Integrand size = 127, antiderivative size = 37 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {-x+\frac {x^2}{4}-\log \left (\left (e-\frac {x}{4 (5-x)}\right )^2\right )}{3 x} \]
[Out]
Leaf count is larger than twice the leaf count of optimal. \(213\) vs. \(2(37)=74\).
Time = 0.68 (sec) , antiderivative size = 213, normalized size of antiderivative = 5.76, number of steps used = 15, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6820, 12, 6874, 45, 84, 78, 2554, 2351, 31} \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {e x}{3 (1+4 e)}+\frac {x}{12 (1+4 e)}+\frac {2}{15} \log (5-x)+\frac {\log (x)}{30 e}-\frac {\log \left (-\frac {x}{5-x}\right )}{30 e}-\frac {(1+4 e) \log (20 e-(1+4 e) x)}{30 e}+\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)}-\frac {20 e^2 \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {(20 e-(1+4 e) x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{60 e x} \]
[In]
[Out]
Rule 12
Rule 31
Rule 45
Rule 78
Rule 84
Rule 2351
Rule 2554
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {40 x+4 e (-5+x)^2 x^2-5 x^3+x^4+4 (-5+x) (4 e (-5+x)+x) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )}{12 (5-x) x^2 (20 e-(1+4 e) x)} \, dx \\ & = \frac {1}{12} \int \frac {40 x+4 e (-5+x)^2 x^2-5 x^3+x^4+4 (-5+x) (4 e (-5+x)+x) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )}{(5-x) x^2 (20 e-(1+4 e) x)} \, dx \\ & = \frac {1}{12} \int \left (\frac {4 e (5-x)}{20 e-(1+4 e) x}+\frac {40}{(5-x) x (20 e-(1+4 e) x)}+\frac {5 x}{(-5+x) (20 e-(1+4 e) x)}+\frac {x^2}{(5-x) (20 e-(1+4 e) x)}+\frac {4 \log \left (\frac {(-20 e+(1+4 e) x)^2}{16 (-5+x)^2}\right )}{x^2}\right ) \, dx \\ & = \frac {1}{12} \int \frac {x^2}{(5-x) (20 e-(1+4 e) x)} \, dx+\frac {1}{3} \int \frac {\log \left (\frac {(-20 e+(1+4 e) x)^2}{16 (-5+x)^2}\right )}{x^2} \, dx+\frac {5}{12} \int \frac {x}{(-5+x) (20 e-(1+4 e) x)} \, dx+\frac {10}{3} \int \frac {1}{(5-x) x (20 e-(1+4 e) x)} \, dx+\frac {1}{3} e \int \frac {5-x}{20 e-(1+4 e) x} \, dx \\ & = \frac {1}{12} \int \left (\frac {1}{1+4 e}+\frac {5}{-5+x}+\frac {80 e^2}{(1+4 e) (20 e-(1+4 e) x)}\right ) \, dx+\frac {5}{12} \int \left (\frac {1}{5-x}+\frac {4 e}{-20 e+(1+4 e) x}\right ) \, dx-\frac {5}{3} \text {Subst}\left (\int \frac {\log \left (\frac {x^2}{16}\right )}{(20 e-5 x)^2} \, dx,x,\frac {-20 e+(1+4 e) x}{-5+x}\right )+\frac {10}{3} \int \left (\frac {1}{25 (-5+x)}+\frac {1}{100 e x}+\frac {(1+4 e)^2}{100 e (20 e-(1+4 e) x)}\right ) \, dx+\frac {1}{3} e \int \left (\frac {1}{1+4 e}+\frac {5}{(1+4 e) (20 e-(1+4 e) x)}\right ) \, dx \\ & = \frac {x}{12 (1+4 e)}+\frac {e x}{3 (1+4 e)}+\frac {2}{15} \log (5-x)+\frac {\log (x)}{30 e}-\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {20 e^2 \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}+\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)}-\frac {(1+4 e) \log (20 e-(1+4 e) x)}{30 e}-\frac {(20 e-(1+4 e) x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{60 e x}+\frac {\text {Subst}\left (\int \frac {1}{20 e-5 x} \, dx,x,\frac {-20 e+(1+4 e) x}{-5+x}\right )}{6 e} \\ & = \frac {x}{12 (1+4 e)}+\frac {e x}{3 (1+4 e)}+\frac {2}{15} \log (5-x)+\frac {\log (x)}{30 e}-\frac {\log \left (-\frac {x}{5-x}\right )}{30 e}-\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {20 e^2 \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}+\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)}-\frac {(1+4 e) \log (20 e-(1+4 e) x)}{30 e}-\frac {(20 e-(1+4 e) x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{60 e x} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(323\) vs. \(2(37)=74\).
