Optimal. Leaf size=113 \[ -\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 x}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3} \]
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Rubi [A] time = 0.0675483, antiderivative size = 113, 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 = {2168, 2161} \[ -\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 x}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2161
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^4} \, dx &=-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3}+\frac{1}{6} (5 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^3} \, dx\\ &=-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3}+\frac{1}{8} \left (5 b^2\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^2} \, dx\\ &=-\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 x}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3}+\frac{1}{16} \left (5 b^3\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 x}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3}\\ \end{align*}
Mathematica [A] time = 0.0703738, size = 107, normalized size = 0.95 \[ \frac{1}{24} \left (-\frac{15 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac{15 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x}-\frac{10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^2}-\frac{8 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^3}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.118, size = 144, normalized size = 1.3 \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{x}^{3}{b}^{3}} \left ( -{\frac{11\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{16}}+ \left ( 5/6\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5/6\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ( -{\frac{5\,{a}^{2}}{16}}-5/8\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) -{\frac{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{16}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-{\frac{5}{16\,\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^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.12887, size = 350, normalized size = 3.1 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{3} x^{3} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a x^{3}}, \frac{15 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a x^{3}}\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 = 1.17132, size = 119, normalized size = 1.05 \begin{align*} \frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{\sqrt{2}{\left (33 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 40 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} + 15 \, \sqrt{b x + a} a^{2} b^{4}\right )}}{b^{3} x^{3}}\right )}}{48 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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