Optimal. Leaf size=110 \[ 5 b \text {ArcTan}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}-5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {5}{3} b \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x} \]
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Rubi [A]
time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2199, 2190,
2192} \begin {gather*} 5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2} \text {ArcTan}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x}+\frac {5}{3} b \tanh ^{-1}(\tanh (a+b x))^{3/2}-5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 2190
Rule 2192
Rule 2199
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^2} \, dx &=-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x}+\frac {1}{2} (5 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x} \, dx\\ &=\frac {5}{3} b \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x}-\frac {1}{2} \left (5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x} \, dx\\ &=-5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {5}{3} b \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x}+\frac {1}{2} \left (5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=5 b \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}-5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}+\frac {5}{3} b \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x}\\ \end {align*}
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Mathematica [A]
time = 0.04, size = 106, normalized size = 0.96 \begin {gather*} -5 b \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {-b x+\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}+\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\frac {2 b^2 x}{3}+\frac {14}{3} b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}{x}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.07, size = 193, normalized size = 1.75
method | result | size |
default | \(2 b \left (\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\, a +2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\left (-\frac {a^{2}}{2}-a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{2}\right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b x}-\frac {5 \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )\) | \(193\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 126, normalized size = 1.15 \begin {gather*} \left [\frac {15 \, a^{\frac {3}{2}} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{6 \, x}, \frac {15 \, \sqrt {-a} a b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{3 \, x}\right ] \end {gather*}
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 \begin {gather*} \int \frac {\operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 89, normalized size = 0.81 \begin {gather*} \frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} a^{2} b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {2} {\left (b x + a\right )}^{\frac {3}{2}} b^{2} + 12 \, \sqrt {2} \sqrt {b x + a} a b^{2} - \frac {3 \, \sqrt {2} \sqrt {b x + a} a^{2} b}{x}\right )}}{6 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.97, size = 616, normalized size = 5.60 \begin {gather*} \frac {2\,b^2\,x\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{3}-\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,x}-\frac {\left (3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-\frac {4\,b^2\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{3}\right )\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{b}+\frac {\sqrt {2}\,b\,\ln \left (\frac {\left (\sqrt {2}\,b\,x-\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )+\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\,2{}\mathrm {i}\right )\,16{}\mathrm {i}}{x\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^{3/2}\,5{}\mathrm {i}}{8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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