Integrand size = 60, antiderivative size = 27 \[ \int \frac {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=\frac {64 x}{\left (5-e^2\right )^2 (3-x) \log \left (\frac {x}{4}\right )} \]
Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=-\frac {64 x}{\left (-5+e^2\right )^2 (-3+x) \log \left (\frac {x}{4}\right )} \]
Integrate[(-192 + 64*x + 192*Log[x/4])/((225 - 150*x + 25*x^2 + E^2*(-90 + 60*x - 10*x^2) + E^4*(9 - 6*x + x^2))*Log[x/4]^2),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.
\(\displaystyle \int \frac {64 x+192 \log \left (\frac {x}{4}\right )-192}{\left (25 x^2+e^2 \left (-10 x^2+60 x-90\right )+e^4 \left (x^2-6 x+9\right )-150 x+225\right ) \log ^2\left (\frac {x}{4}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {64 \left (x+3 \log \left (\frac {x}{4}\right )-3\right )}{\left (5-e^2\right )^2 (3-x)^2 \log ^2\left (\frac {x}{4}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {64 \int -\frac {-x-3 \log \left (\frac {x}{4}\right )+3}{(3-x)^2 \log ^2\left (\frac {x}{4}\right )}dx}{\left (5-e^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {64 \int \frac {-x-3 \log \left (\frac {x}{4}\right )+3}{(3-x)^2 \log ^2\left (\frac {x}{4}\right )}dx}{\left (5-e^2\right )^2}\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle -\frac {256 \int \frac {-x-3 \log \left (\frac {x}{4}\right )+3}{(3-x)^2 \log ^2\left (\frac {x}{4}\right )}d\frac {x}{4}}{\left (5-e^2\right )^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {256 \int \left (-\frac {1}{\log ^2\left (\frac {x}{4}\right ) (x-3)}-\frac {3}{\log \left (\frac {x}{4}\right ) (x-3)^2}\right )d\frac {x}{4}}{\left (5-e^2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {256 \left (-\int \frac {1}{(x-3) \log ^2\left (\frac {x}{4}\right )}d\frac {x}{4}-3 \int \frac {1}{(x-3)^2 \log \left (\frac {x}{4}\right )}d\frac {x}{4}\right )}{\left (5-e^2\right )^2}\) |
Int[(-192 + 64*x + 192*Log[x/4])/((225 - 150*x + 25*x^2 + E^2*(-90 + 60*x - 10*x^2) + E^4*(9 - 6*x + x^2))*Log[x/4]^2),x]
3.17.77.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Time = 2.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78
method | result | size |
norman | \(-\frac {64 x}{\left ({\mathrm e}^{2}-5\right )^{2} \left (-3+x \right ) \ln \left (\frac {x}{4}\right )}\) | \(21\) |
parallelrisch | \(-\frac {64 x}{\left (-10 \,{\mathrm e}^{2}+{\mathrm e}^{4}+25\right ) \ln \left (\frac {x}{4}\right ) \left (-3+x \right )}\) | \(27\) |
risch | \(-\frac {64 x}{\left (x \,{\mathrm e}^{4}-3 \,{\mathrm e}^{4}-10 \,{\mathrm e}^{2} x +30 \,{\mathrm e}^{2}+25 x -75\right ) \ln \left (\frac {x}{4}\right )}\) | \(34\) |
derivativedivides | \(-\frac {64}{\ln \left (\frac {x}{4}\right ) \left (-10 \,{\mathrm e}^{2}+{\mathrm e}^{4}+25\right )}-\frac {192}{\left (-3+x \right ) \ln \left (\frac {x}{4}\right ) \left (-10 \,{\mathrm e}^{2}+{\mathrm e}^{4}+25\right )}\) | \(47\) |
default | \(-\frac {64}{\ln \left (\frac {x}{4}\right ) \left (-10 \,{\mathrm e}^{2}+{\mathrm e}^{4}+25\right )}-\frac {192}{\left (-3+x \right ) \ln \left (\frac {x}{4}\right ) \left (-10 \,{\mathrm e}^{2}+{\mathrm e}^{4}+25\right )}\) | \(47\) |
