Integrand size = 43, antiderivative size = 21 \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=x-\left (\frac {11}{6}-\frac {5 x}{4}\right ) x (5+x) \log (4+x) \]
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Leaf count is larger than twice the leaf count of optimal. \(43\) vs. \(2(21)=42\).
Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.05, number of steps used = 14, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6874, 1864, 2464, 2436, 2332, 2442, 45} \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=\frac {5}{4} x^3 \log (x+4)+\frac {53}{12} x^2 \log (x+4)+x-\frac {55}{6} (x+4) \log (x+4)+\frac {110}{3} \log (x+4) \]
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Rule 45
Rule 1864
Rule 2332
Rule 2436
Rule 2442
Rule 2464
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {48-98 x+53 x^2+15 x^3}{12 (4+x)}+\frac {1}{12} \left (-110+106 x+45 x^2\right ) \log (4+x)\right ) \, dx \\ & = \frac {1}{12} \int \frac {48-98 x+53 x^2+15 x^3}{4+x} \, dx+\frac {1}{12} \int \left (-110+106 x+45 x^2\right ) \log (4+x) \, dx \\ & = \frac {1}{12} \int \left (-70-7 x+15 x^2+\frac {328}{4+x}\right ) \, dx+\frac {1}{12} \int \left (-110 \log (4+x)+106 x \log (4+x)+45 x^2 \log (4+x)\right ) \, dx \\ & = -\frac {35 x}{6}-\frac {7 x^2}{24}+\frac {5 x^3}{12}+\frac {82}{3} \log (4+x)+\frac {15}{4} \int x^2 \log (4+x) \, dx+\frac {53}{6} \int x \log (4+x) \, dx-\frac {55}{6} \int \log (4+x) \, dx \\ & = -\frac {35 x}{6}-\frac {7 x^2}{24}+\frac {5 x^3}{12}+\frac {82}{3} \log (4+x)+\frac {53}{12} x^2 \log (4+x)+\frac {5}{4} x^3 \log (4+x)-\frac {5}{4} \int \frac {x^3}{4+x} \, dx-\frac {53}{12} \int \frac {x^2}{4+x} \, dx-\frac {55}{6} \text {Subst}(\int \log (x) \, dx,x,4+x) \\ & = \frac {10 x}{3}-\frac {7 x^2}{24}+\frac {5 x^3}{12}+\frac {82}{3} \log (4+x)+\frac {53}{12} x^2 \log (4+x)+\frac {5}{4} x^3 \log (4+x)-\frac {55}{6} (4+x) \log (4+x)-\frac {5}{4} \int \left (16-4 x+x^2-\frac {64}{4+x}\right ) \, dx-\frac {53}{12} \int \left (-4+x+\frac {16}{4+x}\right ) \, dx \\ & = x+\frac {110}{3} \log (4+x)+\frac {53}{12} x^2 \log (4+x)+\frac {5}{4} x^3 \log (4+x)-\frac {55}{6} (4+x) \log (4+x) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=\frac {1}{12} \left (12 x-110 x \log (4+x)+53 x^2 \log (4+x)+15 x^3 \log (4+x)\right ) \]
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Time = 0.86 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05
method | result | size |
risch | \(\left (\frac {5}{4} x^{3}+\frac {53}{12} x^{2}-\frac {55}{6} x \right ) \ln \left (4+x \right )+x\) | \(22\) |
norman | \(x +\frac {53 x^{2} \ln \left (4+x \right )}{12}+\frac {5 x^{3} \ln \left (4+x \right )}{4}-\frac {55 \ln \left (4+x \right ) x}{6}\) | \(28\) |
parallelrisch | \(\frac {5 x^{3} \ln \left (4+x \right )}{4}+\frac {53 x^{2} \ln \left (4+x \right )}{12}-\frac {55 \ln \left (4+x \right ) x}{6}+x -8\) | \(29\) |
derivativedivides | \(\frac {5 \ln \left (4+x \right ) \left (4+x \right )^{3}}{4}-\frac {127 \ln \left (4+x \right ) \left (4+x \right )^{2}}{12}+\frac {31 \left (4+x \right ) \ln \left (4+x \right )}{2}+4+x +\frac {82 \ln \left (4+x \right )}{3}\) | \(41\) |
default | \(\frac {5 \ln \left (4+x \right ) \left (4+x \right )^{3}}{4}-\frac {127 \ln \left (4+x \right ) \left (4+x \right )^{2}}{12}+\frac {31 \left (4+x \right ) \ln \left (4+x \right )}{2}+4+x +\frac {82 \ln \left (4+x \right )}{3}\) | \(41\) |
parts | \(\frac {5 \ln \left (4+x \right ) \left (4+x \right )^{3}}{4}+x -4-\frac {127 \ln \left (4+x \right ) \left (4+x \right )^{2}}{12}+\frac {31 \left (4+x \right ) \ln \left (4+x \right )}{2}+\frac {82 \ln \left (4+x \right )}{3}\) | \(41\) |
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=\frac {1}{12} \, {\left (15 \, x^{3} + 53 \, x^{2} - 110 \, x\right )} \log \left (x + 4\right ) + x \]
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Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=x + \left (\frac {5 x^{3}}{4} + \frac {53 x^{2}}{12} - \frac {55 x}{6}\right ) \log {\left (x + 4 \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (17) = 34\).
Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 3.19 \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=\frac {5}{4} \, {\left (x^{3} - 6 \, x^{2} + 48 \, x - 192 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) + \frac {143}{12} \, {\left (x^{2} - 8 \, x + 32 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) + \frac {157}{6} \, {\left (x - 4 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - \frac {110}{3} \, \log \left (x + 4\right )^{2} + x \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=\frac {1}{12} \, {\left (15 \, x^{3} + 53 \, x^{2} - 110 \, x\right )} \log \left (x + 4\right ) + x \]
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Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {48-98 x+53 x^2+15 x^3+\left (-440+314 x+286 x^2+45 x^3\right ) \log (4+x)}{48+12 x} \, dx=\frac {53\,x^2\,\ln \left (x+4\right )}{12}+\frac {5\,x^3\,\ln \left (x+4\right )}{4}-x\,\left (\frac {55\,\ln \left (x+4\right )}{6}-1\right ) \]
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