Optimal. Leaf size=60 \[ -\frac {1}{12 x^2}-\frac {\tan ^{-1}(x)}{6 x^3}-\frac {\tan ^{-1}(x)}{2 x}-\frac {\left (1+x^2\right )^2 \tan ^{-1}(x)^2}{4 x^4}+\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^2\right ) \]
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Rubi [A]
time = 0.05, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {5064, 5070,
4946, 272, 46, 36, 29, 31} \begin {gather*} -\frac {\text {ArcTan}(x)}{6 x^3}-\frac {\left (x^2+1\right )^2 \text {ArcTan}(x)^2}{4 x^4}-\frac {\text {ArcTan}(x)}{2 x}-\frac {1}{12 x^2}-\frac {1}{6} \log \left (x^2+1\right )+\frac {\log (x)}{3} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 272
Rule 4946
Rule 5064
Rule 5070
Rubi steps
\begin {align*} \int \frac {\left (1+x^2\right ) \tan ^{-1}(x)^2}{x^5} \, dx &=-\frac {\left (1+x^2\right )^2 \tan ^{-1}(x)^2}{4 x^4}+\frac {1}{2} \int \frac {\left (1+x^2\right ) \tan ^{-1}(x)}{x^4} \, dx\\ &=-\frac {\left (1+x^2\right )^2 \tan ^{-1}(x)^2}{4 x^4}+\frac {1}{2} \int \frac {\tan ^{-1}(x)}{x^4} \, dx+\frac {1}{2} \int \frac {\tan ^{-1}(x)}{x^2} \, dx\\ &=-\frac {\tan ^{-1}(x)}{6 x^3}-\frac {\tan ^{-1}(x)}{2 x}-\frac {\left (1+x^2\right )^2 \tan ^{-1}(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 {\tan ^{-1}(x)}{6 x^3}-\frac {\tan ^{-1}(x)}{2 x}-\frac {\left (1+x^2\right )^2 \tan ^{-1}(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 {\tan ^{-1}(x)}{6 x^3}-\frac {\tan ^{-1}(x)}{2 x}-\frac {\left (1+x^2\right )^2 \tan ^{-1}(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 {\tan ^{-1}(x)}{6 x^3}-\frac {\tan ^{-1}(x)}{2 x}-\frac {\left (1+x^2\right )^2 \tan ^{-1}(x)^2}{4 x^4}+\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^2\right )\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 56, normalized size = 0.93 \begin {gather*} \frac {-2 \left (x+3 x^3\right ) \tan ^{-1}(x)-3 \left (1+x^2\right )^2 \tan ^{-1}(x)^2+x^2 \left (-1+4 x^2 \log (x)-2 x^2 \log \left (1+x^2\right )\right )}{12 x^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 57, normalized size = 0.95
method | result | size |
default | \(-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}-\frac {\arctan \left (x \right )}{6 x^{3}}-\frac {\arctan \left (x \right )}{2 x}-\frac {\arctan \left (x \right )^{2}}{4}-\frac {1}{12 x^{2}}+\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x^{2}+1\right )}{6}\) | \(57\) |
risch | \(\frac {\left (x^{4}+2 x^{2}+1\right ) \ln \left (i x +1\right )^{2}}{16 x^{4}}-\frac {\left (3 x^{4} \ln \left (-i x +1\right )-6 i x^{3}+6 x^{2} \ln \left (-i x +1\right )-2 i x +3 \ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{24 x^{4}}+\frac {3 x^{4} \ln \left (-i x +1\right )^{2}-12 i x^{3} \ln \left (-i x +1\right )+16 x^{4} \ln \left (x \right )-8 x^{4} \ln \left (x^{2}+1\right )+6 x^{2} \ln \left (-i x +1\right )^{2}-4 i x \ln \left (-i x +1\right )-4 x^{2}+3 \ln \left (-i x +1\right )^{2}}{48 x^{4}}\) | \(174\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 3.24, size = 71, normalized size = 1.18 \begin {gather*} -\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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.67, size = 54, normalized size = 0.90 \begin {gather*} -\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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.27, size = 61, normalized size = 1.02 \begin {gather*} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.12, size = 51, normalized size = 0.85 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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