Optimal. Leaf size=63 \[ -16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+\frac {32}{5} b^3 x^{5/2} \]
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Rubi [A] time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+\frac {32}{5} b^3 x^{5/2} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2168
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^{3/2}} \, dx &=-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+(6 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{\sqrt {x}} \, dx\\ &=12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}-\left (24 b^2\right ) \int \sqrt {x} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=-16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}+\left (16 b^3\right ) \int x^{3/2} \, dx\\ &=\frac {32}{5} b^3 x^{5/2}-16 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))+12 b \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{\sqrt {x}}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 57, normalized size = 0.90 \[ \frac {2 \left (-40 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+30 b x \tanh ^{-1}(\tanh (a+b x))^2-5 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{5 \sqrt {x}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 34, normalized size = 0.54 \[ \frac {2 \, {\left (b^{3} x^{3} + 5 \, a b^{2} x^{2} + 15 \, a^{2} b x - 5 \, a^{3}\right )}}{5 \, \sqrt {x}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 35, normalized size = 0.56 \[ \frac {2}{5} \, b^{3} x^{\frac {5}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} + 6 \, a^{2} b \sqrt {x} - \frac {2 \, a^{3}}{\sqrt {x}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 64, normalized size = 1.02 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{\sqrt {x}}+12 b \left (\frac {b^{2} x^{\frac {5}{2}}}{5}+\frac {2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b \,x^{\frac {3}{2}}}{3}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 55, normalized size = 0.87 \[ 12 \, b \sqrt {x} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{\sqrt {x}} + \frac {16}{5} \, {\left (2 \, b^{2} x^{\frac {5}{2}} - 5 \, b x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.21, size = 182, normalized size = 2.89 \[ \frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,\sqrt {x}}+\frac {2\,b^3\,x^{5/2}}{5}+\frac {3\,b\,\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2}-b^2\,x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right ) \]
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 \[ \int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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