Time = 0.35 (sec) , antiderivative size = 323, normalized size of antiderivative = 8.73 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {5 e x^2-(1+4 e)^2 x \log ^2(-4 e (-5+x)-x)+2 (1+4 e)^2 x \log \left (-\frac {1}{5} (1+4 e) (-5+x)\right ) \log \left (\frac {5}{4 e (-5+x)+x}\right )+x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )+8 e x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )+16 e^2 x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )-20 e \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+8 e x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+16 e^2 x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+(1+4 e)^2 x \log (-4 e (-5+x)-x) \left (2 \log \left (-\frac {1}{5} (1+4 e) (-5+x)\right )+\log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )\right )}{60 e x} \]
[In]
[Out]
Time = 0.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43
method | result | size |
risch | \(-\frac {\ln \left (\frac {\left (16 x^{2}-160 x +400\right ) {\mathrm e}^{2}+\left (8 x^{2}-40 x \right ) {\mathrm e}+x^{2}}{16 x^{2}-160 x +400}\right )}{3 x}+\frac {x}{12}\) | \(53\) |
norman | \(\frac {\frac {x^{2}}{12}-\frac {\ln \left (\frac {\left (16 x^{2}-160 x +400\right ) {\mathrm e}^{2}+\left (8 x^{2}-40 x \right ) {\mathrm e}+x^{2}}{16 x^{2}-160 x +400}\right )}{3}}{x}\) | \(58\) |
parallelrisch | \(-\frac {\left (-10000 x^{2} {\mathrm e}^{2}+40000 \,{\mathrm e}^{2} \ln \left (\frac {\left (16 x^{2}-160 x +400\right ) {\mathrm e}^{2}+\left (8 x^{2}-40 x \right ) {\mathrm e}+x^{2}}{16 x^{2}-160 x +400}\right )\right ) {\mathrm e}^{-2}}{120000 x}\) | \(70\) |
derivativedivides | \(-\frac {4 \ln \left (2\right )}{15 \left (1+\frac {5}{-5+x}\right )}-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{60}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{12 \left (-5+x \right )}}{1+\frac {5}{-5+x}}-\frac {5}{12}+\frac {x}{12}-\frac {\left (8 \,{\mathrm e}+2\right ) {\mathrm e}^{-1} \ln \left (4 \,{\mathrm e}+\frac {5}{-5+x}+1\right )}{60}\) | \(156\) |
default | \(-\frac {4 \ln \left (2\right )}{15 \left (1+\frac {5}{-5+x}\right )}-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{60}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{12 \left (-5+x \right )}}{1+\frac {5}{-5+x}}-\frac {5}{12}+\frac {x}{12}-\frac {\left (8 \,{\mathrm e}+2\right ) {\mathrm e}^{-1} \ln \left (4 \,{\mathrm e}+\frac {5}{-5+x}+1\right )}{60}\) | \(156\) |
parts | \(-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{20}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{4 \left (-5+x \right )}}{3 \left (1+\frac {5}{-5+x}\right )}-\frac {{\mathrm e}^{-1} \ln \left (1+\frac {5}{-5+x}\right )}{30}-\frac {4 \ln \left (2\right )}{15 \left (1+\frac {5}{-5+x}\right )}+\frac {x}{12}+\frac {\ln \left (x \right ) {\mathrm e}^{-1}}{30}+\frac {2 \ln \left (-5+x \right )}{15}-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (4 x \,{\mathrm e}-20 \,{\mathrm e}+x \right )}{30}\) | \(183\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.38 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x^{2} - 4 \, \log \left (\frac {x^{2} + 16 \, {\left (x^{2} - 10 \, x + 25\right )} e^{2} + 8 \, {\left (x^{2} - 5 \, x\right )} e}{16 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )}{12 \, x} \]
[In]
[Out]
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.24 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x}{12} - \frac {\log {\left (\frac {x^{2} + e \left (8 x^{2} - 40 x\right ) + \left (16 x^{2} - 160 x + 400\right ) e^{2}}{16 x^{2} - 160 x + 400} \right )}}{3 x} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (31) = 62\).
Time = 0.37 (sec) , antiderivative size = 269, normalized size of antiderivative = 7.27 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=-\frac {1}{30} \, {\left (4 \, e + 1\right )} e^{\left (-1\right )} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - \frac {1}{3} \, {\left (\frac {80 \, e^{2} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{16 \, e^{2} + 8 \, e + 1} - \frac {x}{4 \, e + 1} - 5 \, \log \left (x - 5\right )\right )} e + \frac {10}{3} \, {\left (\frac {4 \, e \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{4 \, e + 1} - \log \left (x - 5\right )\right )} e - \frac {5}{3} \, {\left (\log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - \log \left (x - 5\right )\right )} e + \frac {{\left (40 \, e \log \left (2\right ) + {\left (x {\left (4 \, e + 1\right )} - 20 \, e\right )} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - 4 \, {\left (x e - 5 \, e\right )} \log \left (x - 5\right )\right )} e^{\left (-1\right )}}{30 \, x} - \frac {20 \, e^{2} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{3 \, {\left (16 \, e^{2} + 8 \, e + 1\right )}} + \frac {5 \, e \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{3 \, {\left (4 \, e + 1\right )}} + \frac {x}{12 \, {\left (4 \, e + 1\right )}} + \frac {2}{15} \, \log \left (x - 5\right ) \]
[In]
[Out]
none
Time = 0.48 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x^{2} - 4 \, \log \left (\frac {16 \, x^{2} e^{2} + 8 \, x^{2} e + x^{2} - 160 \, x e^{2} - 40 \, x e + 400 \, e^{2}}{16 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )}{12 \, x} \]
[In]
[Out]
Time = 3.78 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x}{12}-\frac {\ln \left (\frac {{\mathrm {e}}^2\,\left (16\,x^2-160\,x+400\right )-\mathrm {e}\,\left (40\,x-8\,x^2\right )+x^2}{16\,x^2-160\,x+400}\right )}{3\,x} \]
[In]
[Out]