Optimal. Leaf size=245 \[ -\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{35 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
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Rubi [A] time = 0.182859, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ -\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{35 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2163
Rule 2161
Rubi steps
\begin{align*} \int \frac{1}{x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{2} b \int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{1}{8} \left (5 b^2\right ) \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{16} \left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{\left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{\left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{\left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\left (35 b^3\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{35 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}
Mathematica [A] time = 0.113197, size = 133, normalized size = 0.54 \[ \frac{35 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}}-\frac{87 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-38 b x \tanh ^{-1}(\tanh (a+b x))^2+8 \tanh ^{-1}(\tanh (a+b x))^3+48 b^3 x^3}{24 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.128, size = 186, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ( -{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}} \left ({\frac{1}{{x}^{3}{b}^{3}} \left ({\frac{19\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{16}}+ \left ( -{\frac{17\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{6}}+{\frac{17\,bx}{6}} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ({\frac{29\,{a}^{2}}{16}}+{\frac{29\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{8}}+{\frac{29\, \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{35}{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 ) }-{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \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^{4} \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 = 2.12223, size = 478, normalized size = 1.95 \begin{align*} \left [\frac{105 \,{\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{b x + a}}{48 \,{\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}, -\frac{105 \,{\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{b x + a}}{24 \,{\left (a^{5} b x^{4} + a^{6} x^{3}\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^{4} \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 = 1.1462, size = 128, normalized size = 0.52 \begin{align*} -\frac{35 \, b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{4}} - \frac{2 \, b^{3}}{\sqrt{b x + a} a^{4}} - \frac{57 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{3} - 136 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{3} + 87 \, \sqrt{b x + a} a^{2} b^{3}}{24 \, a^{4} b^{3} x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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