Optimal. Leaf size=50 \[ \frac {2 x}{b^2}-\frac {x^2}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3} \]
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Rubi [A]
time = 0.02, antiderivative size = 50, 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, 2189,
2188, 29} \begin {gather*} \frac {2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^2}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {2 x}{b^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 2188
Rule 2189
Rule 2199
Rubi steps
\begin {align*} \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {x^2}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {2 \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {2 x}{b^2}-\frac {x^2}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {2 x}{b^2}-\frac {x^2}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}\\ &=\frac {2 x}{b^2}-\frac {x^2}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 56, normalized size = 1.12 \begin {gather*} \frac {b x-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}{\tanh ^{-1}(\tanh (a+b x))}+2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(126\) vs.
\(2(50)=100\).
time = 0.24, size = 127, normalized size = 2.54
method | result | size |
default | \(\frac {x}{b^{2}}-\frac {2 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a}{b^{3}}-\frac {2 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3}}-\frac {a^{2}}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{3} \arctanh \left (\tanh \left (b x +a \right )\right )}\) | \(127\) |
risch | \(\text {Expression too large to display}\) | \(3133\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.68, size = 44, normalized size = 0.88 \begin {gather*} \frac {b^{2} x^{2} + a b x - a^{2}}{b^{4} x + a b^{3}} - \frac {2 \, a \log \left (b x + a\right )}{b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 47, normalized size = 0.94 \begin {gather*} \frac {b^{2} x^{2} + a b x - a^{2} - 2 \, {\left (a b x + a^{2}\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \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^{2}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.39, size = 34, normalized size = 0.68 \begin {gather*} \frac {x}{b^{2}} - \frac {2 \, a \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac {a^{2}}{{\left (b x + a\right )} b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.06, size = 302, normalized size = 6.04 \begin {gather*} \frac {x}{b^2}-\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2-4\,a\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )+4\,a^2}{2\,b\,\left (2\,a\,b^2+2\,b^3\,x-b^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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