Integrand size = 76, antiderivative size = 26 \[ \int \frac {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx=\frac {5}{2 x \left (\frac {21}{5}-3 x+\frac {3}{\log \left (\frac {x}{e}\right )}\right )} \]
Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx=\frac {25 (1-\log (x))}{6 x (2-5 x+(-7+5 x) \log (x))} \]
Integrate[(125 - 125*Log[x/E] + (-175 + 250*x)*Log[x/E]^2)/(150*x^2 + (420 *x^2 - 300*x^3)*Log[x/E] + (294*x^2 - 420*x^3 + 150*x^4)*Log[x/E]^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.
\(\displaystyle \int \frac {(250 x-175) \log ^2\left (\frac {x}{e}\right )-125 \log \left (\frac {x}{e}\right )+125}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (150 x^4-420 x^3+294 x^2\right ) \log ^2\left (\frac {x}{e}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {25 \left (10 x+(10 x-7) \log ^2(x)+(9-20 x) \log (x)+3\right )}{6 x^2 (-5 x+(5 x-7) \log (x)+2)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {25}{6} \int \frac {-\left ((7-10 x) \log ^2(x)\right )+(9-20 x) \log (x)+10 x+3}{x^2 (-5 x-(7-5 x) \log (x)+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {25}{6} \int \left (\frac {10 x-7}{x^2 (5 x-7)^2}+\frac {5 (15 x-7)}{x^2 (5 x-7)^2 (5 \log (x) x-5 x-7 \log (x)+2)}+\frac {5 \left (25 x^2-45 x+49\right )}{x^2 (5 x-7)^2 (5 \log (x) x-5 x-7 \log (x)+2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {25}{6} \left (5 \int \frac {1}{x^2 (5 \log (x) x-5 x-7 \log (x)+2)^2}dx-\frac {5}{7} \int \frac {1}{x^2 (5 \log (x) x-5 x-7 \log (x)+2)}dx+\frac {125}{49} \int \frac {1}{x (5 \log (x) x-5 x-7 \log (x)+2)^2}dx+\frac {625}{7} \int \frac {1}{(5 x-7)^2 (5 \log (x) x-5 x-7 \log (x)+2)^2}dx-\frac {625}{49} \int \frac {1}{(5 x-7) (5 \log (x) x-5 x-7 \log (x)+2)^2}dx+\frac {25}{49} \int \frac {1}{x (5 \log (x) x-5 x-7 \log (x)+2)}dx+\frac {250}{7} \int \frac {1}{(5 x-7)^2 (5 \log (x) x-5 x-7 \log (x)+2)}dx-\frac {125}{49} \int \frac {1}{(5 x-7) (5 \log (x) x-5 x-7 \log (x)+2)}dx+\frac {1}{(7-5 x) x}\right )\) |
Int[(125 - 125*Log[x/E] + (-175 + 250*x)*Log[x/E]^2)/(150*x^2 + (420*x^2 - 300*x^3)*Log[x/E] + (294*x^2 - 420*x^3 + 150*x^4)*Log[x/E]^2),x]
3.11.2.3.1 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]]
Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {\frac {25}{6}-\frac {25 \ln \left (x \right )}{6}}{x \left (5 x \ln \left (x \right )-5 x -7 \ln \left (x \right )+2\right )}\) | \(28\) |
norman | \(-\frac {25 \ln \left ({\mathrm e}^{-1} x \right )}{6 \left (5 \ln \left ({\mathrm e}^{-1} x \right ) x -7 \ln \left ({\mathrm e}^{-1} x \right )-5\right ) x}\) | \(36\) |
parallelrisch | \(-\frac {25 \ln \left ({\mathrm e}^{-1} x \right )}{6 \left (5 \ln \left ({\mathrm e}^{-1} x \right ) x -7 \ln \left ({\mathrm e}^{-1} x \right )-5\right ) x}\) | \(36\) |
risch | \(-\frac {25}{6 x \left (5 x -7\right )}-\frac {125}{6 x \left (5 x -7\right ) \left (5 \ln \left ({\mathrm e}^{-1} x \right ) x -7 \ln \left ({\mathrm e}^{-1} x \right )-5\right )}\) | \(45\) |
int(((250*x-175)*ln(x/exp(1))^2-125*ln(x/exp(1))+125)/((150*x^4-420*x^3+29 4*x^2)*ln(x/exp(1))^2+(-300*x^3+420*x^2)*ln(x/exp(1))+150*x^2),x,method=_R ETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx=-\frac {25 \, \log \left (x e^{\left (-1\right )}\right )}{6 \, {\left ({\left (5 \, x^{2} - 7 \, x\right )} \log \left (x e^{\left (-1\right )}\right ) - 5 \, x\right )}} \]
integrate(((250*x-175)*log(x/exp(1))^2-125*log(x/exp(1))+125)/((150*x^4-42 0*x^3+294*x^2)*log(x/exp(1))^2+(-300*x^3+420*x^2)*log(x/exp(1))+150*x^2),x , algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (19) = 38\).
