Optimal. Leaf size=179 \[ \frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
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Rubi [A] time = 0.12, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2161} \[ \frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3} \]
Antiderivative was successfully verified.
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Rule 2161
Rule 2163
Rule 2168
Rubi steps
\begin {align*} \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^4} \, dx &=-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}+\frac {1}{6} b \int \frac {1}{x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac {1}{24} b^2 \int \frac {1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}+\frac {1}{16} b^3 \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac {b^3 \int \frac {1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac {b^3 \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {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 )^{5/2}}+\frac {b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {b}{12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 115, normalized size = 0.64 \[ \frac {1}{24} \left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (14 b x \tanh ^{-1}(\tanh (a+b x))-8 \tanh ^{-1}(\tanh (a+b x))^2-3 b^2 x^2\right )}{x^3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac {3 b^3 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.51, size = 145, normalized size = 0.81 \[ \left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{3} x^{3}}, \frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{3} x^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.92, size = 93, normalized size = 0.52 \[ \frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {\sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x + a} a^{2} b^{4}\right )}}{a^{2} b^{3} x^{3}}\right )}}{48 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 185, normalized size = 1.03 \[ 2 b^{3} \left (\frac {\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{16 a^{2}+32 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+16 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{16}}{b^{3} x^{3}}-\frac {\arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{16 \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}{x^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.51, size = 964, normalized size = 5.39 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{x^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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