Optimal. Leaf size=221 \[ -\frac{b^3}{64 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}+\frac{3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b^5 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5} \]
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Rubi [A] time = 0.174042, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ -\frac{b^3}{64 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}+\frac{3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b^5 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5} \]
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))^{5/2}}{x^6} \, dx &=-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac{1}{2} b \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^5} \, dx\\ &=-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac{1}{16} \left (3 b^2\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^4} \, dx\\ &=-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac{1}{32} b^3 \int \frac{1}{x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{b^3}{64 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}-\frac{1}{128} b^4 \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac{b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^3}{64 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}+\frac{1}{256} \left (3 b^5\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac{b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^3}{64 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}-\frac{\left (3 b^5\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{256 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^3}{64 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}-\frac{\left (3 b^5\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{256 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{3 b^5 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac{b^4}{128 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^3}{64 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{3 b^5}{128 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{16 x^3}-\frac{b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{8 x^4}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^5}\\ \end{align*}
Mathematica [A] time = 0.12318, size = 150, normalized size = 0.68 \[ \frac{1}{640} \left (-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (10 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+8 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-176 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+15 b^4 x^4\right )}{x^5 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac{15 b^5 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.122, size = 262, normalized size = 1.2 \begin{align*} 2\,{b}^{5} \left ({\frac{1}{{b}^{5}{x}^{5}} \left ({\frac{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{9/2}}{256\,{a}^{2}+512\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +256\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}}-{\frac{7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}}{128\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -128\,bx}}-1/10\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}+ \left ({\frac{7\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{128}}-{\frac{7\,bx}{128}} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ( -{\frac{3\,{a}^{2}}{256}}-{\frac{3\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{128}}-{\frac{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{256}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-{\frac{3}{ \left ( 256\,{a}^{2}+512\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +256\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2} \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 ) \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{5}{2}}}{x^{6}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.28065, size = 462, normalized size = 2.09 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{5} x^{5} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt{b x + a}}{1280 \, a^{3} x^{5}}, \frac{15 \, \sqrt{-a} b^{5} x^{5} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt{b x + a}}{640 \, a^{3} x^{5}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.16527, size = 166, normalized size = 0.75 \begin{align*} \frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} b^{6} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{\sqrt{2}{\left (15 \,{\left (b x + a\right )}^{\frac{9}{2}} b^{6} - 70 \,{\left (b x + a\right )}^{\frac{7}{2}} a b^{6} - 128 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} b^{6} + 70 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3} b^{6} - 15 \, \sqrt{b x + a} a^{4} b^{6}\right )}}{a^{2} b^{5} x^{5}}\right )}}{1280 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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