Optimal. Leaf size=69 \[ -\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{32 b^3 \sqrt{x}}{5} \]
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Rubi [A] time = 0.0376807, antiderivative size = 69, normalized size of antiderivative = 1., 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} \[ -\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{32 b^3 \sqrt{x}}{5} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 30
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^3}{x^{7/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{1}{5} (6 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^2}{x^{5/2}} \, dx\\ &=-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{1}{5} \left (8 b^2\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))}{x^{3/2}} \, dx\\ &=-\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{1}{5} \left (16 b^3\right ) \int \frac{1}{\sqrt{x}} \, dx\\ &=\frac{32 b^3 \sqrt{x}}{5}-\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.0308873, size = 57, normalized size = 0.83 \[ \frac{2 \left (-8 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-2 b x \tanh ^{-1}(\tanh (a+b x))^2-\tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{5 x^{5/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.044, size = 56, normalized size = 0.8 \begin{align*} -{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{5}{x}^{-{\frac{5}{2}}}}+{\frac{12\,b}{5} \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{3}{x}^{-{\frac{3}{2}}}}+{\frac{4\,b}{3} \left ( -{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ){\frac{1}{\sqrt{x}}}}+2\,b\sqrt{x} \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.08763, size = 74, normalized size = 1.07 \begin{align*} \frac{16}{5} \,{\left (2 \, b^{2} \sqrt{x} - \frac{b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{\sqrt{x}}\right )} b - \frac{4 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac{3}{2}}} - \frac{2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.96389, size = 78, normalized size = 1.13 \begin{align*} \frac{2 \,{\left (5 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} - 5 \, a^{2} b x - a^{3}\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.12904, size = 46, normalized size = 0.67 \begin{align*} 2 \, b^{3} \sqrt{x} - \frac{2 \,{\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x + a^{3}\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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