parts | \(\frac {32}{{\mathrm e}^{4} \ln \left (2\right )-\frac {{\mathrm e}^{4} \ln \left (x \right )}{2}-10 \,{\mathrm e}^{2} \ln \left (2\right )+5 \,{\mathrm e}^{2} \ln \left (x \right )+25 \ln \left (2\right )-\frac {25 \ln \left (x \right )}{2}}+\frac {192}{\left (-3+x \right ) \left (-10 \,{\mathrm e}^{2}+{\mathrm e}^{4}+25\right ) \left (2 \ln \left (2\right )-\ln \left (x \right )\right )}\) | \(72\) |
int((192*ln(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90)*exp(2 )+25*x^2-150*x+225)/ln(1/4*x)^2,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=-\frac {64 \, x}{{\left ({\left (x - 3\right )} e^{4} - 10 \, {\left (x - 3\right )} e^{2} + 25 \, x - 75\right )} \log \left (\frac {1}{4} \, x\right )} \]
integrate((192*log(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90 )*exp(2)+25*x^2-150*x+225)/log(1/4*x)^2,x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=- \frac {64 x}{\left (- 10 x e^{2} + 25 x + x e^{4} - 3 e^{4} - 75 + 30 e^{2}\right ) \log {\left (\frac {x}{4} \right )}} \]
integrate((192*ln(1/4*x)+64*x-192)/((x**2-6*x+9)*exp(2)**2+(-10*x**2+60*x- 90)*exp(2)+25*x**2-150*x+225)/ln(1/4*x)**2,x)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.41 \[ \int \frac {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=\frac {64 \, x}{2 \, {\left (e^{4} \log \left (2\right ) - 10 \, e^{2} \log \left (2\right ) + 25 \, \log \left (2\right )\right )} x - 6 \, e^{4} \log \left (2\right ) + 60 \, e^{2} \log \left (2\right ) - {\left (x {\left (e^{4} - 10 \, e^{2} + 25\right )} - 3 \, e^{4} + 30 \, e^{2} - 75\right )} \log \left (x\right ) - 150 \, \log \left (2\right )} \]
integrate((192*log(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90 )*exp(2)+25*x^2-150*x+225)/log(1/4*x)^2,x, algorithm=\
64*x/(2*(e^4*log(2) - 10*e^2*log(2) + 25*log(2))*x - 6*e^4*log(2) + 60*e^2 *log(2) - (x*(e^4 - 10*e^2 + 25) - 3*e^4 + 30*e^2 - 75)*log(x) - 150*log(2 ))
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=-\frac {64 \, x}{x e^{4} \log \left (\frac {1}{4} \, x\right ) - 10 \, x e^{2} \log \left (\frac {1}{4} \, x\right ) + 25 \, x \log \left (\frac {1}{4} \, x\right ) - 3 \, e^{4} \log \left (\frac {1}{4} \, x\right ) + 30 \, e^{2} \log \left (\frac {1}{4} \, x\right ) - 75 \, \log \left (\frac {1}{4} \, x\right )} \]
integrate((192*log(1/4*x)+64*x-192)/((x^2-6*x+9)*exp(2)^2+(-10*x^2+60*x-90 )*exp(2)+25*x^2-150*x+225)/log(1/4*x)^2,x, algorithm=\
-64*x/(x*e^4*log(1/4*x) - 10*x*e^2*log(1/4*x) + 25*x*log(1/4*x) - 3*e^4*lo g(1/4*x) + 30*e^2*log(1/4*x) - 75*log(1/4*x))
Time = 11.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-192+64 x+192 \log \left (\frac {x}{4}\right )}{\left (225-150 x+25 x^2+e^2 \left (-90+60 x-10 x^2\right )+e^4 \left (9-6 x+x^2\right )\right ) \log ^2\left (\frac {x}{4}\right )} \, dx=-\frac {64\,x}{\ln \left (\frac {x}{4}\right )\,{\left ({\mathrm {e}}^2-5\right )}^2\,\left (x-3\right )} \]