Optimal. Leaf size=115 \[ -\frac {3}{8 a \left (1-a^2 x^2\right )}+\frac {3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)^2}{8 a}-\frac {3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{8 a} \]
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Rubi [A]
time = 0.07, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6103, 6141,
267} \begin {gather*} -\frac {3}{8 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{8 a}+\frac {3 \tanh ^{-1}(a x)^2}{8 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 267
Rule 6103
Rule 6141
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^2} \, dx &=\frac {x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{8 a}-\frac {1}{2} (3 a) \int \frac {x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac {3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{8 a}+\frac {3}{2} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)^2}{8 a}-\frac {3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{8 a}-\frac {1}{4} (3 a) \int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac {3}{8 a \left (1-a^2 x^2\right )}+\frac {3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)^2}{8 a}-\frac {3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{8 a}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 71, normalized size = 0.62 \begin {gather*} \frac {3-6 a x \tanh ^{-1}(a x)+3 \left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^2-4 a x \tanh ^{-1}(a x)^3+\left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^4}{8 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 = 123.62, size = 1357, normalized size = 11.80
method | result | size |
risch | \(\frac {\ln \left (a x +1\right )^{4}}{128 a}-\frac {\left (x^{2} \ln \left (-a x +1\right ) a^{2}+2 a x -\ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )^{3}}{32 \left (a^{2} x^{2}-1\right ) a}+\frac {3 \left (a^{2} x^{2} \ln \left (-a x +1\right )^{2}+2 a^{2} x^{2}+4 a x \ln \left (-a x +1\right )-\ln \left (-a x +1\right )^{2}+2\right ) \ln \left (a x +1\right )^{2}}{64 a \left (a x -1\right ) \left (a x +1\right )}-\frac {\left (a^{2} x^{2} \ln \left (-a x +1\right )^{3}+6 x^{2} \ln \left (-a x +1\right ) a^{2}+6 a \ln \left (-a x +1\right )^{2} x -\ln \left (-a x +1\right )^{3}+12 a x +6 \ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )}{32 a \left (a x -1\right ) \left (a x +1\right )}+\frac {a^{2} x^{2} \ln \left (-a x +1\right )^{4}+12 a^{2} x^{2} \ln \left (-a x +1\right )^{2}+8 a x \ln \left (-a x +1\right )^{3}-\ln \left (-a x +1\right )^{4}+48 a x \ln \left (-a x +1\right )+12 \ln \left (-a x +1\right )^{2}+48}{128 a \left (a x -1\right ) \left (a x +1\right )}\) | \(336\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1357\) |
default | \(\text {Expression too large to display}\) | \(1357\) |
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. 459 vs.
\(2 (99) = 198\).
time = 0.28, size = 459, normalized size = 3.99 \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 )^{3} - \frac {3 \, {\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 )^{2}}{16 \, {\left (a^{4} x^{2} - a^{2}\right )}} - \frac {1}{128} \, {\left (\frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{4} - 4 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{4} + 6 \, {\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 )^{2} + 12 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4 \, {\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} + 6 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} \log \left (a x + 1\right ) - 48\right )} a^{2}}{a^{6} x^{2} - a^{4}} - \frac {8 \, {\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 \operatorname {artanh}\left (a x\right )}{a^{5} x^{2} - a^{3}}\right )} a \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 113, normalized size = 0.98 \begin {gather*} -\frac {8 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} - {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} + 48 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 12 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 48}{128 \, {\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}^{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 = 1.52, size = 122, normalized size = 1.06 \begin {gather*} \frac {1}{32} \, a^{2} {\left (\frac {{\left (a x - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3}}{{\left (a x + 1\right )} a^{4}} + \frac {3 \, {\left (a x - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{{\left (a x + 1\right )} a^{4}} + \frac {6 \, {\left (a x - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{{\left (a x + 1\right )} a^{4}} + \frac {6 \, {\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 = 1.72, size = 378, normalized size = 3.29 \begin {gather*} \frac {3\,{\ln \left (a\,x+1\right )}^2}{32\,a}-\frac {3}{2\,\left (4\,a-4\,a^3\,x^2\right )}-\frac {3\,{\ln \left (1-a\,x\right )}^2}{16\,a-16\,a^3\,x^2}+\frac {3\,{\ln \left (1-a\,x\right )}^2}{32\,a}+\frac {{\ln \left (a\,x+1\right )}^4}{128\,a}+\frac {{\ln \left (1-a\,x\right )}^4}{128\,a}-\frac {3\,{\ln \left (a\,x+1\right )}^2}{16\,\left (a-a^3\,x^2\right )}-\frac {3\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{16\,a}-\frac {\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^3}{32\,a}-\frac {{\ln \left (a\,x+1\right )}^3\,\ln \left (1-a\,x\right )}{32\,a}-\frac {3\,x\,\ln \left (a\,x+1\right )}{8\,\left (a^2\,x^2-1\right )}+\frac {6\,x\,\ln \left (1-a\,x\right )}{16\,a^2\,x^2-16}+\frac {3\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{8\,a-8\,a^3\,x^2}+\frac {3\,{\ln \left (a\,x+1\right )}^2\,{\ln \left (1-a\,x\right )}^2}{64\,a}-\frac {x\,{\ln \left (a\,x+1\right )}^3}{16\,\left (a^2\,x^2-1\right )}+\frac {x\,{\ln \left (1-a\,x\right )}^3}{2\,\left (8\,a^2\,x^2-8\right )}-\frac {6\,x\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2}{32\,a^2\,x^2-32}+\frac {6\,x\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )}{32\,a^2\,x^2-32} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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