Optimal. Leaf size=121 \[ 2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{2}{3} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}-2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \]
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Rubi [A] time = 0.0702375, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2159, 2161} \[ 2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{2}{3} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}-2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \]
Antiderivative was successfully verified.
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Rule 2159
Rule 2161
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x} \, dx &=\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x} \, dx\\ &=-\frac{2}{3} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\left (\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x} \, dx\\ &=2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{2}{3} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}+\left (\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}+2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{2}{3} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}\\ \end{align*}
Mathematica [A] time = 0.0699247, size = 99, normalized size = 0.82 \[ \frac{2}{15} \left (15 b^2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}+23 \tanh ^{-1}(\tanh (a+b x))^{5/2}-35 b x \tanh ^{-1}(\tanh (a+b x))^{3/2}-15 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right ) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.109, size = 222, normalized size = 1.8 \begin{align*}{\frac{2}{5} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,a}{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -2\,bx-2\,a}{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+2\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{a}^{2}+4\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-2\,{\frac{{a}^{3}+3\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +3\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}+ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{\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 ) } \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}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.0917, size = 288, normalized size = 2.38 \begin{align*} \left [a^{\frac{5}{2}} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + \frac{2}{15} \,{\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt{b x + a}, 2 \, \sqrt{-a} a^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) + \frac{2}{15} \,{\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt{b x + a}\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 = 1.17054, size = 99, normalized size = 0.82 \begin{align*} \frac{1}{15} \, \sqrt{2}{\left (\frac{15 \, \sqrt{2} a^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 3 \, \sqrt{2}{\left (b x + a\right )}^{\frac{5}{2}} + 5 \, \sqrt{2}{\left (b x + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{2} \sqrt{b x + a} a^{2}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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