Integrand size = 57, antiderivative size = 21 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x+\frac {3+\frac {e^x}{x}+\log (x)}{8+2 x} \]
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Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(21)=42\).
Time = 0.47 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43, number of steps used = 21, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {1608, 27, 12, 6874, 46, 45, 2208, 2209, 2351, 31} \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x-\frac {e^x}{8 (x+4)}+\frac {3}{2 (x+4)}+\frac {e^x}{8 x}-\frac {x \log (x)}{8 (x+4)}+\frac {\log (x)}{8} \]
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Rule 12
Rule 27
Rule 31
Rule 45
Rule 46
Rule 1608
Rule 2208
Rule 2209
Rule 2351
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{x^2 \left (32+16 x+2 x^2\right )} \, dx \\ & = \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{2 x^2 (4+x)^2} \, dx \\ & = \frac {1}{2} \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{x^2 (4+x)^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {30}{(4+x)^2}+\frac {4}{x (4+x)^2}+\frac {16 x}{(4+x)^2}+\frac {2 x^2}{(4+x)^2}+\frac {e^x \left (-4+2 x+x^2\right )}{x^2 (4+x)^2}-\frac {\log (x)}{(4+x)^2}\right ) \, dx \\ & = -\frac {15}{4+x}+\frac {1}{2} \int \frac {e^x \left (-4+2 x+x^2\right )}{x^2 (4+x)^2} \, dx-\frac {1}{2} \int \frac {\log (x)}{(4+x)^2} \, dx+2 \int \frac {1}{x (4+x)^2} \, dx+8 \int \frac {x}{(4+x)^2} \, dx+\int \frac {x^2}{(4+x)^2} \, dx \\ & = -\frac {15}{4+x}-\frac {x \log (x)}{8 (4+x)}+\frac {1}{8} \int \frac {1}{4+x} \, dx+\frac {1}{2} \int \left (-\frac {e^x}{4 x^2}+\frac {e^x}{4 x}+\frac {e^x}{4 (4+x)^2}-\frac {e^x}{4 (4+x)}\right ) \, dx+2 \int \left (\frac {1}{16 x}-\frac {1}{4 (4+x)^2}-\frac {1}{16 (4+x)}\right ) \, dx+8 \int \left (-\frac {4}{(4+x)^2}+\frac {1}{4+x}\right ) \, dx+\int \left (1+\frac {16}{(4+x)^2}-\frac {8}{4+x}\right ) \, dx \\ & = x+\frac {3}{2 (4+x)}+\frac {\log (x)}{8}-\frac {x \log (x)}{8 (4+x)}-\frac {1}{8} \int \frac {e^x}{x^2} \, dx+\frac {1}{8} \int \frac {e^x}{x} \, dx+\frac {1}{8} \int \frac {e^x}{(4+x)^2} \, dx-\frac {1}{8} \int \frac {e^x}{4+x} \, dx \\ & = \frac {e^x}{8 x}+x+\frac {3}{2 (4+x)}-\frac {e^x}{8 (4+x)}+\frac {\operatorname {ExpIntegralEi}(x)}{8}-\frac {\operatorname {ExpIntegralEi}(4+x)}{8 e^4}+\frac {\log (x)}{8}-\frac {x \log (x)}{8 (4+x)}-\frac {1}{8} \int \frac {e^x}{x} \, dx+\frac {1}{8} \int \frac {e^x}{4+x} \, dx \\ & = \frac {e^x}{8 x}+x+\frac {3}{2 (4+x)}-\frac {e^x}{8 (4+x)}+\frac {\log (x)}{8}-\frac {x \log (x)}{8 (4+x)} \\ \end{align*}
Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=\frac {e^x+x \left (3+8 x+2 x^2\right )+x \log (x)}{2 x (4+x)} \]
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Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
norman | \(\frac {x^{3}-\frac {29 x}{2}+\frac {x \ln \left (x \right )}{2}+\frac {{\mathrm e}^{x}}{2}}{\left (4+x \right ) x}\) | \(26\) |
parallelrisch | \(-\frac {-8 x^{3}-4 x \ln \left (x \right )+116 x -4 \,{\mathrm e}^{x}}{8 x \left (4+x \right )}\) | \(29\) |
risch | \(\frac {\ln \left (x \right )}{2 x +8}+\frac {2 x^{3}+8 x^{2}+3 x +{\mathrm e}^{x}}{2 \left (4+x \right ) x}\) | \(37\) |
default | \(-\frac {{\mathrm e}^{x}}{8 \left (4+x \right )}+\frac {{\mathrm e}^{x}}{8 x}-\frac {\ln \left (x \right ) x}{8 \left (4+x \right )}+x +\frac {\ln \left (x \right )}{8}+\frac {3}{2 \left (4+x \right )}\) | \(40\) |
parts | \(-\frac {{\mathrm e}^{x}}{8 \left (4+x \right )}+\frac {{\mathrm e}^{x}}{8 x}-\frac {\ln \left (x \right ) x}{8 \left (4+x \right )}+x +\frac {\ln \left (x \right )}{8}+\frac {3}{2 \left (4+x \right )}\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=\frac {2 \, x^{3} + 8 \, x^{2} + x \log \left (x\right ) + 3 \, x + e^{x}}{2 \, {\left (x^{2} + 4 \, x\right )}} \]
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Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x + \frac {e^{x}}{2 x^{2} + 8 x} + \frac {\log {\left (x \right )}}{2 x + 8} + \frac {3}{2 x + 8} \]
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Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x + \frac {x \log \left (x\right ) + e^{x}}{2 \, {\left (x^{2} + 4 \, x\right )}} + \frac {3}{2 \, {\left (x + 4\right )}} \]
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Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=\frac {2 \, x^{3} + 8 \, x^{2} + x \log \left (x\right ) + 3 \, x + e^{x}}{2 \, {\left (x^{2} + 4 \, x\right )}} \]
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Time = 9.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {4 x+30 x^2+16 x^3+2 x^4+e^x \left (-4+2 x+x^2\right )-x^2 \log (x)}{32 x^2+16 x^3+2 x^4} \, dx=x+\frac {\frac {{\mathrm {e}}^x}{2}+x\,\left (\frac {\ln \left (x\right )}{2}+\frac {3}{2}\right )}{x\,\left (x+4\right )} \]
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