Integrand size = 13, antiderivative size = 63 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=-\frac {1}{12 x^3}-\frac {3}{4 x}-\frac {3 \arctan (x)}{4}-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x) \]
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Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5068, 4946, 331, 209, 4940, 2438} \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}-\frac {3 \arctan (x)}{4}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x)-\frac {1}{12 x^3}-\frac {3}{4 x} \]
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Rule 209
Rule 331
Rule 2438
Rule 4940
Rule 4946
Rule 5068
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\arctan (x)}{x^5}+\frac {2 \arctan (x)}{x^3}+\frac {\arctan (x)}{x}\right ) \, dx \\ & = 2 \int \frac {\arctan (x)}{x^3} \, dx+\int \frac {\arctan (x)}{x^5} \, dx+\int \frac {\arctan (x)}{x} \, dx \\ & = -\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}+\frac {1}{2} i \int \frac {\log (1-i x)}{x} \, dx-\frac {1}{2} i \int \frac {\log (1+i x)}{x} \, dx+\frac {1}{4} \int \frac {1}{x^4 \left (1+x^2\right )} \, dx+\int \frac {1}{x^2 \left (1+x^2\right )} \, dx \\ & = -\frac {1}{12 x^3}-\frac {1}{x}-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x)-\frac {1}{4} \int \frac {1}{x^2 \left (1+x^2\right )} \, dx-\int \frac {1}{1+x^2} \, dx \\ & = -\frac {1}{12 x^3}-\frac {3}{4 x}-\arctan (x)-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x)+\frac {1}{4} \int \frac {1}{1+x^2} \, dx \\ & = -\frac {1}{12 x^3}-\frac {3}{4 x}-\frac {3 \arctan (x)}{4}-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x) \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.00 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.29 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=-\frac {\arctan (x)}{4 x^4}-\frac {\arctan (x)}{x^2}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-x^2\right )}{12 x^3}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-x^2\right )}{x}+\frac {1}{2} i \operatorname {PolyLog}(2,-i x)-\frac {1}{2} i \operatorname {PolyLog}(2,i x) \]
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Time = 0.36 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25
method | result | size |
default | \(\arctan \left (x \right ) \ln \left (x \right )-\frac {\arctan \left (x \right )}{x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}+\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}-\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}-\frac {1}{12 x^{3}}-\frac {3}{4 x}-\frac {3 \arctan \left (x \right )}{4}\) | \(79\) |
parts | \(\arctan \left (x \right ) \ln \left (x \right )-\frac {\arctan \left (x \right )}{x^{2}}-\frac {\arctan \left (x \right )}{4 x^{4}}+\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}-\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}-\frac {1}{12 x^{3}}-\frac {3}{4 x}-\frac {3 \arctan \left (x \right )}{4}\) | \(79\) |
meijerg | \(-\frac {1}{12 x^{3}}-\frac {3}{4 x}-\frac {2 \left (-\frac {3 x^{4}}{8}+\frac {3}{8}\right ) \arctan \left (\sqrt {x^{2}}\right )}{3 x^{3} \sqrt {x^{2}}}-\frac {i x \,\operatorname {Li}_{2}\left (i \sqrt {x^{2}}\right )}{2 \sqrt {x^{2}}}+\frac {i x \,\operatorname {Li}_{2}\left (-i \sqrt {x^{2}}\right )}{2 \sqrt {x^{2}}}-\frac {\left (x^{2}+1\right ) \arctan \left (x \right )}{x^{2}}\) | \(85\) |
risch | \(-\frac {3}{4 x}-\frac {1}{12 x^{3}}+\frac {3 i \ln \left (-i x \right )}{8}-\frac {3 \arctan \left (x \right )}{4}-\frac {i \ln \left (-i x +1\right )}{8 x^{4}}-\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}-\frac {i \ln \left (-i x +1\right )}{2 x^{2}}-\frac {3 i \ln \left (i x \right )}{8}+\frac {i \ln \left (i x +1\right )}{8 x^{4}}+\frac {i \operatorname {dilog}\left (i x +1\right )}{2}+\frac {i \ln \left (i x +1\right )}{2 x^{2}}\) | \(104\) |
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\[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\int { \frac {{\left (x^{2} + 1\right )}^{2} \arctan \left (x\right )}{x^{5}} \,d x } \]
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\[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\int \frac {\left (x^{2} + 1\right )^{2} \operatorname {atan}{\left (x \right )}}{x^{5}}\, dx \]
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none
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=-\frac {3 \, \pi x^{4} \log \left (x^{2} + 1\right ) - 12 \, x^{4} \arctan \left (x\right ) \log \left (x\right ) + 6 i \, x^{4} {\rm Li}_2\left (i \, x + 1\right ) - 6 i \, x^{4} {\rm Li}_2\left (-i \, x + 1\right ) + 9 \, x^{3} + 3 \, {\left (3 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \left (x\right ) + x}{12 \, x^{4}} \]
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\[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\int { \frac {{\left (x^{2} + 1\right )}^{2} \arctan \left (x\right )}{x^{5}} \,d x } \]
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Time = 0.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1+x^2\right )^2 \arctan (x)}{x^5} \, dx=\frac {x^2-\frac {1}{3}}{4\,x^3}-\frac {\mathrm {atan}\left (x\right )}{x^2}-\frac {\mathrm {atan}\left (x\right )}{4\,x^4}-\frac {3\,\mathrm {atan}\left (x\right )}{4}-\frac {1}{x}-\frac {{\mathrm {Li}}_{\mathrm {2}}\left (1-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {\mathrm {polylog}\left (2,-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]
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