Optimal. Leaf size=28 \[ -\frac {x}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2} \]
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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2199, 2188, 29}
\begin {gather*} \frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}-\frac {x}{b \tanh ^{-1}(\tanh (a+b x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 2188
Rule 2199
Rubi steps
\begin {align*} \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {x}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {\int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=-\frac {x}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ &=-\frac {x}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 27, normalized size = 0.96 \begin {gather*} \frac {1-\frac {b x}{\tanh ^{-1}(\tanh (a+b x))}+\log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 56, normalized size = 2.00
method | result | size |
default | \(\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{b^{2}}+\frac {a}{b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a}{b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )}\) | \(56\) |
risch | \(\frac {4 i x}{b \left (-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-4 i \ln \left ({\mathrm e}^{b x +a}\right )\right )}+\frac {\ln \left (\ln \left ({\mathrm e}^{b x +a}\right )+\frac {i \pi \left (-\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+2 \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}+\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-\mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}\right )}{4}\right )}{b^{2}}\) | \(541\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.67, size = 26, normalized size = 0.93 \begin {gather*} \frac {a}{b^{3} x + a b^{2}} + \frac {\log \left (b x + a\right )}{b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 28, normalized size = 1.00 \begin {gather*} \frac {{\left (b x + a\right )} \log \left (b x + a\right ) + a}{b^{3} x + a b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 25.24, size = 36, normalized size = 1.29 \begin {gather*} \begin {cases} - \frac {x}{b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}} + \frac {\log {\left (\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )} \right )}}{b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 \operatorname {atanh}^{2}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.39, size = 24, normalized size = 0.86 \begin {gather*} \frac {\log \left ({\left | b x + a \right |}\right )}{b^{2}} + \frac {a}{{\left (b x + a\right )} b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.08, size = 28, normalized size = 1.00 \begin {gather*} \frac {\ln \left (\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\right )}{b^2}-\frac {x}{b\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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