Optimal. Leaf size=191 \[ -\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]
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Rubi [A] time = 0.127917, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ -\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2163
Rule 2161
Rubi steps
\begin{align*} \int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{4} (3 b) \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{1}{8} \left (15 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{\left (15 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{\left (15 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\left (15 b^2\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}
Mathematica [A] time = 0.116435, size = 115, normalized size = 0.6 \[ \frac{1}{4} \left (\frac{9 b x \tanh ^{-1}(\tanh (a+b x))-2 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2}{x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.127, size = 131, normalized size = 0.7 \begin{align*} 2\,{b}^{2} \left ({\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}+{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ({\frac{7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{8}}+ \left ( -{\frac{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{8}}+{\frac{9\,bx}{8}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-{\frac{15}{8\,\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 ) } \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{1}{x^{3} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.13596, size = 420, normalized size = 2.2 \begin{align*} \left [\frac{15 \,{\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt{a} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x + a}}{8 \,{\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, \frac{15 \,{\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x + a}}{4 \,{\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\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{1}{x^{3} \operatorname{atanh}^{\frac{3}{2}}{\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 = 1.1765, size = 108, normalized size = 0.57 \begin{align*} \frac{15 \, b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a} a^{3}} + \frac{2 \, b^{2}}{\sqrt{b x + a} a^{3}} + \frac{7 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{2} - 9 \, \sqrt{b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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