Optimal. Leaf size=88 \[ \frac {x}{4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{4 a}-\frac {\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{6 a} \]
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Rubi [A]
time = 0.05, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {6103, 6141,
205, 212} \begin {gather*} \frac {x}{4 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{6 a}+\frac {\tanh ^{-1}(a x)}{4 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 212
Rule 6103
Rule 6141
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx &=\frac {x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{6 a}-a \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac {\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{6 a}+\frac {1}{2} \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {x}{4 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{6 a}+\frac {1}{4} \int \frac {1}{1-a^2 x^2} \, dx\\ &=\frac {x}{4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{4 a}-\frac {\tanh ^{-1}(a x)}{2 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{6 a}\\ \end {align*}
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Mathematica [A]
time = 0.04, size = 93, normalized size = 1.06 \begin {gather*} \frac {12 \tanh ^{-1}(a x)-12 a x \tanh ^{-1}(a x)^2+4 \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^3-3 \left (2 a x+\left (-1+a^2 x^2\right ) \log (1-a x)+\left (1-a^2 x^2\right ) \log (1+a x)\right )}{24 a \left (-1+a^2 x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 71.18, size = 1340, normalized size = 15.23
method | result | size |
risch | \(\frac {\ln \left (a x +1\right )^{3}}{48 a}-\frac {\left (x^{2} \ln \left (-a x +1\right ) a^{2}+2 a x -\ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )^{2}}{16 \left (a^{2} x^{2}-1\right ) a}+\frac {\left (a^{2} x^{2} \ln \left (-a x +1\right )^{2}+4 a x \ln \left (-a x +1\right )-\ln \left (-a x +1\right )^{2}+4\right ) \ln \left (a x +1\right )}{16 a \left (a x -1\right ) \left (a x +1\right )}-\frac {a^{2} x^{2} \ln \left (-a x +1\right )^{3}+6 \ln \left (a x -1\right ) a^{2} x^{2}-6 \ln \left (-a x -1\right ) a^{2} x^{2}+6 a \ln \left (-a x +1\right )^{2} x -\ln \left (-a x +1\right )^{3}+12 a x -6 \ln \left (a x -1\right )+6 \ln \left (-a x -1\right )+12 \ln \left (-a x +1\right )}{48 a \left (a x -1\right ) \left (a x +1\right )}\) | \(251\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1340\) |
default | \(\text {Expression too large to display}\) | \(1340\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 268 vs.
\(2 (75) = 150\).
time = 0.27, size = 268, normalized size = 3.05 \begin {gather*} -\frac {1}{4} \, {\left (\frac {2 \, x}{a^{2} x^{2} - 1} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )^{2} + \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} - 3 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} - 12 \, a x + 3 \, {\left (2 \, a^{2} x^{2} + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right ) - 6 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{48 \, {\left (a^{5} x^{2} - a^{3}\right )}} - \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4\right )} a \operatorname {artanh}\left (a x\right )}{8 \, {\left (a^{4} x^{2} - a^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 95, normalized size = 1.08 \begin {gather*} -\frac {6 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 12 \, a x - 6 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{48 \, {\left (a^{3} x^{2} - a\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.42, size = 88, normalized size = 1.00 \begin {gather*} \frac {1}{16} \, a^{2} {\left (\frac {{\left (a x - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{{\left (a x + 1\right )} a^{4}} + \frac {2 \, {\left (a x - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{{\left (a x + 1\right )} a^{4}} + \frac {2 \, {\left (a x - 1\right )}}{{\left (a x + 1\right )} a^{4}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.50, size = 213, normalized size = 2.42 \begin {gather*} \frac {{\ln \left (a\,x+1\right )}^3}{48\,a}-\frac {\ln \left (a\,x+1\right )}{4\,\left (a-a^3\,x^2\right )}-\frac {{\ln \left (1-a\,x\right )}^3}{48\,a}-\frac {x}{4\,a^2\,x^2-4}+\frac {\ln \left (1-a\,x\right )}{4\,a-4\,a^3\,x^2}+\frac {\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2}{16\,a}-\frac {{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )}{16\,a}-\frac {x\,{\ln \left (a\,x+1\right )}^2}{8\,\left (a^2\,x^2-1\right )}-\frac {x\,{\ln \left (1-a\,x\right )}^2}{2\,\left (4\,a^2\,x^2-4\right )}+\frac {x\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{4\,a^2\,x^2-4}-\frac {\mathrm {atan}\left (a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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