Optimal. Leaf size=106 \[ \frac {16 b \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {8 b \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
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Rubi [A] time = 0.05, antiderivative size = 106, 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} \[ \frac {16 b \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {8 b \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2167
Rule 2171
Rubi steps
\begin {align*} \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {(4 b) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=-\frac {8 b \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {(8 b) \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 )^2}\\ &=-\frac {8 b \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b \sqrt {x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 66, normalized size = 0.62 \[ \frac {2 \left (6 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2-b^2 x^2\right )}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.38, size = 58, normalized size = 0.55 \[ -\frac {2 \, {\left (8 \, b^{2} x^{2} + 12 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 62, normalized size = 0.58 \[ -\frac {2 \, \sqrt {x} {\left (\frac {5 \, b^{2} x}{a^{3}} + \frac {6 \, b}{a^{2}}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} + \frac {4 \, \sqrt {b}}{{\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )} a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 104, normalized size = 0.98 \[ -\frac {2}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {8 b \left (\frac {\sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 45, normalized size = 0.42 \[ -\frac {2 \, {\left (8 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 3 \, a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} \sqrt {x}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.69, size = 348, normalized size = 3.28 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {4}{b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {128\,x^2}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}-\frac {32\,x}{b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{5/2}-\frac {x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{b}+\frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2}} \]
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 \[ \int \frac {1}{x^{\frac {3}{2}} \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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