Integrand size = 50, antiderivative size = 27 \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=\frac {1}{8} \left (x-(3-e) \left (x-\left (-5+\frac {\log (x)}{x}\right )^2\right )\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(27)=54\).
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.70, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {12, 14, 37, 2372, 45, 2342, 2341} \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=\frac {(3-e) \log ^2(x)}{8 x^2}-\frac {(3-e) (5 x+1)^2 \log (x)}{8 x^2}+\frac {(3-e) \log (x)}{8 x^2}-\frac {1}{8} (2-e) x+\frac {25}{8} (3-e) \log (x) \]
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Rule 12
Rule 14
Rule 37
Rule 45
Rule 2341
Rule 2342
Rule 2372
Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{x^3} \, dx \\ & = \frac {1}{8} \int \left (\frac {-10 (3-e)-(2-e) x^2}{x^2}-\frac {2 (-3+e) (1+5 x) \log (x)}{x^3}+\frac {2 (-3+e) \log ^2(x)}{x^3}\right ) \, dx \\ & = \frac {1}{8} \int \frac {-10 (3-e)-(2-e) x^2}{x^2} \, dx+\frac {1}{4} (3-e) \int \frac {(1+5 x) \log (x)}{x^3} \, dx+\frac {1}{4} (-3+e) \int \frac {\log ^2(x)}{x^3} \, dx \\ & = -\frac {(3-e) (1+5 x)^2 \log (x)}{8 x^2}+\frac {(3-e) \log ^2(x)}{8 x^2}+\frac {1}{8} \int \left (-2 \left (1-\frac {e}{2}\right )+\frac {10 (-3+e)}{x^2}\right ) \, dx+\frac {1}{4} (-3+e) \int -\frac {(1+5 x)^2}{2 x^3} \, dx+\frac {1}{4} (-3+e) \int \frac {\log (x)}{x^3} \, dx \\ & = \frac {3-e}{16 x^2}+\frac {5 (3-e)}{4 x}-\frac {1}{8} (2-e) x+\frac {(3-e) \log (x)}{8 x^2}-\frac {(3-e) (1+5 x)^2 \log (x)}{8 x^2}+\frac {(3-e) \log ^2(x)}{8 x^2}+\frac {1}{8} (3-e) \int \frac {(1+5 x)^2}{x^3} \, dx \\ & = \frac {3-e}{16 x^2}+\frac {5 (3-e)}{4 x}-\frac {1}{8} (2-e) x+\frac {(3-e) \log (x)}{8 x^2}-\frac {(3-e) (1+5 x)^2 \log (x)}{8 x^2}+\frac {(3-e) \log ^2(x)}{8 x^2}+\frac {1}{8} (3-e) \int \left (\frac {1}{x^3}+\frac {10}{x^2}+\frac {25}{x}\right ) \, dx \\ & = -\frac {1}{8} (2-e) x+\frac {25}{8} (3-e) \log (x)+\frac {(3-e) \log (x)}{8 x^2}-\frac {(3-e) (1+5 x)^2 \log (x)}{8 x^2}+\frac {(3-e) \log ^2(x)}{8 x^2} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=\frac {(-2+e) x^3+10 (-3+e) x \log (x)-(-3+e) \log ^2(x)}{8 x^2} \]
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Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
method | result | size |
risch | \(-\frac {\left ({\mathrm e}-3\right ) \ln \left (x \right )^{2}}{8 x^{2}}+\frac {5 \left ({\mathrm e}-3\right ) \ln \left (x \right )}{4 x}-\frac {x}{4}+\frac {x \,{\mathrm e}}{8}\) | \(34\) |
norman | \(\frac {\left (-\frac {1}{4}+\frac {{\mathrm e}}{8}\right ) x^{3}+\left (-\frac {{\mathrm e}}{8}+\frac {3}{8}\right ) \ln \left (x \right )^{2}+\left (\frac {5 \,{\mathrm e}}{4}-\frac {15}{4}\right ) x \ln \left (x \right )}{x^{2}}\) | \(37\) |
parallelrisch | \(\frac {x^{3} {\mathrm e}+10 x \,{\mathrm e} \ln \left (x \right )-{\mathrm e} \ln \left (x \right )^{2}-2 x^{3}-30 x \ln \left (x \right )+3 \ln \left (x \right )^{2}}{8 x^{2}}\) | \(44\) |
default | \(\frac {x \,{\mathrm e}}{8}+\frac {{\mathrm e} \left (-\frac {\ln \left (x \right )^{2}}{2 x^{2}}-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )}{4}-\frac {5 \,{\mathrm e} \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{4}-\frac {x}{4}+\frac {3 \ln \left (x \right )^{2}}{8 x^{2}}-\frac {{\mathrm e} \left (-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )}{4}-\frac {15 \ln \left (x \right )}{4 x}-\frac {5 \,{\mathrm e}}{4 x}\) | \(93\) |
parts | \(\frac {x \,{\mathrm e}}{8}-\frac {x}{4}-\frac {10 \,{\mathrm e}-30}{8 x}+\left (\frac {{\mathrm e}}{4}-\frac {3}{4}\right ) \left (-\frac {\ln \left (x \right )^{2}}{2 x^{2}}-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )-\frac {5 \,{\mathrm e} \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{4}-\frac {{\mathrm e} \left (-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )}{4}-\frac {15 \ln \left (x \right )}{4 x}-\frac {15}{4 x}-\frac {3 \ln \left (x \right )}{8 x^{2}}-\frac {3}{16 x^{2}}\) | \(108\) |
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Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=\frac {x^{3} e - 2 \, x^{3} - {\left (e - 3\right )} \log \left (x\right )^{2} + 10 \, {\left (x e - 3 \, x\right )} \log \left (x\right )}{8 \, x^{2}} \]
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Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=x \left (- \frac {1}{4} + \frac {e}{8}\right ) + \frac {\left (-15 + 5 e\right ) \log {\left (x \right )}}{4 x} + \frac {\left (3 - e\right ) \log {\left (x \right )}^{2}}{8 x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (24) = 48\).
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.70 \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=\frac {1}{8} \, x e + \frac {5}{4} \, {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} e + \frac {1}{16} \, {\left (\frac {2 \, \log \left (x\right )}{x^{2}} + \frac {1}{x^{2}}\right )} e - \frac {1}{4} \, x - \frac {{\left (2 \, \log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 1\right )} e}{16 \, x^{2}} - \frac {5 \, e}{4 \, x} - \frac {15 \, \log \left (x\right )}{4 \, x} + \frac {3 \, {\left (2 \, \log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 1\right )}}{16 \, x^{2}} - \frac {3 \, \log \left (x\right )}{8 \, x^{2}} - \frac {3}{16 \, x^{2}} \]
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Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=\frac {x^{3} e - 2 \, x^{3} + 10 \, x e \log \left (x\right ) - e \log \left (x\right )^{2} - 30 \, x \log \left (x\right ) + 3 \, \log \left (x\right )^{2}}{8 \, x^{2}} \]
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Time = 8.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {-30 x-2 x^3+e \left (10 x+x^3\right )+(6+e (-2-10 x)+30 x) \log (x)+(-6+2 e) \log ^2(x)}{8 x^3} \, dx=\frac {\frac {x^2\,\ln \left (x\right )\,\left (10\,\mathrm {e}-30\right )}{8}-\frac {x\,{\ln \left (x\right )}^2\,\left (\mathrm {e}-3\right )}{8}}{x^3}+x\,\left (\frac {\mathrm {e}}{8}-\frac {1}{4}\right ) \]
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