Optimal. Leaf size=125 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {1}{4 b^2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{4 b^2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]
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Rubi [A] time = 0.07, antiderivative size = 125, 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 = {2168, 2163, 2162} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {1}{4 b^2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{4 b^2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))}-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]
Antiderivative was successfully verified.
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Rule 2162
Rule 2163
Rule 2168
Rubi steps
\begin {align*} \int \frac {\sqrt {x}}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {\int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {1}{4 b^2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))}-\frac {\int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=-\frac {1}{4 b^2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {1}{4 b^2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))}-\frac {\int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac {1}{4 b^2 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {\sqrt {x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {1}{4 b^2 \sqrt {x} \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}
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Mathematica [A] time = 0.15, size = 107, normalized size = 0.86 \[ \frac {1}{4} \left (\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}+\frac {\sqrt {x}}{b \tanh ^{-1}(\tanh (a+b x))^2-b^2 x \tanh ^{-1}(\tanh (a+b x))}-\frac {2 \sqrt {x}}{b \tanh ^{-1}(\tanh (a+b x))^2}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 186, normalized size = 1.49 \[ \left [-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}, -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) - {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 52, normalized size = 0.42 \[ \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} + \frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 98, normalized size = 0.78 \[ \frac {\frac {2 x^{\frac {3}{2}}}{8 \arctanh \left (\tanh \left (b x +a \right )\right )-8 b x}-\frac {\sqrt {x}}{4 b}}{\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 64, normalized size = 0.51 \[ \frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.84, size = 580, normalized size = 4.64 \[ \frac {\sqrt {2}\,\ln \left (-\frac {4\,\sqrt {b}\,\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}\,\left (\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 )+4\,\sqrt {b}\,\sqrt {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}+2\,\sqrt {2}\,b\,x\right )\,\left (b^3\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+4\,a^2\,b^3-4\,a\,b^3\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\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 )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}\right )}{4\,b^{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 )}^{3/2}}-\frac {2\,\sqrt {x}}{b\,{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}^2}-\frac {\sqrt {x}}{b\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\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 )} \]
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 \[ \int \frac {\sqrt {x}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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