Optimal. Leaf size=158 \[ \frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{1}{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b}{4 x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
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Rubi [A] time = 0.0962233, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ \frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{1}{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b}{4 x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2163
Rule 2161
Rubi steps
\begin{align*} \int \frac{1}{x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx &=-\frac{1}{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{1}{4} b \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac{b}{4 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{1}{8} \left (3 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac{b}{4 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (3 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b}{4 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (3 b^2\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{3 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 )^{5/2}}+\frac{b}{4 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}
Mathematica [A] time = 0.0841114, size = 98, normalized size = 0.62 \[ \frac{1}{4} \left (\frac{\left (5 b x-2 \tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.147, size = 148, normalized size = 0.9 \begin{align*} 2\,{b}^{2} \left ({\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ){b}^{2}{x}^{2}}}+3\,{\frac{1}{-4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx} \left ( 2\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) bx}}-2\,{\frac{1}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}{\it Artanh} \left ({\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}} \right ) } \right ) } \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.20169, size = 301, normalized size = 1.91 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt{b x + a}}{8 \, a^{3} x^{2}}, \frac{3 \, \sqrt{-a} b^{2} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt{b x + a}}{4 \, a^{3} x^{2}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.16351, size = 93, normalized size = 0.59 \begin{align*} \frac{\frac{3 \, b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{3} - 5 \, \sqrt{b x + a} a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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