Optimal. Leaf size=110 \[ -\frac {16 b^2 \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
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Rubi [A] time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ -\frac {16 b^2 \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
Antiderivative was successfully verified.
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Rule 2167
Rule 2171
Rubi steps
\begin {align*} \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {(4 b) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (8 b^2\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {16 b^2 \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 b}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 64, normalized size = 0.58 \[ \frac {2 \left (-6 b x \tanh ^{-1}(\tanh (a+b x))+\tanh ^{-1}(\tanh (a+b x))^2-3 b^2 x^2\right )}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 49, normalized size = 0.45 \[ \frac {2 \, {\left (8 \, b^{2} x^{2} + 4 \, a b x - a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 107, normalized size = 0.97 \[ \frac {2 \, b^{2} \sqrt {x}}{\sqrt {b x + a} a^{3}} - \frac {4 \, {\left (3 \, b^{\frac {3}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 12 \, a b^{\frac {3}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + 5 \, a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{3} a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 105, normalized size = 0.95 \[ -\frac {2}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {8 b \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {2 b \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\right )}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 45, normalized size = 0.41 \[ \frac {2 \, {\left (8 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} x^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.69, size = 286, normalized size = 2.60 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {32\,x}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {4}{3\,b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {128\,b\,x^2}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}\right )}{x^{5/2}-\frac {x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2\,b}} \]
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 \[ \int \frac {1}{x^{\frac {5}{2}} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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