Optimal. Leaf size=119 \[ -\frac {3 x}{8 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{8 a^2}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )} \]
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Rubi [A]
time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6141, 6103,
205, 212} \begin {gather*} -\frac {3 x}{8 a \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}-\frac {3 \tanh ^{-1}(a x)}{8 a^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 212
Rule 6103
Rule 6141
Rubi steps
\begin {align*} \int \frac {x \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^2} \, dx &=\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx}{2 a}\\ &=-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}+\frac {3}{2} \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx}{4 a}\\ &=-\frac {3 x}{8 a \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 \int \frac {1}{1-a^2 x^2} \, dx}{8 a}\\ &=-\frac {3 x}{8 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{8 a^2}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 91, normalized size = 0.76 \begin {gather*} \frac {6 a x-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 (-1+a^2 x^2\right ) \log (1-a x)-3 \left (-1+a^2 x^2\right ) \log (1+a x)}{16 a^2 \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 = 166.97, size = 1359, normalized size = 11.42
method | result | size |
risch | \(-\frac {\left (a^{2} x^{2}+1\right ) \ln \left (a x +1\right )^{3}}{32 a^{2} \left (a x -1\right ) \left (a x +1\right )}+\frac {3 \left (x^{2} \ln \left (-a x +1\right ) a^{2}+2 a x +\ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )^{2}}{32 a^{2} \left (a x -1\right ) \left (a x +1\right )}-\frac {3 \left (a^{2} x^{2} \ln \left (-a x +1\right )^{2}+4 a x \ln \left (-a x +1\right )+\ln \left (-a x +1\right )^{2}+4\right ) \ln \left (a x +1\right )}{32 a^{2} \left (a x -1\right ) \left (a x +1\right )}+\frac {a^{2} x^{2} \ln \left (-a x +1\right )^{3}+6 x^{2} \ln \left (-a x +1\right ) a^{2}-6 a^{2} x^{2} \ln \left (a x +1\right )+6 a \ln \left (-a x +1\right )^{2} x +\ln \left (-a x +1\right )^{3}+12 a x +6 \ln \left (-a x +1\right )+6 \ln \left (a x +1\right )}{32 a^{2} \left (a x -1\right ) \left (a x +1\right )}\) | \(262\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1359\) |
default | \(\text {Expression too large to display}\) | \(1359\) |
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. 298 vs.
\(2 (103) = 206\).
time = 0.27, size = 298, normalized size = 2.50 \begin {gather*} \frac {3 \, {\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}}{8 \, a} - \frac {\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}}{a^{5} x^{2} - a^{3}} - \frac {6 \, {\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 )}{a^{4} x^{2} - a^{2}}}{32 \, a} - \frac {\operatorname {artanh}\left (a x\right )^{3}}{2 \, {\left (a^{2} x^{2} - 1\right )} a^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.45, size = 97, normalized size = 0.82 \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 )}{32 \, {\left (a^{4} x^{2} - a^{2}\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 {x \operatorname {atanh}^{3}{\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 = 0.41, size = 192, normalized size = 1.61 \begin {gather*} -\frac {1}{64} \, {\left ({\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} - 3 \, {\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} - \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 6 \, {\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - \frac {6 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{3}} + \frac {6 \, {\left (a x - 1\right )}}{{\left (a x + 1\right )} a^{3}}\right )} a \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.76, size = 239, normalized size = 2.01 \begin {gather*} -\frac {6\,\ln \left (1-a\,x\right )-6\,\ln \left (a\,x+1\right )+12\,a\,x-{\ln \left (a\,x+1\right )}^3+{\ln \left (1-a\,x\right )}^3-3\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2+3\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )-a^2\,x^2\,\left (6\,\ln \left (a\,x+1\right )-6\,\ln \left (1-a\,x\right )\right )-a^2\,x^2\,{\ln \left (a\,x+1\right )}^3+a^2\,x^2\,{\ln \left (1-a\,x\right )}^3+6\,a\,x\,{\ln \left (a\,x+1\right )}^2+6\,a\,x\,{\ln \left (1-a\,x\right )}^2-12\,a\,x\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )-3\,a^2\,x^2\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2+3\,a^2\,x^2\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )}{32\,a^2-32\,a^4\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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