Integrand size = 13, antiderivative size = 47 \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=\frac {1}{6 x^2}-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {4 \log (x)}{3}-\frac {2}{3} \log \left (1+x^2\right ) \]
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Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {5039, 4947, 272, 46, 36, 29, 31, 5005} \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {1}{6 x^2}-\frac {2}{3} \log \left (x^2+1\right )+\frac {4 \log (x)}{3}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {\cot ^{-1}(x)}{x} \]
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 272
Rule 4947
Rule 5005
Rule 5039
Rubi steps \begin{align*} \text {integral}& = \int \frac {\cot ^{-1}(x)}{x^4} \, dx-\int \frac {\cot ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx \\ & = -\frac {\cot ^{-1}(x)}{3 x^3}-\frac {1}{3} \int \frac {1}{x^3 \left (1+x^2\right )} \, dx-\int \frac {\cot ^{-1}(x)}{x^2} \, dx+\int \frac {\cot ^{-1}(x)}{1+x^2} \, dx \\ & = -\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2-\frac {1}{6} \text {Subst}\left (\int \frac {1}{x^2 (1+x)} \, dx,x,x^2\right )+\int \frac {1}{x \left (1+x^2\right )} \, dx \\ & = -\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2-\frac {1}{6} \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,x^2\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right ) \\ & = \frac {1}{6 x^2}-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^2\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right ) \\ & = \frac {1}{6 x^2}-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {4 \log (x)}{3}-\frac {2}{3} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=\frac {1}{6 x^2}-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {4 \log (x)}{3}-\frac {2}{3} \log \left (1+x^2\right ) \]
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Time = 0.42 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91
method | result | size |
default | \(-\frac {\operatorname {arccot}\left (x \right )}{3 x^{3}}+\frac {\operatorname {arccot}\left (x \right )}{x}+\operatorname {arccot}\left (x \right ) \arctan \left (x \right )+\frac {1}{6 x^{2}}+\frac {4 \ln \left (x \right )}{3}-\frac {2 \ln \left (x^{2}+1\right )}{3}+\frac {\arctan \left (x \right )^{2}}{2}\) | \(43\) |
parts | \(-\frac {\operatorname {arccot}\left (x \right )}{3 x^{3}}+\frac {\operatorname {arccot}\left (x \right )}{x}+\operatorname {arccot}\left (x \right ) \arctan \left (x \right )+\frac {1}{6 x^{2}}+\frac {4 \ln \left (x \right )}{3}-\frac {2 \ln \left (x^{2}+1\right )}{3}+\frac {\arctan \left (x \right )^{2}}{2}\) | \(43\) |
parallelrisch | \(\frac {-3 \operatorname {arccot}\left (x \right )^{2} x^{3}+8 \ln \left (x \right ) x^{3}-4 \ln \left (x^{2}+1\right ) x^{3}+6 x^{2} \operatorname {arccot}\left (x \right )+x -2 \,\operatorname {arccot}\left (x \right )}{6 x^{3}}\) | \(46\) |
risch | \(\frac {\ln \left (i x +1\right )^{2}}{8}-\frac {\left (3 \ln \left (-i x +1\right ) x^{3}-6 i x^{2}+2 i\right ) \ln \left (i x +1\right )}{12 x^{3}}+\frac {-6 i \ln \left (\left (-\pi +8 i\right ) x +8+i \pi \right ) \pi \,x^{3}+6 i \ln \left (\left (-\pi -8 i\right ) x +8-i \pi \right ) \pi \,x^{3}+3 \ln \left (-i x +1\right )^{2} x^{3}-12 i x^{2} \ln \left (-i x +1\right )-16 \ln \left (\left (-\pi +8 i\right ) x +8+i \pi \right ) x^{3}-16 \ln \left (\left (-\pi -8 i\right ) x +8-i \pi \right ) x^{3}+32 \ln \left (-x \right ) x^{3}+12 \pi \,x^{2}+4 i \ln \left (-i x +1\right )-4 \pi +4 x}{24 x^{3}}\) | \(194\) |
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Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=-\frac {3 \, x^{3} \operatorname {arccot}\left (x\right )^{2} + 4 \, x^{3} \log \left (x^{2} + 1\right ) - 8 \, x^{3} \log \left (x\right ) - 2 \, {\left (3 \, x^{2} - 1\right )} \operatorname {arccot}\left (x\right ) - x}{6 \, x^{3}} \]
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=\frac {4 \log {\left (x \right )}}{3} - \frac {2 \log {\left (x^{2} + 1 \right )}}{3} - \frac {\operatorname {acot}^{2}{\left (x \right )}}{2} + \frac {\operatorname {acot}{\left (x \right )}}{x} + \frac {1}{6 x^{2}} - \frac {\operatorname {acot}{\left (x \right )}}{3 x^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.17 \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=\frac {1}{3} \, {\left (\frac {3 \, x^{2} - 1}{x^{3}} + 3 \, \arctan \left (x\right )\right )} \operatorname {arccot}\left (x\right ) + \frac {3 \, x^{2} \arctan \left (x\right )^{2} - 4 \, x^{2} \log \left (x^{2} + 1\right ) + 8 \, x^{2} \log \left (x\right ) + 1}{6 \, x^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (\frac {1}{x}\right )^{2} + \frac {\arctan \left (\frac {1}{x}\right )}{x} + \frac {1}{6 \, x^{2}} - \frac {\arctan \left (\frac {1}{x}\right )}{3 \, x^{3}} - \frac {2}{3} \, \log \left (\frac {1}{x^{2}} + 1\right ) \]
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Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\cot ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx=\frac {4\,\ln \left (x\right )}{3}-\frac {2\,\ln \left (x^2+1\right )}{3}-\frac {{\mathrm {acot}\left (x\right )}^2}{2}+\frac {1}{6\,x^2}+\frac {\mathrm {acot}\left (x\right )\,\left (x^2-\frac {1}{3}\right )}{x^3} \]
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