Integrand size = 51, antiderivative size = 26 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=4+\frac {10 x^2}{-x+4 x \left (-e^{2 x}+\log (3)\right )} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.40 (sec) , antiderivative size = 250, normalized size of antiderivative = 9.62, number of steps used = 18, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.255, Rules used = {6820, 12, 6874, 2215, 2221, 2317, 2438, 2216, 2222, 2320, 36, 29, 31} \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=\frac {5 (2-\log (6561)) \operatorname {PolyLog}\left (2,-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{2 (1-\log (81))^2}-\frac {5 \operatorname {PolyLog}\left (2,-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}+\frac {5 (1-2 x)^2}{2 (1-4 \log (3))}+\frac {5 (1-2 x) \log \left (\frac {4 e^{2 x}}{1-4 \log (3)}+1\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (\frac {4 e^{2 x}}{1-4 \log (3)}+1\right )}{(1-\log (81))^2}-\frac {5 (2-\log (6561)) \log \left (4 e^{2 x}+1-4 \log (3)\right )}{2 (1-\log (81))^2}-\frac {5 x (2-\log (6561))}{(1-\log (81)) \left (4 e^{2 x}+1-4 \log (3)\right )}+\frac {5 x (2-\log (6561))}{(1-\log (81))^2} \]
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Rule 12
Rule 29
Rule 31
Rule 36
Rule 2215
Rule 2216
Rule 2221
Rule 2222
Rule 2317
Rule 2320
Rule 2438
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 \left (-1+e^{2 x} (-4+8 x)+\log (81)\right )}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx \\ & = 10 \int \frac {-1+e^{2 x} (-4+8 x)+\log (81)}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx \\ & = 10 \int \left (\frac {-1+2 x}{1+4 e^{2 x}-4 \log (3)}+\frac {x (-2+\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right )^2}\right ) \, dx \\ & = 10 \int \frac {-1+2 x}{1+4 e^{2 x}-4 \log (3)} \, dx-(10 (2-\log (6561))) \int \frac {x}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx \\ & = \frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {40 \int \frac {e^{2 x} (-1+2 x)}{1+4 e^{2 x}-4 \log (3)} \, dx}{1-4 \log (3)}-\frac {(10 (2-\log (6561))) \int \frac {x}{1+4 e^{2 x}-4 \log (3)} \, dx}{1-\log (81)}+\frac {(40 (2-\log (6561))) \int \frac {e^{2 x} x}{\left (1+4 e^{2 x}-4 \log (3)\right )^2} \, dx}{1-\log (81)} \\ & = \frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {10 \int \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right ) \, dx}{1-4 \log (3)}+\frac {(40 (2-\log (6561))) \int \frac {e^{2 x} x}{1+4 e^{2 x}-4 \log (3)} \, dx}{(1-\log (81))^2}+\frac {(5 (2-\log (6561))) \int \frac {1}{1+4 e^{2 x}-4 \log (3)} \, dx}{1-\log (81)} \\ & = \frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{(1-\log (81))^2}+\frac {5 \text {Subst}\left (\int \frac {\log \left (1+\frac {4 x}{1-4 \log (3)}\right )}{x} \, dx,x,e^{2 x}\right )}{1-4 \log (3)}-\frac {(5 (2-\log (6561))) \int \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right ) \, dx}{(1-\log (81))^2}+\frac {(5 (2-\log (6561))) \text {Subst}\left (\int \frac {1}{x (1+4 x-4 \log (3))} \, dx,x,e^{2 x}\right )}{2 (1-\log (81))} \\ & = \frac {5 (1-2 x)^2}{2 (1-4 \log (3))}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{(1-\log (81))^2}-\frac {5 \operatorname {PolyLog}\left (2,-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {(5 (2-\log (6561))) \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )}{2 (1-\log (81))^2}-\frac {(5 (2-\log (6561))) \text {Subst}\left (\int \frac {\log \left (1+\frac {4 x}{1-4 \log (3)}\right )}{x} \, dx,x,e^{2 x}\right )}{2 (1-\log (81))^2}-\frac {(10 (2-\log (6561))) \text {Subst}\left (\int \frac {1}{1+4 x-4 \log (3)} \, dx,x,e^{2 x}\right )}{(1-\log (81))^2} \\ & = \frac {5 (1-2 x)^2}{2 (1-4 \log (3))}+\frac {5 x (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x^2 (2-\log (6561))}{(1-\log (81))^2}-\frac {5 x (2-\log (6561))}{\left (1+4 e^{2 x}-4 \log (3)\right ) (1-\log (81))}+\frac {5 (1-2 x) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 x (2-\log (6561)) \log \left (1+\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{(1-\log (81))^2}-\frac {5 (2-\log (6561)) \log \left (1+4 e^{2 x}-4 \log (3)\right )}{2 (1-\log (81))^2}-\frac {5 \operatorname {PolyLog}\left (2,-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{1-4 \log (3)}+\frac {5 (2-\log (6561)) \operatorname {PolyLog}\left (2,-\frac {4 e^{2 x}}{1-4 \log (3)}\right )}{2 (1-\log (81))^2} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {10 x}{1+4 e^{2 x}-\log (81)} \]
