ln \left (2\right )\right )+2 \,{\mathrm e}^{2} x}\right )\) | \(29\) |
parts | \(-4 x +\ln \left (\frac {5}{2}\right ) \ln \left (\ln \left (\frac {5}{2}\right )+2 x \right )+2 x \ln \left (\frac {x^{2}-12}{\ln \left (5\right )-\ln \left (2\right )+2 x}\right )-4 \left (\frac {\ln \left (5\right )}{4}-\frac {\ln \left (2\right )}{4}\right ) \ln \left (\ln \left (5\right )-\ln \left (2\right )+2 x \right )\) | \(59\) |
int((((2*x^2-24)*ln(5/2)+4*x^3-48*x)*ln((x^2-12)/(exp(2)*ln(5/2)+2*exp(2)* x))+4*x^2*ln(5/2)+4*x^3+48*x)/((x^2-12)*ln(5/2)+2*x^3-24*x),x,method=_RETU RNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 \, x \log \left (\frac {x^{2} - 12}{2 \, x e^{2} + e^{2} \log \left (\frac {5}{2}\right )}\right ) \]
integrate((((2*x^2-24)*log(5/2)+4*x^3-48*x)*log((x^2-12)/(exp(2)*log(5/2)+ 2*exp(2)*x))+4*x^2*log(5/2)+4*x^3+48*x)/((x^2-12)*log(5/2)+2*x^3-24*x),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 x \log {\left (\frac {x^{2} - 12}{2 x e^{2} + e^{2} \log {\left (\frac {5}{2} \right )}} \right )} \]
integrate((((2*x**2-24)*ln(5/2)+4*x**3-48*x)*ln((x**2-12)/(exp(2)*ln(5/2)+ 2*exp(2)*x))+4*x**2*ln(5/2)+4*x**3+48*x)/((x**2-12)*ln(5/2)+2*x**3-24*x),x )
Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 233, normalized size of antiderivative = 8.32 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=-\frac {\log \left (\frac {5}{2}\right )^{3} \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} + 2 \, {\left (\frac {\log \left (\frac {5}{2}\right )^{2} \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} + \frac {2 \, \sqrt {3} \log \left (\frac {5}{2}\right ) \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {24 \, \log \left (x^{2} - 12\right )}{\log \left (\frac {5}{2}\right )^{2} - 48}\right )} \log \left (\frac {5}{2}\right ) + 2 \, x \log \left (x^{2} - 12\right ) - 2 \, x \log \left (2 \, x + \log \left (5\right ) - \log \left (2\right )\right ) - {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (2 \, x + \log \left (5\right ) - \log \left (2\right )\right ) - 4 \, \sqrt {3} \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right ) - 4 \, x + \frac {48 \, \log \left (\frac {5}{2}\right ) \log \left (x^{2} - 12\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {48 \, \log \left (\frac {5}{2}\right ) \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {192 \, \sqrt {3} \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} \]
integrate((((2*x^2-24)*log(5/2)+4*x^3-48*x)*log((x^2-12)/(exp(2)*log(5/2)+ 2*exp(2)*x))+4*x^2*log(5/2)+4*x^3+48*x)/((x^2-12)*log(5/2)+2*x^3-24*x),x, algorithm=\
-log(5/2)^3*log(2*x + log(5/2))/(log(5/2)^2 - 48) + 2*(log(5/2)^2*log(2*x + log(5/2))/(log(5/2)^2 - 48) + 2*sqrt(3)*log(5/2)*log((x - 2*sqrt(3))/(x + 2*sqrt(3)))/(log(5/2)^2 - 48) - 24*log(x^2 - 12)/(log(5/2)^2 - 48))*log( 5/2) + 2*x*log(x^2 - 12) - 2*x*log(2*x + log(5) - log(2)) - (log(5) - log( 2))*log(2*x + log(5) - log(2)) - 4*sqrt(3)*log((x - 2*sqrt(3))/(x + 2*sqrt (3))) - 4*x + 48*log(5/2)*log(x^2 - 12)/(log(5/2)^2 - 48) - 48*log(5/2)*lo g(2*x + log(5/2))/(log(5/2)^2 - 48) - 192*sqrt(3)*log((x - 2*sqrt(3))/(x + 2*sqrt(3)))/(log(5/2)^2 - 48)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 \, x \log \left (x^{2} - 12\right ) + {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (2 \, x + \log \left (5\right ) - \log \left (2\right )\right ) - 2 \, x \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right ) - \log \left (\frac {5}{2}\right ) \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right ) - 4 \, x \]
integrate((((2*x^2-24)*log(5/2)+4*x^3-48*x)*log((x^2-12)/(exp(2)*log(5/2)+ 2*exp(2)*x))+4*x^2*log(5/2)+4*x^3+48*x)/((x^2-12)*log(5/2)+2*x^3-24*x),x, algorithm=\
2*x*log(x^2 - 12) + (log(5) - log(2))*log(2*x + log(5) - log(2)) - 2*x*log (2*x + log(5/2)) - log(5/2)*log(2*x + log(5/2)) - 4*x
Timed out. \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=\text {Hanged} \]