Optimal. Leaf size=116 \[ \frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{7/2}}+\frac{2 x^{5/2}}{5 b} \]
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Rubi [A] time = 0.0791174, antiderivative size = 116, 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, 2162} \[ \frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{7/2}}+\frac{2 x^{5/2}}{5 b} \]
Antiderivative was successfully verified.
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Rule 2159
Rule 2162
Rubi steps
\begin{align*} \int \frac{x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac{2 x^{5/2}}{5 b}-\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right ) \int \frac{x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac{2 x^{5/2}}{5 b}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2 \int \frac{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac{2 x^{5/2}}{5 b}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}-\frac{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3 \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac{2 x^{5/2}}{5 b}+\frac{2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}+\frac{2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}{b^{7/2}}\\ \end{align*}
Mathematica [A] time = 0.105778, size = 108, normalized size = 0.93 \[ \frac{2 \left (-35 b^{3/2} x^{3/2} \tanh ^{-1}(\tanh (a+b x))+15 \sqrt{b} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2-15 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )+23 b^{5/2} x^{5/2}\right )}{15 b^{7/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.126, size = 330, normalized size = 2.8 \begin{align*}{\frac{2}{5\,b}{x}^{{\frac{5}{2}}}}-{\frac{2\,a}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}-{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -2\,bx-2\,a}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}+2\,{\frac{\sqrt{x}{a}^{2}}{{b}^{3}}}+4\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{x}}{{b}^{3}}}+2\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}\sqrt{x}}{{b}^{3}}}-2\,{\frac{{a}^{3}}{{b}^{3}\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}} \right ) }-6\,{\frac{{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{b}^{3}\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}} \right ) }-6\,{\frac{a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{{b}^{3}\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}} \right ) }-2\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{{b}^{3}\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.20097, size = 308, normalized size = 2.66 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt{x}}{15 \, b^{3}}, -\frac{2 \,{\left (15 \, a^{2} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) -{\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt{x}\right )}}{15 \, b^{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{x^{\frac{5}{2}}}{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.13205, size = 80, normalized size = 0.69 \begin{align*} -\frac{2 \, a^{3} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} + \frac{2 \,{\left (3 \, b^{4} x^{\frac{5}{2}} - 5 \, a b^{3} x^{\frac{3}{2}} + 15 \, a^{2} b^{2} \sqrt{x}\right )}}{15 \, b^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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