Optimal. Leaf size=125 \[ \frac {b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac {b}{4 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
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Rubi [A] time = 0.07, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ \frac {b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac {b}{4 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
Antiderivative was successfully verified.
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Rule 2161
Rule 2163
Rule 2168
Rubi steps
\begin {align*} \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx &=-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}+\frac {1}{4} b \int \frac {1}{x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {b}{4 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac {1}{8} b^2 \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=-\frac {b}{4 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac {b^2 \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {b}{4 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 89, normalized size = 0.71 \[ \frac {1}{4} \left (\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}+\frac {\left (\frac {b x}{b x-\tanh ^{-1}(\tanh (a+b x))}-2\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^2}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 119, normalized size = 0.95 \[ \left [\frac {\sqrt {a} b^{2} x^{2} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a^{2} x^{2}}, -\frac {\sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a^{2} x^{2}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 75, normalized size = 0.60 \[ -\frac {\sqrt {2} {\left (\frac {\sqrt {2} b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {2} {\left ({\left (b x + a\right )}^{\frac {3}{2}} b^{3} + \sqrt {b x + a} a b^{3}\right )}}{a b^{2} x^{2}}\right )}}{8 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 92, normalized size = 0.74 \[ 2 b^{2} \left (\frac {-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{8 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{8}}{b^{2} x^{2}}+\frac {\arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{8 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{\frac {3}{2}}}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.74, size = 741, normalized size = 5.93 \[ \frac {b\,\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}}}{2\,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 )}-\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 (\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 )}{x^2\,\left (2\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+4\,b\,x\right )}+\frac {\sqrt {2}\,b^2\,\ln \left (\frac {\left (2\,\sqrt {2}\,a-\sqrt {2}\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )+\sqrt {2}\,b\,x+\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}}\,\sqrt {\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}\,2{}\mathrm {i}\right )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )\,4{}\mathrm {i}}{x\,\sqrt {\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 )\,1{}\mathrm {i}}{4\,{\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 )}^{3/2}} \]
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 \[ \int \frac {\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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