Integrand size = 8, antiderivative size = 61 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=-\frac {1}{12 x^2}-\frac {\arctan (x)}{6 x^3}+\frac {\arctan (x)}{2 x}+\frac {\arctan (x)^2}{4}-\frac {\arctan (x)^2}{4 x^4}-\frac {2 \log (x)}{3}+\frac {1}{3} \log \left (1+x^2\right ) \]
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Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4946, 5038, 272, 46, 36, 29, 31, 5004} \[ \int \frac {\arctan (x)^2}{x^5} \, dx=-\frac {\arctan (x)^2}{4 x^4}-\frac {\arctan (x)}{6 x^3}+\frac {\arctan (x)^2}{4}+\frac {\arctan (x)}{2 x}-\frac {1}{12 x^2}+\frac {1}{3} \log \left (x^2+1\right )-\frac {2 \log (x)}{3} \]
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 272
Rule 4946
Rule 5004
Rule 5038
Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan (x)^2}{4 x^4}+\frac {1}{2} \int \frac {\arctan (x)}{x^4 \left (1+x^2\right )} \, dx \\ & = -\frac {\arctan (x)^2}{4 x^4}+\frac {1}{2} \int \frac {\arctan (x)}{x^4} \, dx-\frac {1}{2} \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx \\ & = -\frac {\arctan (x)}{6 x^3}-\frac {\arctan (x)^2}{4 x^4}+\frac {1}{6} \int \frac {1}{x^3 \left (1+x^2\right )} \, dx-\frac {1}{2} \int \frac {\arctan (x)}{x^2} \, dx+\frac {1}{2} \int \frac {\arctan (x)}{1+x^2} \, dx \\ & = -\frac {\arctan (x)}{6 x^3}+\frac {\arctan (x)}{2 x}+\frac {\arctan (x)^2}{4}-\frac {\arctan (x)^2}{4 x^4}+\frac {1}{12} \text {Subst}\left (\int \frac {1}{x^2 (1+x)} \, dx,x,x^2\right )-\frac {1}{2} \int \frac {1}{x \left (1+x^2\right )} \, dx \\ & = -\frac {\arctan (x)}{6 x^3}+\frac {\arctan (x)}{2 x}+\frac {\arctan (x)^2}{4}-\frac {\arctan (x)^2}{4 x^4}+\frac {1}{12} \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right ) \\ & = -\frac {1}{12 x^2}-\frac {\arctan (x)}{6 x^3}+\frac {\arctan (x)}{2 x}+\frac {\arctan (x)^2}{4}-\frac {\arctan (x)^2}{4 x^4}-\frac {\log (x)}{6}+\frac {1}{12} \log \left (1+x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right ) \\ & = -\frac {1}{12 x^2}-\frac {\arctan (x)}{6 x^3}+\frac {\arctan (x)}{2 x}+\frac {\arctan (x)^2}{4}-\frac {\arctan (x)^2}{4 x^4}-\frac {2 \log (x)}{3}+\frac {1}{3} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=-\frac {1}{12 x^2}+\frac {\left (-1+3 x^2\right ) \arctan (x)}{6 x^3}+\frac {\left (-1+x^4\right ) \arctan (x)^2}{4 x^4}-\frac {2 \log (x)}{3}+\frac {1}{3} \log \left (1+x^2\right ) \]
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Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79
method | result | size |
default | \(-\frac {1}{12 x^{2}}-\frac {\arctan \left (x \right )}{6 x^{3}}+\frac {\arctan \left (x \right )}{2 x}+\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {2 \ln \left (x \right )}{3}+\frac {\ln \left (x^{2}+1\right )}{3}\) | \(48\) |
parts | \(-\frac {1}{12 x^{2}}-\frac {\arctan \left (x \right )}{6 x^{3}}+\frac {\arctan \left (x \right )}{2 x}+\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {2 \ln \left (x \right )}{3}+\frac {\ln \left (x^{2}+1\right )}{3}\) | \(48\) |
parallelrisch | \(-\frac {-3 x^{4} \arctan \left (x \right )^{2}+8 x^{4} \ln \left (x \right )-4 \ln \left (x^{2}+1\right ) x^{4}-6 x^{3} \arctan \left (x \right )+x^{2}+2 x \arctan \left (x \right )+3 \arctan \left (x \right )^{2}}{12 x^{4}}\) | \(55\) |
risch | \(-\frac {\left (x^{4}-1\right ) \ln \left (i x +1\right )^{2}}{16 x^{4}}+\frac {\left (3 x^{4} \ln \left (-i x +1\right )-6 i x^{3}+2 i x -3 \ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{24 x^{4}}-\frac {3 x^{4} \ln \left (-i x +1\right )^{2}+32 x^{4} \ln \left (x \right )-16 \ln \left (x^{2}+1\right ) x^{4}-12 i x^{3} \ln \left (-i x +1\right )+4 i x \ln \left (-i x +1\right )+4 x^{2}-3 \ln \left (-i x +1\right )^{2}}{48 x^{4}}\) | \(143\) |
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Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=\frac {4 \, x^{4} \log \left (x^{2} + 1\right ) - 8 \, x^{4} \log \left (x\right ) + 3 \, {\left (x^{4} - 1\right )} \arctan \left (x\right )^{2} - x^{2} + 2 \, {\left (3 \, x^{3} - x\right )} \arctan \left (x\right )}{12 \, x^{4}} \]
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Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=- \frac {2 \log {\left (x \right )}}{3} + \frac {\log {\left (x^{2} + 1 \right )}}{3} + \frac {\operatorname {atan}^{2}{\left (x \right )}}{4} + \frac {\operatorname {atan}{\left (x \right )}}{2 x} - \frac {1}{12 x^{2}} - \frac {\operatorname {atan}{\left (x \right )}}{6 x^{3}} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4 x^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=\frac {1}{6} \, {\left (\frac {3 \, x^{2} - 1}{x^{3}} + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {3 \, x^{2} \arctan \left (x\right )^{2} - 4 \, x^{2} \log \left (x^{2} + 1\right ) + 8 \, x^{2} \log \left (x\right ) + 1}{12 \, x^{2}} - \frac {\arctan \left (x\right )^{2}}{4 \, x^{4}} \]
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\[ \int \frac {\arctan (x)^2}{x^5} \, dx=\int { \frac {\arctan \left (x\right )^{2}}{x^{5}} \,d x } \]
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Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {\arctan (x)^2}{x^5} \, dx=\frac {\ln \left (x^2+1\right )}{3}-\frac {2\,\ln \left (x\right )}{3}-{\mathrm {atan}\left (x\right )}^2\,\left (\frac {1}{4\,x^4}-\frac {1}{4}\right )-\frac {1}{12\,x^2}+\frac {\mathrm {atan}\left (x\right )\,\left (\frac {x^2}{2}-\frac {1}{6}\right )}{x^3} \]
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