Integrand size = 12, antiderivative size = 82 \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {5 x^2}{18}+\frac {2}{3} x \arctan (x)-\frac {2}{9} x^3 \arctan (x)-\frac {\arctan (x)^2}{3}-\frac {11}{18} \log \left (1+x^2\right )-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right ) \]
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Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4946, 272, 45, 5141, 6857, 5036, 4930, 266, 5004, 2525, 2437, 2338} \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=-\frac {2}{9} x^3 \arctan (x)+\frac {1}{3} x^3 \arctan (x) \log \left (x^2+1\right )+\frac {2}{3} x \arctan (x)-\frac {\arctan (x)^2}{3}+\frac {5 x^2}{18}+\frac {1}{12} \log ^2\left (x^2+1\right )-\frac {1}{6} x^2 \log \left (x^2+1\right )-\frac {11}{18} \log \left (x^2+1\right ) \]
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Rule 45
Rule 266
Rule 272
Rule 2338
Rule 2437
Rule 2525
Rule 4930
Rule 4946
Rule 5004
Rule 5036
Rule 5141
Rule 6857
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-2 \int \left (\frac {x^3 (-1+2 x \arctan (x))}{6 \left (1+x^2\right )}+\frac {x \log \left (1+x^2\right )}{6 \left (1+x^2\right )}\right ) \, dx \\ & = -\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-\frac {1}{3} \int \frac {x^3 (-1+2 x \arctan (x))}{1+x^2} \, dx-\frac {1}{3} \int \frac {x \log \left (1+x^2\right )}{1+x^2} \, dx \\ & = -\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-\frac {1}{6} \text {Subst}\left (\int \frac {\log (1+x)}{1+x} \, dx,x,x^2\right )-\frac {1}{3} \int \left (-\frac {x^3}{1+x^2}+\frac {2 x^4 \arctan (x)}{1+x^2}\right ) \, dx \\ & = -\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-\frac {1}{6} \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )+\frac {1}{3} \int \frac {x^3}{1+x^2} \, dx-\frac {2}{3} \int \frac {x^4 \arctan (x)}{1+x^2} \, dx \\ & = -\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{6} \text {Subst}\left (\int \frac {x}{1+x} \, dx,x,x^2\right )-\frac {2}{3} \int x^2 \arctan (x) \, dx+\frac {2}{3} \int \frac {x^2 \arctan (x)}{1+x^2} \, dx \\ & = -\frac {2}{9} x^3 \arctan (x)-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{6} \text {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,x^2\right )+\frac {2}{9} \int \frac {x^3}{1+x^2} \, dx+\frac {2}{3} \int \arctan (x) \, dx-\frac {2}{3} \int \frac {\arctan (x)}{1+x^2} \, dx \\ & = \frac {x^2}{6}+\frac {2}{3} x \arctan (x)-\frac {2}{9} x^3 \arctan (x)-\frac {\arctan (x)^2}{3}-\frac {1}{6} \log \left (1+x^2\right )-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{9} \text {Subst}\left (\int \frac {x}{1+x} \, dx,x,x^2\right )-\frac {2}{3} \int \frac {x}{1+x^2} \, dx \\ & = \frac {x^2}{6}+\frac {2}{3} x \arctan (x)-\frac {2}{9} x^3 \arctan (x)-\frac {\arctan (x)^2}{3}-\frac {1}{2} \log \left (1+x^2\right )-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{9} \text {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,x^2\right ) \\ & = \frac {5 x^2}{18}+\frac {2}{3} x \arctan (x)-\frac {2}{9} x^3 \arctan (x)-\frac {\arctan (x)^2}{3}-\frac {11}{18} \log \left (1+x^2\right )-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \arctan (x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{36} \left (10 x^2-12 \arctan (x)^2-2 \left (11+3 x^2\right ) \log \left (1+x^2\right )+3 \log ^2\left (1+x^2\right )+4 x \arctan (x) \left (6-2 x^2+3 x^2 \log \left (1+x^2\right )\right )\right ) \]
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Time = 1.66 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83
method | result | size |
parallelrisch | \(\frac {x^{3} \arctan \left (x \right ) \ln \left (x^{2}+1\right )}{3}-\frac {2 x^{3} \arctan \left (x \right )}{9}-\frac {x^{2} \ln \left (x^{2}+1\right )}{6}+\frac {5 x^{2}}{18}+\frac {2 x \arctan \left (x \right )}{3}-\frac {\arctan \left (x \right )^{2}}{3}+\frac {\ln \left (x^{2}+1\right )^{2}}{12}-\frac {11 \ln \left (x^{2}+1\right )}{18}-\frac {5}{18}\) | \(68\) |
default | \(\text {Expression too large to display}\) | \(3041\) |
risch | \(\text {Expression too large to display}\) | \(5252\) |
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Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.67 \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {5}{18} \, x^{2} - \frac {2}{9} \, {\left (x^{3} - 3 \, x\right )} \arctan \left (x\right ) - \frac {1}{3} \, \arctan \left (x\right )^{2} + \frac {1}{18} \, {\left (6 \, x^{3} \arctan \left (x\right ) - 3 \, x^{2} - 11\right )} \log \left (x^{2} + 1\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right )^{2} \]
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Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {x^{3} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{3} - \frac {2 x^{3} \operatorname {atan}{\left (x \right )}}{9} - \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{6} + \frac {5 x^{2}}{18} + \frac {2 x \operatorname {atan}{\left (x \right )}}{3} + \frac {\log {\left (x^{2} + 1 \right )}^{2}}{12} - \frac {11 \log {\left (x^{2} + 1 \right )}}{18} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{3} \]
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Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {5}{18} \, x^{2} + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (x^{2} + 1\right ) - 2 \, x^{3} + 6 \, x - 6 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{3} \, \arctan \left (x\right )^{2} - \frac {1}{18} \, {\left (3 \, x^{2} + 11\right )} \log \left (x^{2} + 1\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right )^{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (66) = 132\).
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.65 \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {1}{6} \, \pi x^{3} \log \left (x^{2} + 1\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{3} \, x^{3} \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{9} \, \pi x^{3} \mathrm {sgn}\left (x\right ) + \frac {2}{9} \, x^{3} \arctan \left (\frac {1}{x}\right ) - \frac {1}{6} \, x^{2} \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \pi ^{2} \mathrm {sgn}\left (x\right ) + \frac {1}{3} \, \pi x \mathrm {sgn}\left (x\right ) + \frac {1}{3} \, \pi \arctan \left (\frac {1}{x}\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{6} \, \pi ^{2} + \frac {5}{18} \, x^{2} - \frac {1}{3} \, \pi \arctan \left (x\right ) - \frac {1}{3} \, \pi \arctan \left (\frac {1}{x}\right ) - \frac {2}{3} \, x \arctan \left (\frac {1}{x}\right ) - \frac {1}{3} \, \arctan \left (\frac {1}{x}\right )^{2} + \frac {1}{12} \, \log \left (x^{2} + 1\right )^{2} - \frac {11}{18} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.53 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int x^2 \arctan (x) \log \left (1+x^2\right ) \, dx=\frac {{\ln \left (x^2+1\right )}^2}{12}-\frac {11\,\ln \left (x^2+1\right )}{18}-\frac {{\mathrm {atan}\left (x\right )}^2}{3}-x^2\,\left (\frac {\ln \left (x^2+1\right )}{6}-\frac {5}{18}\right )-x^3\,\left (\frac {2\,\mathrm {atan}\left (x\right )}{9}-\frac {\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )}{3}\right )+\frac {2\,x\,\mathrm {atan}\left (x\right )}{3} \]
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