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx=- \frac {125}{- 150 x^{2} + 210 x + \left (150 x^{3} - 420 x^{2} + 294 x\right ) \log {\left (\frac {x}{e} \right )}} - \frac {25}{30 x^{2} - 42 x} \]
integrate(((250*x-175)*ln(x/exp(1))**2-125*ln(x/exp(1))+125)/((150*x**4-42 0*x**3+294*x**2)*ln(x/exp(1))**2+(-300*x**3+420*x**2)*ln(x/exp(1))+150*x** 2),x)
-125/(-150*x**2 + 210*x + (150*x**3 - 420*x**2 + 294*x)*log(x*exp(-1))) - 25/(30*x**2 - 42*x)
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx=\frac {25 \, {\left (\log \left (x\right ) - 1\right )}}{6 \, {\left (5 \, x^{2} - {\left (5 \, x^{2} - 7 \, x\right )} \log \left (x\right ) - 2 \, x\right )}} \]
integrate(((250*x-175)*log(x/exp(1))^2-125*log(x/exp(1))+125)/((150*x^4-42 0*x^3+294*x^2)*log(x/exp(1))^2+(-300*x^3+420*x^2)*log(x/exp(1))+150*x^2),x , algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (21) = 42\).
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx=-\frac {125}{6 \, {\left (25 \, x^{3} \log \left (x\right ) - 25 \, x^{3} - 70 \, x^{2} \log \left (x\right ) + 45 \, x^{2} + 49 \, x \log \left (x\right ) - 14 \, x\right )}} - \frac {125}{42 \, {\left (5 \, x - 7\right )}} + \frac {25}{42 \, x} \]
integrate(((250*x-175)*log(x/exp(1))^2-125*log(x/exp(1))+125)/((150*x^4-42 0*x^3+294*x^2)*log(x/exp(1))^2+(-300*x^3+420*x^2)*log(x/exp(1))+150*x^2),x , algorithm=\
-125/6/(25*x^3*log(x) - 25*x^3 - 70*x^2*log(x) + 45*x^2 + 49*x*log(x) - 14 *x) - 125/42/(5*x - 7) + 25/42/x
Timed out. \[ \int \frac {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx=\int \frac {\left (250\,x-175\right )\,{\ln \left (x\,{\mathrm {e}}^{-1}\right )}^2-125\,\ln \left (x\,{\mathrm {e}}^{-1}\right )+125}{{\ln \left (x\,{\mathrm {e}}^{-1}\right )}^2\,\left (150\,x^4-420\,x^3+294\,x^2\right )+\ln \left (x\,{\mathrm {e}}^{-1}\right )\,\left (420\,x^2-300\,x^3\right )+150\,x^2} \,d x \]
int((log(x*exp(-1))^2*(250*x - 175) - 125*log(x*exp(-1)) + 125)/(log(x*exp (-1))^2*(294*x^2 - 420*x^3 + 150*x^4) + log(x*exp(-1))*(420*x^2 - 300*x^3) + 150*x^2),x)