Optimal. Leaf size=92 \[ \frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]
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Rubi [A]
time = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps
used = 6, 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 {b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b}{x \left (b x-\coth ^{-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^3 \coth ^{-1}(\tanh (a+b x))} \, dx &=\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))}\\ &=\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \int \frac {1}{x \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 {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b^2 \int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^3 \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac {b^2 \text {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 )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}+\frac {b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 66, normalized size = 0.72 \begin {gather*} \frac {-4 b x \coth ^{-1}(\tanh (a+b x))+\coth ^{-1}(\tanh (a+b x))^2+b^2 x^2 \left (3-2 \log (x)+2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{3} \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains complex when optimal does not.
time = 0.53, size = 106, normalized size = 1.15 \begin {gather*} \frac {8 \, b^{2} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac {8 \, b^{2} \log \left (x\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac {i \, \pi + 4 \, b x - 2 \, a}{{\left (\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}\right )} x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 199 vs.
\(2 (90) = 180\).
time = 0.39, size = 199, normalized size = 2.16 \begin {gather*} -\frac {2 \, {\left (\pi ^{4} a + 8 \, \pi ^{2} a^{3} + 16 \, a^{5} - 8 \, {\left (\pi ^{3} b^{2} - 12 \, \pi a^{2} b^{2}\right )} x^{2} \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 ) - 4 \, {\left (3 \, \pi ^{2} a b^{2} - 4 \, a^{3} b^{2}\right )} x^{2} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 8 \, {\left (3 \, \pi ^{2} a b^{2} - 4 \, a^{3} b^{2}\right )} x^{2} \log \left (x\right ) + 2 \, {\left (\pi ^{4} b - 16 \, a^{4} b\right )} x\right )}}{{\left (\pi ^{6} + 12 \, \pi ^{4} a^{2} + 48 \, \pi ^{2} a^{4} + 64 \, a^{6}\right )} x^{2}} \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^{3} \operatorname {acoth}{\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 [C] Result contains complex when optimal does not.
time = 0.41, size = 109, normalized size = 1.18 \begin {gather*} -\frac {8 \, b^{2} \log \left (\pi - 2 i \, b x - 2 i \, a\right )}{-i \, \pi ^{3} - 6 \, \pi ^{2} a + 12 i \, \pi a^{2} + 8 \, a^{3}} + \frac {8 i \, b^{2} \log \left (x\right )}{\pi ^{3} - 6 i \, \pi ^{2} a - 12 \, \pi a^{2} + 8 i \, a^{3}} - \frac {-i \, \pi + 4 \, b x - 2 \, a}{\pi ^{2} x^{2} - 4 i \, \pi a x^{2} - 4 \, a^{2} x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.60, size = 300, normalized size = 3.26 \begin {gather*} \frac {{\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 )\,\left (2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+8\,b\,x\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+12\,b^2\,x^2+8\,b\,x\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+b^2\,x^2\,\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 )\,16{}\mathrm {i}}{x^2\,{\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 )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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