Integrand size = 10, antiderivative size = 34 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=-\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x)+\frac {1}{4} a \log \left (1+a^2 x^4\right ) \]
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Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4947, 272, 36, 29, 31} \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {1}{4} a \log \left (a^2 x^4+1\right )-\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x) \]
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Rule 29
Rule 31
Rule 36
Rule 272
Rule 4947
Rubi steps \begin{align*} \text {integral}& = -\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \int \frac {1}{x \left (1+a^2 x^4\right )} \, dx \\ & = -\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-\frac {1}{4} a \text {Subst}\left (\int \frac {1}{x \left (1+a^2 x\right )} \, dx,x,x^4\right ) \\ & = -\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-\frac {1}{4} a \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^4\right )+\frac {1}{4} a^3 \text {Subst}\left (\int \frac {1}{1+a^2 x} \, dx,x,x^4\right ) \\ & = -\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x)+\frac {1}{4} a \log \left (1+a^2 x^4\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=-\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x)+\frac {1}{4} a \log \left (1+a^2 x^4\right ) \]
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Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91
method | result | size |
default | \(-\frac {\operatorname {arccot}\left (a \,x^{2}\right )}{2 x^{2}}-a \left (\ln \left (x \right )-\frac {\ln \left (a^{2} x^{4}+1\right )}{4}\right )\) | \(31\) |
parts | \(-\frac {\operatorname {arccot}\left (a \,x^{2}\right )}{2 x^{2}}-a \left (\ln \left (x \right )-\frac {\ln \left (a^{2} x^{4}+1\right )}{4}\right )\) | \(31\) |
parallelrisch | \(-\frac {4 a \ln \left (x \right ) x^{2}-a \ln \left (a^{2} x^{4}+1\right ) x^{2}+2 \,\operatorname {arccot}\left (a \,x^{2}\right )}{4 x^{2}}\) | \(39\) |
risch | \(-\frac {i \ln \left (i a \,x^{2}+1\right )}{4 x^{2}}-\frac {4 a \ln \left (x \right ) x^{2}-a \ln \left (a^{2} x^{4}+1\right ) x^{2}-i \ln \left (-i a \,x^{2}+1\right )+\pi }{4 x^{2}}\) | \(62\) |
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Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {a x^{2} \log \left (a^{2} x^{4} + 1\right ) - 4 \, a x^{2} \log \left (x\right ) - 2 \, \operatorname {arccot}\left (a x^{2}\right )}{4 \, x^{2}} \]
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Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=- a \log {\left (x \right )} + \frac {a \log {\left (a^{2} x^{4} + 1 \right )}}{4} - \frac {\operatorname {acot}{\left (a x^{2} \right )}}{2 x^{2}} \]
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Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {1}{4} \, a {\left (\log \left (a^{2} x^{4} + 1\right ) - \log \left (x^{4}\right )\right )} - \frac {\operatorname {arccot}\left (a x^{2}\right )}{2 \, x^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=-\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {1}{a x^{2}}\right )}{a x^{2}} - \log \left (\frac {1}{a^{2} x^{4}} + 1\right )\right )} \]
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Time = 0.77 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {a\,\ln \left (-a^2\,x^4-1\right )}{4}-\frac {\mathrm {acot}\left (a\,x^2\right )}{2\,x^2}-a\,\ln \left (x\right ) \]
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