Integrand size = 64, antiderivative size = 22 \[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=x \left (-1-e^{\frac {1}{-4+\frac {5}{x}}}+2 x+\log (5)\right ) \]
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\[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=\int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{(-5+4 x)^2} \, dx \\ & = \int \left (-1+4 x+\frac {e^{\frac {x}{5-4 x}} \left (-25+35 x-16 x^2\right )}{(5-4 x)^2}+\log (5)\right ) \, dx \\ & = 2 x^2-x (1-\log (5))+\int \frac {e^{\frac {x}{5-4 x}} \left (-25+35 x-16 x^2\right )}{(5-4 x)^2} \, dx \\ & = 2 x^2-x (1-\log (5))+\int \left (-e^{\frac {x}{5-4 x}}-\frac {25 e^{\frac {x}{5-4 x}}}{4 (-5+4 x)^2}-\frac {5 e^{\frac {x}{5-4 x}}}{4 (-5+4 x)}\right ) \, dx \\ & = 2 x^2-x (1-\log (5))-\frac {5}{4} \int \frac {e^{\frac {x}{5-4 x}}}{-5+4 x} \, dx-\frac {25}{4} \int \frac {e^{\frac {x}{5-4 x}}}{(-5+4 x)^2} \, dx-\int e^{\frac {x}{5-4 x}} \, dx \\ & = 2 x^2-x (1-\log (5))-\frac {5}{4} \int \frac {e^{-\frac {1}{4}+\frac {5}{4 (5-4 x)}}}{-5+4 x} \, dx-\frac {25}{4} \int \frac {e^{-\frac {1}{4}+\frac {5}{4 (5-4 x)}}}{(-5+4 x)^2} \, dx-\int e^{\frac {x}{5-4 x}} \, dx \\ & = -\frac {5}{4} e^{-\frac {1}{4}+\frac {5}{4 (5-4 x)}}+2 x^2+\frac {5 \operatorname {ExpIntegralEi}\left (\frac {5}{4 (5-4 x)}\right )}{16 \sqrt [4]{e}}-x (1-\log (5))-\int e^{\frac {x}{5-4 x}} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=x \left (-1-e^{\frac {x}{5-4 x}}+2 x+\log (5)\right ) \]
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Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27
method | result | size |
risch | \(x \ln \left (5\right )+2 x^{2}-x \,{\mathrm e}^{-\frac {x}{-5+4 x}}-x\) | \(28\) |
parallelrisch | \(x \ln \left (5\right )+2 x^{2}-x \,{\mathrm e}^{-\frac {x}{-5+4 x}}+5 \ln \left (5\right )-x -\frac {95}{4}\) | \(33\) |
parts | \(2 x^{2}-x -\frac {{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (-5+4 x \right )}} \left (-5+4 x \right )}{4}-\frac {5 \,{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (-5+4 x \right )}}}{4}+x \ln \left (5\right )\) | \(47\) |
derivativedivides | \(\frac {\ln \left (5\right ) \left (-5+4 x \right )}{4}+\frac {\left (-5+4 x \right )^{2}}{8}-5+4 x -\frac {{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (-5+4 x \right )}} \left (-5+4 x \right )}{4}-\frac {5 \,{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (-5+4 x \right )}}}{4}\) | \(57\) |
default | \(\frac {\ln \left (5\right ) \left (-5+4 x \right )}{4}+\frac {\left (-5+4 x \right )^{2}}{8}-5+4 x -\frac {{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (-5+4 x \right )}} \left (-5+4 x \right )}{4}-\frac {5 \,{\mathrm e}^{-\frac {1}{4}-\frac {5}{4 \left (-5+4 x \right )}}}{4}\) | \(57\) |
norman | \(\frac {\left (-14+4 \ln \left (5\right )\right ) x^{2}+8 x^{3}+5 x \,{\mathrm e}^{-\frac {x}{-5+4 x}}-4 x^{2} {\mathrm e}^{-\frac {x}{-5+4 x}}+\frac {25}{4}-\frac {25 \ln \left (5\right )}{4}}{-5+4 x}\) | \(60\) |
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Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=2 \, x^{2} - x e^{\left (-\frac {x}{4 \, x - 5}\right )} + x \log \left (5\right ) - x \]
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Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=2 x^{2} + x \left (-1 + \log {\left (5 \right )}\right ) - x e^{- \frac {x}{4 x - 5}} \]
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\[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=\int { \frac {64 \, x^{3} - 176 \, x^{2} - {\left (16 \, x^{2} - 35 \, x + 25\right )} e^{\left (-\frac {x}{4 \, x - 5}\right )} + {\left (16 \, x^{2} - 40 \, x + 25\right )} \log \left (5\right ) + 140 \, x - 25}{16 \, x^{2} - 40 \, x + 25} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (21) = 42\).
Time = 0.33 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.55 \[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=-\frac {5 \, {\left (\frac {8 \, x e^{\left (-\frac {x}{4 \, x - 5}\right )}}{4 \, x - 5} - \frac {32 \, x^{2} e^{\left (-\frac {x}{4 \, x - 5}\right )}}{{\left (4 \, x - 5\right )}^{2}} + \frac {8 \, x \log \left (5\right )}{4 \, x - 5} + \frac {32 \, x}{4 \, x - 5} - 2 \, \log \left (5\right ) - 3\right )}}{8 \, {\left (\frac {8 \, x}{4 \, x - 5} - \frac {16 \, x^{2}}{{\left (4 \, x - 5\right )}^{2}} - 1\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-25+140 x-176 x^2+64 x^3+e^{-\frac {x}{-5+4 x}} \left (-25+35 x-16 x^2\right )+\left (25-40 x+16 x^2\right ) \log (5)}{25-40 x+16 x^2} \, dx=2\,x^2-x\,\left ({\mathrm {e}}^{-\frac {x}{4\,x-5}}-\ln \left (5\right )+1\right ) \]
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