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Time = 1.38 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69
method | result | size |
norman | \(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1}\) | \(18\) |
risch | \(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1}\) | \(18\) |
parallelrisch | \(\frac {10 x}{-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1}\) | \(18\) |
default | \(-\frac {10 \ln \left ({\mathrm e}^{x}\right )}{\left (-1+4 \ln \left (3\right )\right )^{2}}-\frac {40 \left (-\frac {\ln \left (4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1\right )}{8}-\frac {\frac {\ln \left (3\right )}{2}-\frac {1}{8}}{4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1}\right )}{\left (-1+4 \ln \left (3\right )\right )^{2}}+\frac {5}{4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1}+40 \ln \left (3\right ) \left (\frac {\ln \left ({\mathrm e}^{x}\right )}{\left (-1+4 \ln \left (3\right )\right )^{2}}+\frac {-\frac {\ln \left (4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1\right )}{2}-\frac {4 \left (\frac {\ln \left (3\right )}{2}-\frac {1}{8}\right )}{4 \,{\mathrm e}^{2 x}-4 \ln \left (3\right )+1}}{\left (-1+4 \ln \left (3\right )\right )^{2}}\right )+\frac {5 \ln \left (-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1\right )}{-1+4 \ln \left (3\right )}+\frac {40 x \,{\mathrm e}^{2 x}}{\left (-1+4 \ln \left (3\right )\right ) \left (-4 \,{\mathrm e}^{2 x}+4 \ln \left (3\right )-1\right )}\) | \(196\) |
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {10 \, x}{4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1} \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=- \frac {10 x}{4 e^{2 x} - 4 \log {\left (3 \right )} + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (23) = 46\).
Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 8.62 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=20 \, {\left (\frac {2 \, x}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} - \frac {\log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right )}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} - \frac {1}{4 \, {\left (4 \, \log \left (3\right ) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \left (3\right )^{2} + 8 \, \log \left (3\right ) - 1}\right )} \log \left (3\right ) - \frac {40 \, x e^{\left (2 \, x\right )}}{4 \, {\left (4 \, \log \left (3\right ) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \left (3\right )^{2} + 8 \, \log \left (3\right ) - 1} - \frac {10 \, x}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} + \frac {5 \, \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right )}{16 \, \log \left (3\right )^{2} - 8 \, \log \left (3\right ) + 1} + \frac {5 \, \log \left (e^{\left (2 \, x\right )} - \log \left (3\right ) + \frac {1}{4}\right )}{4 \, \log \left (3\right ) - 1} + \frac {5}{4 \, {\left (4 \, \log \left (3\right ) - 1\right )} e^{\left (2 \, x\right )} - 16 \, \log \left (3\right )^{2} + 8 \, \log \left (3\right ) - 1} + \frac {5}{4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (23) = 46\).
Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 5.35 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {5 \, {\left (8 \, x \log \left (3\right ) + 4 \, e^{\left (2 \, x\right )} \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right ) - 4 \, \log \left (3\right ) \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right ) - 4 \, e^{\left (2 \, x\right )} \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \left (3\right ) - 1\right ) + 4 \, \log \left (3\right ) \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \left (3\right ) - 1\right ) - 2 \, x + \log \left (4 \, e^{\left (2 \, x\right )} - 4 \, \log \left (3\right ) + 1\right ) - \log \left (-4 \, e^{\left (2 \, x\right )} + 4 \, \log \left (3\right ) - 1\right )\right )}}{16 \, e^{\left (2 \, x\right )} \log \left (3\right ) - 16 \, \log \left (3\right )^{2} - 4 \, e^{\left (2 \, x\right )} + 8 \, \log \left (3\right ) - 1} \]
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Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-10+e^{2 x} (-40+80 x)+40 \log (3)}{1+16 e^{4 x}+e^{2 x} (8-32 \log (3))-8 \log (3)+16 \log ^2(3)} \, dx=-\frac {10\,x}{4\,{\mathrm {e}}^{2\,x}-\ln \left (81\right )+1} \]
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