Optimal. Leaf size=75 \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
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Rubi [A] time = 0.0419205, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2168, 2165} \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2165
Rubi steps
\begin{align*} \int \frac{x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{\int \frac{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{b}\\ &=-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}
Mathematica [A] time = 0.0640778, size = 78, normalized size = 1.04 \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.118, size = 59, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,b}{x}^{{\frac{3}{2}}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{\sqrt{x}}{{b}^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}+2\,{\frac{\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }{{b}^{5/2}}} \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{x^{\frac{3}{2}}}{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.12632, size = 451, normalized size = 6.01 \begin{align*} \left [\frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac{2 \,{\left (3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}\right )}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\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 = 1.28144, size = 66, normalized size = 0.88 \begin{align*} -\frac{2 \, \sqrt{x}{\left (\frac{4 \, x}{b} + \frac{3 \, a}{b^{2}}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}}} - \frac{2 \, \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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