Optimal. Leaf size=75 \[ -\frac {3 x}{32 a \left (1-a^2 x^2\right )}-\frac {x}{16 a \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \tanh ^{-1}(a x)}{32 a^2} \]
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Rubi [A] time = 0.05, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5994, 199, 206} \[ -\frac {3 x}{32 a \left (1-a^2 x^2\right )}-\frac {x}{16 a \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \tanh ^{-1}(a x)}{32 a^2} \]
Antiderivative was successfully verified.
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Rule 199
Rule 206
Rule 5994
Rubi steps
\begin {align*} \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx &=\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {\int \frac {1}{\left (1-a^2 x^2\right )^3} \, dx}{4 a}\\ &=-\frac {x}{16 a \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx}{16 a}\\ &=-\frac {x}{16 a \left (1-a^2 x^2\right )^2}-\frac {3 x}{32 a \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}-\frac {3 \int \frac {1}{1-a^2 x^2} \, dx}{32 a}\\ &=-\frac {x}{16 a \left (1-a^2 x^2\right )^2}-\frac {3 x}{32 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{32 a^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )^2}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 88, normalized size = 1.17 \[ \frac {3 x}{32 a \left (a^2 x^2-1\right )}-\frac {x}{16 a \left (a^2 x^2-1\right )^2}+\frac {\tanh ^{-1}(a x)}{4 a^2 \left (a^2 x^2-1\right )^2}+\frac {3 \log (1-a x)}{64 a^2}-\frac {3 \log (a x+1)}{64 a^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 71, normalized size = 0.95 \[ \frac {6 \, a^{3} x^{3} - 10 \, a x - {\left (3 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 5\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{64 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.25, size = 239, normalized size = 3.19 \[ -\frac {1}{256} \, {\left (2 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {4 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )}}{{\left (a x + 1\right )}^{2} a^{3}} - \frac {\frac {{\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {4 \, {\left (a x + 1\right )} a^{3}}{a x - 1}}{a^{6}}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right ) + \frac {{\left (a x - 1\right )}^{2} {\left (\frac {8 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )}}{{\left (a x + 1\right )}^{2} a^{3}} + \frac {\frac {{\left (a x + 1\right )}^{2} a^{3}}{{\left (a x - 1\right )}^{2}} - \frac {8 \, {\left (a x + 1\right )} a^{3}}{a x - 1}}{a^{6}}\right )} a \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 92, normalized size = 1.23 \[ \frac {\arctanh \left (a x \right )}{4 a^{2} \left (a^{2} x^{2}-1\right )^{2}}-\frac {1}{64 a^{2} \left (a x -1\right )^{2}}+\frac {3}{64 a^{2} \left (a x -1\right )}+\frac {3 \ln \left (a x -1\right )}{64 a^{2}}+\frac {1}{64 a^{2} \left (a x +1\right )^{2}}+\frac {3}{64 a^{2} \left (a x +1\right )}-\frac {3 \ln \left (a x +1\right )}{64 a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 82, normalized size = 1.09 \[ \frac {\frac {2 \, {\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac {3 \, \log \left (a x + 1\right )}{a} + \frac {3 \, \log \left (a x - 1\right )}{a}}{64 \, a} + \frac {\operatorname {artanh}\left (a x\right )}{4 \, {\left (a^{2} x^{2} - 1\right )}^{2} a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.08, size = 105, normalized size = 1.40 \[ \frac {\frac {3\,\ln \left (a\,x-1\right )}{64}-\frac {3\,\ln \left (a\,x+1\right )}{64}}{a^2}+\frac {\frac {\mathrm {atanh}\left (a\,x\right )}{4}-x^2\,\left (a^2\,\left (\frac {3\,\ln \left (a\,x-1\right )}{32}-\frac {3\,\ln \left (a\,x+1\right )}{32}\right )-2\,a^2\,\left (\frac {3\,\ln \left (a\,x-1\right )}{64}-\frac {3\,\ln \left (a\,x+1\right )}{64}\right )\right )-\frac {5\,a\,x}{32}+\frac {3\,a^3\,x^3}{32}}{a^2\,{\left (a^2\,x^2-1\right )}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.41, size = 158, normalized size = 2.11 \[ \begin {cases} - \frac {3 a^{4} x^{4} \operatorname {atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac {3 a^{3} x^{3}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac {6 a^{2} x^{2} \operatorname {atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} - \frac {5 a x}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} + \frac {5 \operatorname {atanh}{\left (a x \right )}}{32 a^{6} x^{4} - 64 a^{4} x^{2} + 32 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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