Optimal. Leaf size=212 \[ \frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]
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Rubi [A] time = 0.150657, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ \frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2163
Rule 2161
Rubi steps
\begin{align*} \int \frac{1}{x^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx &=-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{1}{6} b \int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{1}{8} b^2 \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{1}{16} \left (5 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\left (5 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 )}\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (5 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 )^2}\\ &=-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (5 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 )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\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 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{5 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{3 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}
Mathematica [A] time = 0.107486, size = 117, normalized size = 0.55 \[ \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-26 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+33 b^2 x^2\right )}{24 x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.148, size = 200, normalized size = 0.9 \begin{align*} 2\,{b}^{3} \left ( 2/3\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ){b}^{3}{x}^{3}}}+10/3\,{\frac{1}{-4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx} \left ({\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ){b}^{2}{x}^{2}}}+3\,{\frac{1}{-4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx} \left ( 2\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) bx}}-2\,{\frac{1}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) \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 ) } \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} \sqrt{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.15512, size = 356, normalized size = 1.68 \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 (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a^{4} x^{3}}, -\frac{15 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a^{4} x^{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{1}{x^{4} \sqrt{\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 = 1.14601, size = 113, normalized size = 0.53 \begin{align*} -\frac{\frac{15 \, b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{15 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 40 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} + 33 \, \sqrt{b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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