Optimal. Leaf size=68 \[ -\frac {4 b \sqrt {x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
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Rubi [A]
time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2202, 2198}
\begin {gather*} \frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {4 b \sqrt {x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2198
Rule 2202
Rubi steps
\begin {align*} \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {(2 b) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac {4 b \sqrt {x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 43, normalized size = 0.63 \begin {gather*} -\frac {2 \left (b x+\tanh ^{-1}(\tanh (a+b x))\right )}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b 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 = 0.12, size = 59, normalized size = 0.87
method | result | size |
derivativedivides | \(-\frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {4 b \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\) | \(59\) |
default | \(-\frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {4 b \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\) | \(59\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.47, size = 32, normalized size = 0.47 \begin {gather*} -\frac {2 \, {\left (2 \, b^{2} x^{2} + 3 \, a b x + a^{2}\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} a^{2} \sqrt {x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.32, size = 34, normalized size = 0.50 \begin {gather*} -\frac {2 \, {\left (2 \, b x + a\right )} \sqrt {b x + a} \sqrt {x}}{a^{2} b x^{2} + a^{3} 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^{\frac {3}{2}} \operatorname {atanh}^{\frac {3}{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 = 50, normalized size = 0.74 \begin {gather*} -\frac {2 \, b \sqrt {x}}{\sqrt {b x + a} a^{2}} + \frac {4 \, \sqrt {b}}{{\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )} a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.49, size = 281, normalized size = 4.13 \begin {gather*} -\frac {\sqrt {\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 )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,x}{{\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 )}^2}-\frac {8\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-8\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+16\,b\,x}{2\,b\,{\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 )}^2}\right )}{x^{3/2}-\frac {\sqrt {x}\,\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 )}{2\,b}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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