Optimal. Leaf size=68 \[ -3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {3}{2} b \tanh ^{-1}(\tanh (a+b x))^2-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+3 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log (x) \]
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Rubi [A]
time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2199, 2190,
2189, 29} \begin {gather*} -3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+\frac {3}{2} b \tanh ^{-1}(\tanh (a+b x))^2+3 b \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 2189
Rule 2190
Rule 2199
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^2} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+(3 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x} \, dx\\ &=\frac {3}{2} b \tanh ^{-1}(\tanh (a+b x))^2-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}-\left (3 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x} \, dx\\ &=-3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {3}{2} b \tanh ^{-1}(\tanh (a+b x))^2-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+\left (3 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{x} \, dx\\ &=-3 b^2 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac {3}{2} b \tanh ^{-1}(\tanh (a+b x))^2-\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}+3 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log (x)\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 62, normalized size = 0.91 \begin {gather*} -\frac {\tanh ^{-1}(\tanh (a+b x))^3}{x}-6 b^2 x \tanh ^{-1}(\tanh (a+b x)) \log (x)+3 b \tanh ^{-1}(\tanh (a+b x))^2 (1+\log (x))+\frac {3}{2} b^3 x^2 (-1+2 \log (x)) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.24, size = 96, normalized size = 1.41
method | result | size |
default | \(-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{x}+3 b \left (\ln \left (x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{2}-2 b \left (\frac {b \,x^{2} \ln \left (x \right )}{2}-\frac {b \,x^{2}}{4}+\ln \left (x \right ) x a -x a +\ln \left (x \right ) x \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )-x \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )\right )\right )\) | \(96\) |
risch | \(\text {Expression too large to display}\) | \(4727\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.69, size = 65, normalized size = 0.96 \begin {gather*} 3 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) + \frac {3}{2} \, {\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \left (x\right ) - 2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right )\right )} b - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.33, size = 36, normalized size = 0.53 \begin {gather*} \frac {b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \left (x\right ) - 2 \, a^{3}}{2 \, x} \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 (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.38, size = 33, normalized size = 0.49 \begin {gather*} \frac {1}{2} \, b^{3} x^{2} + 3 \, a b^{2} x + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) - \frac {a^{3}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.07, size = 415, normalized size = 6.10 \begin {gather*} \frac {3\,b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4}-\frac {{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{8\,x}+\frac {3\,b\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4}+\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{8\,x}-\frac {3\,b^3\,x^2}{2}+\frac {3\,b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (x\right )}{4}+\frac {3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{8\,x}-\frac {3\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{8\,x}+\frac {3\,b\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (x\right )}{4}-\frac {3\,b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+3\,b^3\,x^2\,\ln \left (x\right )-\frac {3\,b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )}{2}+3\,b^2\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )-3\,b^2\,x\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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