Optimal. Leaf size=146 \[ \frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2}{8 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3} \]
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Rubi [A] time = 0.0935789, antiderivative size = 146, 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} \[ \frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2}{8 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2163
Rule 2161
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^4} \, dx &=-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}+\frac{1}{2} b \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx\\ &=-\frac{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}+\frac{1}{8} b^2 \int \frac{1}{x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{b^2}{8 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}-\frac{1}{16} b^3 \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=-\frac{b^2}{8 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}-\frac{b^3 \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{b^2}{8 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^3}\\ \end{align*}
Mathematica [A] time = 0.0918898, size = 117, normalized size = 0.8 \[ \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-\frac{b^2}{8 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}-\frac{\tanh ^{-1}(\tanh (a+b x))-b x}{3 x^3}-\frac{7 b}{12 x^2}\right )+\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.115, size = 116, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{x}^{3}{b}^{3}} \left ( -1/16\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}-1/6\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ( 1/16\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -1/16\,bx \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }+1/16\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3/2}}{\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 ) \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{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{3}{2}}}{x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.18842, size = 351, normalized size = 2.4 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{3} x^{3} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a^{2} x^{3}}, -\frac{3 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a^{2} x^{3}}\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{\operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}}{x^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.20262, size = 126, normalized size = 0.86 \begin{align*} -\frac{\sqrt{2}{\left (\frac{3 \, \sqrt{2} b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{\sqrt{2}{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} + 8 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} - 3 \, \sqrt{b x + a} a^{2} b^{4}\right )}}{a b^{3} x^{3}}\right )}}{48 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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