Integrand size = 13, antiderivative size = 60 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=-\frac {1}{12 x^2}-\frac {\arctan (x)}{6 x^3}-\frac {\arctan (x)}{2 x}-\frac {\left (1+x^2\right )^2 \arctan (x)^2}{4 x^4}+\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^2\right ) \]
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Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {5064, 5070, 4946, 272, 46, 36, 29, 31} \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=-\frac {\arctan (x)}{6 x^3}-\frac {\left (x^2+1\right )^2 \arctan (x)^2}{4 x^4}-\frac {\arctan (x)}{2 x}-\frac {1}{12 x^2}-\frac {1}{6} \log \left (x^2+1\right )+\frac {\log (x)}{3} \]
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 272
Rule 4946
Rule 5064
Rule 5070
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (1+x^2\right )^2 \arctan (x)^2}{4 x^4}+\frac {1}{2} \int \frac {\left (1+x^2\right ) \arctan (x)}{x^4} \, dx \\ & = -\frac {\left (1+x^2\right )^2 \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} \, dx \\ & = -\frac {\arctan (x)}{6 x^3}-\frac {\arctan (x)}{2 x}-\frac {\left (1+x^2\right )^2 \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 {1}{x \left (1+x^2\right )} \, dx \\ & = -\frac {\arctan (x)}{6 x^3}-\frac {\arctan (x)}{2 x}-\frac {\left (1+x^2\right )^2 \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}{4} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right ) \\ & = -\frac {\arctan (x)}{6 x^3}-\frac {\arctan (x)}{2 x}-\frac {\left (1+x^2\right )^2 \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} \, 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 {\left (1+x^2\right )^2 \arctan (x)^2}{4 x^4}+\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\frac {-2 \left (x+3 x^3\right ) \arctan (x)-3 \left (1+x^2\right )^2 \arctan (x)^2+x^2 \left (-1+4 x^2 \log (x)-2 x^2 \log \left (1+x^2\right )\right )}{12 x^4} \]
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Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )}{6 x^{3}}-\frac {\arctan \left (x \right )}{2 x}-\frac {1}{12 x^{2}}+\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x^{2}+1\right )}{6}\) | \(57\) |
parts | \(-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )}{6 x^{3}}-\frac {\arctan \left (x \right )}{2 x}-\frac {1}{12 x^{2}}+\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x^{2}+1\right )}{6}\) | \(57\) |
parallelrisch | \(\frac {-3 x^{4} \arctan \left (x \right )^{2}+4 x^{4} \ln \left (x \right )-2 \ln \left (x^{2}+1\right ) x^{4}-6 x^{3} \arctan \left (x \right )-6 x^{2} \arctan \left (x \right )^{2}-x^{2}-2 x \arctan \left (x \right )-3 \arctan \left (x \right )^{2}}{12 x^{4}}\) | \(66\) |
risch | \(\frac {\left (x^{4}+2 x^{2}+1\right ) \ln \left (i x +1\right )^{2}}{16 x^{4}}-\frac {\left (3 x^{4} \ln \left (-i x +1\right )-6 i x^{3}+6 x^{2} \ln \left (-i x +1\right )-2 i x +3 \ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{24 x^{4}}+\frac {3 x^{4} \ln \left (-i x +1\right )^{2}-12 i x^{3} \ln \left (-i x +1\right )+16 x^{4} \ln \left (x \right )-8 \ln \left (x^{2}+1\right ) x^{4}+6 x^{2} \ln \left (-i x +1\right )^{2}-4 i x \ln \left (-i x +1\right )-4 x^{2}+3 \ln \left (-i x +1\right )^{2}}{48 x^{4}}\) | \(174\) |
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Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=-\frac {2 \, x^{4} \log \left (x^{2} + 1\right ) - 4 \, x^{4} \log \left (x\right ) + 3 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2} + x^{2} + 2 \, {\left (3 \, x^{3} + x\right )} \arctan \left (x\right )}{12 \, x^{4}} \]
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Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\frac {\log {\left (x \right )}}{3} - \frac {\log {\left (x^{2} + 1 \right )}}{6} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4} - \frac {\operatorname {atan}{\left (x \right )}}{2 x} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2 x^{2}} - \frac {1}{12 x^{2}} - \frac {\operatorname {atan}{\left (x \right )}}{6 x^{3}} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4 x^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=-\frac {1}{6} \, {\left (\frac {3 \, x^{2} + 1}{x^{3}} + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {3 \, x^{2} \arctan \left (x\right )^{2} - 2 \, x^{2} \log \left (x^{2} + 1\right ) + 4 \, x^{2} \log \left (x\right ) - 1}{12 \, x^{2}} - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2}}{4 \, x^{4}} \]
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\[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\int { \frac {{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2}}{x^{5}} \,d x } \]
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Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {\left (1+x^2\right ) \arctan (x)^2}{x^5} \, dx=\frac {\ln \left (x\right )}{3}-\frac {\ln \left (x^2+1\right )}{6}-{\mathrm {atan}\left (x\right )}^2\,\left (\frac {\frac {x^2}{2}+\frac {1}{4}}{x^4}+\frac {1}{4}\right )-\frac {1}{12\,x^2}-\frac {\mathrm {atan}\left (x\right )\,\left (\frac {x^2}{2}+\frac {1}{6}\right )}{x^3} \]
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