Optimal. Leaf size=65 \[ \frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]
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Rubi [A]
time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2194, 2191,
2188, 29} \begin {gather*} \frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 2188
Rule 2191
Rule 2194
Rubi steps
\begin {align*} \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))} \, dx}{b x-\tanh ^{-1}(\tanh (a+b x))}\\ &=\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \int \frac {1}{x} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {b^2 \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \text {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {1}{x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 45, normalized size = 0.69 \begin {gather*} \frac {-\tanh ^{-1}(\tanh (a+b x))+b x \left (1-\log (x)+\log \left (\tanh ^{-1}(\tanh (a+b x))\right )\right )}{x \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.02, size = 64, normalized size = 0.98 \[\frac {b \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x}-\frac {b \ln \left (x \right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 28, normalized size = 0.43 \begin {gather*} \frac {b \log \left (b x + a\right )}{a^{2}} - \frac {b \log \left (x\right )}{a^{2}} - \frac {1}{a x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 26, normalized size = 0.40 \begin {gather*} \frac {b x \log \left (b x + a\right ) - b x \log \left (x\right ) - a}{a^{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 {1}{x^{2} \operatorname {atanh}{\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 = 30, normalized size = 0.46 \begin {gather*} \frac {b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {1}{a x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.83, size = 210, normalized size = 3.23 \begin {gather*} \frac {2\,\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 )+4\,b\,x+b\,x\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\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\,b\,x}\right )\,8{}\mathrm {i}}{x\,{\left (\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\,b\,x\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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