Optimal. Leaf size=70 \[ -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]
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Rubi [A] time = 0.05, antiderivative size = 70, 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 = {2163, 2160, 2157, 29} \[ -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]
Antiderivative was successfully verified.
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Rule 29
Rule 2157
Rule 2160
Rule 2163
Rubi steps
\begin {align*} \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\int \frac {1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))}\\ &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac {\int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 53, normalized size = 0.76 \[ \frac {\coth ^{-1}(\tanh (a+b x)) \left (-\log \left (\coth ^{-1}(\tanh (a+b x))\right )+\log (b x)+1\right )-b x}{\coth ^{-1}(\tanh (a+b x)) \left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 306, normalized size = 4.37 \[ -\frac {2 \, {\left (2 \, \pi ^{4} - 32 \, a^{4} - 8 \, {\left (\pi ^{2} a b + 4 \, a^{3} b\right )} x + 16 \, {\left (4 \, \pi a b^{2} x^{2} + 8 \, \pi a^{2} b x + \pi ^{3} a + 4 \, \pi a^{3}\right )} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - {\left (\pi ^{4} - 16 \, a^{4} + 4 \, {\left (\pi ^{2} b^{2} - 4 \, a^{2} b^{2}\right )} x^{2} + 8 \, {\left (\pi ^{2} a b - 4 \, a^{3} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 2 \, {\left (\pi ^{4} - 16 \, a^{4} + 4 \, {\left (\pi ^{2} b^{2} - 4 \, a^{2} b^{2}\right )} x^{2} + 8 \, {\left (\pi ^{2} a b - 4 \, a^{3} b\right )} x\right )} \log \relax (x)\right )}}{\pi ^{6} + 12 \, \pi ^{4} a^{2} + 48 \, \pi ^{2} a^{4} + 64 \, a^{6} + 4 \, {\left (\pi ^{4} b^{2} + 8 \, \pi ^{2} a^{2} b^{2} + 16 \, a^{4} b^{2}\right )} x^{2} + 8 \, {\left (\pi ^{4} a b + 8 \, \pi ^{2} a^{3} b + 16 \, a^{5} b\right )} x} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F(-1)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.77, size = 77, normalized size = 1.10 \[ \frac {4 \, \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} - \frac {4 \, \log \relax (x)}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} - \frac {4}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2} + {\left (2 i \, \pi b - 4 \, a b\right )} x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.03, size = 421, normalized size = 6.01 \[ -\frac {4\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-4\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+8\,b\,x+\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}-\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,\left (\ln \relax (2)+\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,8{}\mathrm {i}}{\left (\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,{\left (\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2} \]
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 \[ \int \frac {1}{x \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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